Assumption 1. There exists a smooth function $v=v(Z)\in (-1,1)$ defined on $Z\in [0,+\infty )$ solving the ODE (2.8) with $v(0)=0$ and $v\in C_{\operatorname {o}}^{\infty }([0,+\infty ))$ .

Proof. By Lemma 2.3 (i) and the definition of $F_n$ , we have $F_n\in {\mathscr X}_{\mu _{n-1}-2}$ , where we have used the fact that $\mu _{n_1}+\cdots +\mu _{n_p}=\mu _{n-1}-2$ if $n_1+\cdots +n_p=n$ and $\mu _{n_1}+\lambda _{n_2}+\lambda _{n_3}-2=\mu _{n-1}-2$ if $n_1+n_2+n_3=n$ . Using Lemma 2.3 (i) and the definition of $G_n$ , we have $G_n\in {\mathscr X}_{(2n-1)(\beta -1)-2\beta /(p-1)-2}$ , where we have used the fact that $\mu _{n_1}+\lambda _{n_2}-2=(2n-1)(\beta -1)-2\beta /(p-1)-2$ if $n_1+n_2=n$ . It follows from Lemma 2.3 (i) that $\rho _0G_n\in {\mathscr X}_{\mu _0+(2n-1)(\beta -1)-2\beta /(p-1)-2}={\mathscr X}_{2n(\beta -1)-\beta \ell -1}$ (recall that $\rho _0\in {\mathscr X}_{\mu _0}$ , $\ell =1+\frac {4}{p-1}$ ). Since $\rho _0^{2-p}\in {\mathscr X}_{(2-p)\mu _0}$ , $F_n\in {\mathscr X}_{\mu _{n-1}-2}$ , by Lemma 2.3 (i) we get $\rho _0^{2-p}F_n\in {\mathscr X}_{\mu _{n-1}-2+(2-p)\mu _0}$ , then by $\phi _0\in {\mathscr X}_{\lambda _0}$ we have

$$\begin{align*}\partial^\alpha\left(\rho_0^{2-p}\partial_\alpha\phi_0 F_n\right)\in {\mathscr X}_{\mu_{n-1}-2+(2-p)\mu_0+\lambda_0-2}={\mathscr X}_{2n(\beta-1)-\beta\ell-1}.\end{align*}$$

Hence by the definition of $H_n$ in (2.18), we have $H_n\in {\mathscr X}_{2n(\beta -1)-\beta \ell -1}={\mathscr X}_{\lambda _n-\gamma }$ (recall that $\gamma =\beta (\ell -1)+2$ ). Moreover, $F_n, G_n, H_n$ are real-valued.

By Proposition 2.4, there exists (real-valued) $\phi _n\in {\mathscr X}_{\lambda _n}$ such that $\mathscr {L}(\phi _n)=H_n$ (otherwise take $\operatorname {Re}\phi _n$ ), then (2.18) holds. Let $\rho _n$ be defined by (2.17). Then $\rho _n$ is real-valued. Moreover, using (i) of Lemma 2.3, $\rho _0^{1-p}\in {\mathscr X}_{(1-p)\mu _0}$ , $\rho _0^{2-p}\in {\mathscr X}_{(2-p)\mu _0}$ , $F_n\in {\mathscr X}_{\mu _{n-1}-2}$ , $\phi _0\in {\mathscr X}_{\lambda _0}$ and $\phi _n\in {\mathscr X}_{\lambda _n}$ , we have

$$ \begin{align*}\rho_0^{1-p}F_n\in{\mathscr X}_{\mu_0(1-p)+\mu_{n-1}-2}={\mathscr X}_{\mu_n},\quad \rho_0^{2-p}\partial^\alpha\phi_0\partial_\alpha\phi_n\in{\mathscr X}_{\mu_0(2-p)+\lambda_0+\lambda_n-2}={\mathscr X}_{\mu_n},\end{align*} $$

hence $\rho _n\in {\mathscr X}_{\mu _n}$ . Now (2.15) follows from (2.17), and (2.16) follows from (2.17) and (2.18).

Proof of Theorem 1.1.

Let $T=b=1$ and $(\rho _*, \phi _*)\in C^\infty (\mathcal C)$ , $c_0\in (0,T)$ be given by Proposition 2.7, and $w_*=\rho _*$ , $\Phi _*=\phi _*$ . Then $w_*, \Phi _*\in C^\infty (\mathcal C)$ , (2.28) is equivalent to (2.25), (2.29) is equivalent to (2.26). The definitions of $E_*$ , $J_*$ in (2.24) and (2.30) are the same, and (2.27) also follows from Proposition 2.7. It follows from Proposition 2.8 that there exist $c_1\in (0,c_0)\subset (0,1)$ , $\widetilde u\in C^2((T-c_1,T)\times \mathbb {R}^d;\mathbb {C})$ such that $\widetilde u(t,\cdot ),\partial _t\widetilde u(t,\cdot )\in C_c^{\infty }(\mathbb {R}^d;\mathbb {C})$ for $t\in (T-c_1,T),$ $\Box \widetilde u=|\widetilde u|^{p-1}\widetilde u$ for $t\in (T-c_1,T), |x|\leq T-t$ , and

(2.32) $$ \begin{align} C_1^{-1}(T-t)^{-\frac{2\beta}{p-1}}\leq |\widetilde u(t,x)|\leq C_1(T-t)^{-\frac{2\beta}{p-1}},\qquad\forall\ t\in(T-c_1,T),\ |x|\leq T-t \end{align} $$

for some constant $ C_1>0$ . Choose initial data $u_0, u_1: \mathbb {R}^d\to \mathbb {C}$ such that

$$\begin{align*}u_0(x)=\widetilde u(T-c_1/2, x), \quad u_1(x)=\partial_t\widetilde u(T-c_1/2, x),\qquad\forall\ x\in\mathbb{R}^d.\end{align*}$$

Then $u_0,u_1\in C_c^{\infty }(\mathbb {R}^d;\mathbb {C})$ . Moreover, let $u_*(t,x)=\widetilde u(t+T-c_1/2,x)$ for $t\in (-c_1/2,c_1/2)$ , $x\in \mathbb {R}^d$ then $u_*(0,x)=u_0(x)$ , $\partial _t u_*(0,x)=u_1(x)$ and $\Box u_*=| u_*|^{p-1} u_*$ for $t\in [0,c_1/2), |x|\leq c_1/2-t$ . Suppose for contradiction that Theorem 1.1 fails for this initial data $u_0, u_1$ , then there exists a smooth function $u: [0,+\infty )\times \mathbb {R}^d\to \mathbb {C}$ such that $\Box u=|u|^{p-1}u$ and $u(0, x)=u_0(x), \partial _tu(0, x)=u_1(x)$ for all $x\in \mathbb {R}^d$ . Finite speed of propagation shows that $u= u_*$ in the region $\{(t,x)\in [0, c_1/2)\times \mathbb {R}^d: |x|\leq c_1/2-t\}$ . Hence by (2.32) we have

$$\begin{align*}|u(t,0)|=|u_*(t,0)|=|\widetilde u(t+T-c_1/2,0)|\geq C_1^{-1}(c_1/2-t)^{-\frac{2\beta}{p-1}},\qquad\forall\ t\in[0, c_1/2).\end{align*}$$

On the other hand, since u is smooth on $[0, +\infty )\times \mathbb {R}^d$ , we have $|u(t,x)|\leq C$ for all $|x|\leq 2T$ and $t\in [0, c_1/2]$ , where $C>0$ is a constant. This reaches a contradiction.

Proof of Corollary 1.2.

Let $T=1$ . If $d=4$ , $k=3$ , $p\geq 29$ , $\ell =1+\frac {4}{p-1}$ , then $1<\ell \leq 1+\frac {4}{29-1}=\frac {8}{7}<\ell ^*(3)$ . If $d=5$ , $k=4$ , $p\geq 17$ , $\ell =1+\frac {4}{p-1}$ , then $1<\ell \leq 1+\frac {4}{17-1}=\frac {5}{4}<\ell _1(4)$ . Thus, if $d=4$ , $p\geq 29$ or $d=5$ , $p\geq 17$ , then (2.33) holds for $k=d-1$ , $\ell =1+\frac {4}{p-1}$ and the result follows from Theorem 1.1.

The remaining case is $d>5$ , $p\geq 17$ . Then Assumption 1 holds with d replaced by $d'=5$ . By the proof of Theorem 1.1, there exists $c_1\in (0,1)$ , $\widetilde u\in C^2((T-c_1,T)\times \mathbb {R}^5;\mathbb {C})$ such that $\widetilde u(t,\cdot ),\partial _t\widetilde u(t,\cdot )\in C_c^{\infty }(\mathbb {R}^5;\mathbb {C})$ for $t\in (T-c_1,T),$ $\Box \widetilde u=|\widetilde u|^{p-1}\widetilde u$ for $t\in (T-c_1,T), |x|\leq T-t$ , and (2.32) holds for some constant $ C_1>0$ . Choose initial data $u_0, u_1: \mathbb {R}^d\to \mathbb {C}$ such that

$$\begin{align*}u_0(x)=\eta(|x|)\widetilde u(T-c_1/2, x_1,\cdots,x_5), \quad u_1(x)=\eta(|x|)\partial_t\widetilde u(T-c_1/2, x_1,\cdots,x_5),\end{align*}$$

for all $x=(x_1,\cdots ,x_d)\in \mathbb {R}^d$ . Then $u_0,u_1\in C_c^{\infty }(\mathbb {R}^d;\mathcal C)$ . Moreover, let $u_*(t,x)=\eta (|x|)\widetilde u(t+T-c_1/2,x_1,\cdots ,x_5)$ for $t\in (-c_1/2,c_1/2)$ , $x=(x_1,\cdots ,x_d)\in \mathbb {R}^d$ . Then $u_*(0,x)=u_0(x)$ , $\partial _t u_*(0,x)=u_1(x)$ and $\Box u_*=| u_*|^{p-1} u_*$ for $t\in [0,c_1/2), |x|\leq c_1/2-t$ . Here we used that $\eta (|x|)=1 $ for $|x|\leq 1$ and that if $t\in [0,c_1/2), |x|\leq c_1/2-t$ then $|x|\leq c_1/2<1$ . Suppose for contradiction that Corollary 1.2 fails for this kind of initial data $u_0, u_1$ , then there exists a smooth function $u: [0,+\infty )\times \mathbb {R}^d\to \mathbb {C}$ such that $\Box u=|u|^{p-1}u$ and $u(0, x)=u_0(x), \partial _tu(0, x)=u_1(x)$ for all $x\in \mathbb {R}^d$ , and we can get a contradiction as in the proof of Theorem 1.1.

Proof of Proposition 2.7.

Step 1. Construction of the sequence $\{{T}_n\}_{n\geq 0}$ . Let $A_N>0$ be given by Lemma 3.1 and the sequence $\{\widetilde {T}_n\}_{n\geq 0}$ be given by Lemma 3.2. Let ${T}_n=T$ for $0\leq n<N_0$ and $T_n=\min (\widetilde {T}_n,A_n^{-1/k_n},T_{n-1}/4)$ for $n\geq N_0$ . Then $0<{T}_n\leq {T}_{n-1}/4$ , $T_n\leq \widetilde {T}_n$ , $A_n{T}_n^{k_n}\leq 1$ , for all $n\geq N_0$ , and $\lim \limits _{N\to +\infty }T_N=0$ . So, there hold (3.2) for $j\in \mathbb {Z}\cap [0,k_N]$ , (3.3), (3.4), and (3.5) with $\widetilde {c}\in (0,T)$ given by Lemma 3.3.

Step 2. Proof of (2.25). As $ \beta>1$ , $(T-t)^{\beta -1}\in L^\infty (\mathcal C)$ , we get by (3.3) that

$$ \begin{align*} &(T-t)^{\beta-1+j}D^j(\phi_*-\phi_0)=(T-t)^{\beta-1}(T-t)^{j}D^j(\phi_*-\phi_0)\in L^\infty(\mathcal C),\quad \forall\ j\in\mathbb{Z}_{\geq0},\\ &(T-t)^{\frac{2\beta}{p-1}+j}D^j(\rho_*-\rho_0)=(T-t)^{\beta-1}(T-t)^{\frac{2\beta}{p-1}+j-\beta+1}D^j(\rho_*-\rho_0)\in L^\infty(\mathcal C),\quad \forall\ j\in\mathbb{Z}_{\geq0}, \end{align*} $$

which, along with with (3.4), implies (2.25).

Step 3. Proof of (2.26). By (3.3), we have (for some $ C_1>0$ )

$$ \begin{align*} &(T-t)(|\partial_t(\phi_*-\phi_0)|+|D_x(\phi_*-\phi_0)|)+(T-t)^{\frac{2\beta}{p-1}-\beta+1}|\rho_*-\rho_0|\leq C_1\quad \text{in}\,\, \mathcal C. \end{align*} $$

Now we take $c_0\in (0,T)$ such that $c_0+c_0^{\beta -1}C_1\leq \widetilde {c}$ , where the existence of such a $c_0$ is ensured by $\beta>1$ and $ \widetilde {c}>0$ . Then for $(t,x)\in \mathcal C, T-t<c_0$ , we get by (3.5) that (as $T-t>0$ )

$$ \begin{align*} (T-t)^{\beta}(\partial_t\phi_*-|D_x\phi_*|) \geq& (T-t)^{\beta}(\partial_t\phi_0-|D_x\phi_0|)-(T-t)^{\beta}(|\partial_t(\phi_*-\phi_0)|+|D_x(\phi_*-\phi_0)|)\\ \geq& \widetilde{c}-(T-t)^{\beta-1}C_1\geq \widetilde{c}-c_0^{\beta-1}C_1\geq c_0, \end{align*} $$

and

$$ \begin{align*} (T-t)^{\frac{2\beta}{p-1}}\rho_*\geq (T-t)^{\frac{2\beta}{p-1}}&\rho_0-(T-t)^{\frac{2\beta}{p-1}}|\rho_*-\rho_0| \geq \widetilde{c}-(T-t)^{\beta-1}C_1\geq \widetilde{c}-c_0^{\beta-1}C_1\geq c_0. \end{align*} $$

Step 4. Proof of (2.27). We fix $ \lambda>0$ , $j\in \mathbb {Z}_{\geq 0} $ . As $\lim \limits _{N\to +\infty }k_N=+\infty $ , there exists $N_1\in \mathbb {Z}_{\geq N_0}$ such that $ k_N>\lambda +j$ for $N\in \mathbb {Z}_{\geq N_1}$ . Then by (3.2) and $A_n{T}_n^{k_n}\leq 1$ , we have

$$ \begin{align*} |D^jE_{*}(t,x)|+|D^jJ_{*}(t,x)|\leq A_N(T-t)^{2k_N}\leq A_N{T}_N^{k_N}(T-t)^{k_N}\leq (T-t)^{k_N}\leq (T-t)^{\lambda}, \end{align*} $$

for $T-t\in [T_{N+1},T_N]$ , $(t,x)\in \mathcal C$ , $N\in \mathbb {Z}_{\geq N_1}$ . As $\lim _{N\to +\infty }T_N=0$ , we have

(3.6) $$ \begin{align} |D^jE_{*}(t,x)|+|D^jJ_{*}(t,x)|\leq (T-t)^{\lambda},\quad \forall\ T-t\in (0,T_{N_1}],\ (t,x)\in \mathcal C. \end{align} $$

As $\rho _*, \phi _*\in C^\infty ([0, T)\times \mathbb {R}^d)$ (see footnote 6), we have $E_*, J_*\in C^\infty ([0,T)\times \mathbb {R}^d)$ by (2.24). Thus, there exists a constant $ C(j,T_{N_1})>0$ such that

$$ \begin{align*} |D^jE_{*}(t,x)|+|D^jJ_{*}(t,x)|\leq C(j,T_{N_1}),\quad \forall\ T-t\in [T_{N_1},T],\ |x|\leq 2T. \end{align*} $$

Then (recall that $\mathcal C=\left \{(t,x)\in [0, T)\times \mathbb {R}^d: |x|<2(T-t)\right \}$ )

$$ \begin{align*} |D^jE_{*}(t,x)|+|D^jJ_{*}(t,x)|\leq C(j,T_{N_1})T_{N_1}^{-\lambda}(T-t)^{\lambda},\quad \forall\ T-t\in [T_{N_1},T],\ (t,x)\in \mathcal C, \end{align*} $$

which along with with (3.6) implies (2.27).

Proof. By the definition of ${\mathscr X}_{\lambda }^*$ , it suffices to prove the result for $f(t,x,s)=f_1(t,x)\eta _1\big (\frac {T-t}{s}\big )$ , $g(t,x,s)=g_1(t,x)\widetilde {\eta }_1\big (\frac {T-t}{s}\big )$ for some $f_1\in {\mathscr X}_{\lambda }, g_1\in {\mathscr X}_\mu $ , $\eta _1,\widetilde {\eta }_1\in \mathscr Y_0$ .

In this case, $\Delta f(t,x,s)=\Delta f_1(t,x)\eta _1\big (\frac {T-t}{s}\big )$ . By Lemma 2.3 (i), we have $\Delta f_1\in {\mathscr X}_{\lambda -2},$ thus $\Delta f\in {\mathscr X}_{\lambda -2}^*$ . We also have

$$ \begin{align*}\partial_t f(t,x,s)=\partial_t f_1(t,x)\eta_1\big((T-t)/{s}\big)+(T-t)^{-1}f_1(t,x)\eta_2\big({(T-t)/{s}}\big)\end{align*} $$

with $ \eta _2(z)=-z\eta _1'(z)\in \mathscr Y_0$ (as $\eta _2=0 $ , $\eta _2'=0 $ in $[0,1]\cup [2,+\infty ) $ ). By Lemma 2.3 (i), we have $\partial _t f_1\in {\mathscr X}_{\lambda -1}, (T-t)^{-1} \in {\mathscr X}_{-1}, (T-t)^{-1} f_1\in {\mathscr X}_{\lambda -1},$ thus $\partial _t f\in {\mathscr X}_{\lambda -1}^*$ .

In this case, $(fg)(t,x,s)=(f_1g_1)(t,x)(\eta _1\widetilde {\eta }_1)\big (\frac {T-t}{s}\big )$ . By Lemma 2.3 (i), we have $f_1g_1\in {\mathscr X}_{\lambda +\mu },$ as $\mathscr Y_0$ is a ring we have $\eta _1\widetilde {\eta }_1\in \mathscr Y_0$ , thus $fg\in {\mathscr X}_{\lambda +\mu }^*$ . This completes the proof of (i).

Assume that $\alpha _1,\cdots ,\alpha _d\in \mathbb {Z}_{\geq 0}$ are such that $\alpha _1+\cdots +\alpha _d=j$ , then we have

$$ \begin{align*} &(T-t)^{-\mu}\partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d}f(t,x,s) =(T-t)^{-\mu}\partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d} f_1(t,x)\eta_1(({T-t})/{s}). \end{align*} $$

Then (ii) follows from Lemma 2.3 (ii) and $ \eta _{1}\in L^{\infty }([0,+\infty ))$ .

Proof of Lemma 3.4.

By Lemma 3.5 (i), we have $fg\in {\mathscr X}_{\lambda +\mu }^*$ , $\Delta f\in {\mathscr X}_{\lambda -2}^*$ , $\partial _t f\in {\mathscr X}_{\lambda -1}^*$ , $\partial _t^2 f\in {\mathscr X}_{\lambda -2}^*$ , thus $\Box f=-\partial _t^2f+\Delta f\in {\mathscr X}_{\lambda -2}^*$ . As a consequence, we have $\Box (fg)\in \mathscr {X}_{(\lambda +\mu )-2}^* $ , $(\Box f)g\in \mathscr {X}_{(\lambda -2)+\mu }^*$ , $f\Box g\in \mathscr {X}_{\lambda +(\mu -2)}^* $ , hence $\partial ^\alpha f\partial _\alpha g=(\Box (fg)-(\Box f)g-f\Box g)/2\in \mathscr {X}_{\lambda +\mu -2}^*$ .

Assume that $\alpha _0,\alpha _1,\cdots ,\alpha _d\in \mathbb {Z}_{\geq 0}$ are such that $\alpha _0+\cdots +\alpha _d=j$ . By Lemma 3.5 (i), we have $\partial _t^{\alpha _0} f\in {\mathscr X}_{\lambda -\alpha _0}^*$ . Then by Lemma 3.5 (ii) and $\lambda -\alpha _0>j'+\mu $ (here $j'=\alpha _1+\cdots +\alpha _d=j-\alpha _0$ ), we have

$$ \begin{align*} &(T-t)^{-\mu}|\partial_t^{\alpha_0}\partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d}f|\leq (T-t)^{-\mu}|D_x^{j'}\partial_t^{\alpha_0}f|\in L^\infty(\mathcal C\times(0,1]). \end{align*} $$

This completes the proof.

Proof of Lemma 3.1.

For $t\in [0, T), x\in \mathbb {R}^d, s\in (0,+\infty )$ and $N\in \mathbb {Z}_{\geq 0}$ , let

$$ \begin{align*} &\rho_{N*}(t,x,s):=\sum_{n=0}^{N}\rho_n(t,x) b^n+\eta\left(\frac{T-t}{s}\right)\rho_{N+1}(t,x) b^{N+1},\\ & \phi_{N*}(t,x,s):=\sum_{n=0}^N \phi_n(t,x)b^n+\eta\left(\frac{T-t}{s}\right)\phi_{N+1}(t,x) b^{N+1}. \end{align*} $$

Then by (2.23), for all $N\in \mathbb {Z}_{\geq N_0-1}$ , we have

$$ \begin{align*} &\rho_{*}(t,x)=\rho_{N*}(t,x,T_{N+1}),\quad \phi_{*}(t,x)=\phi_{N*}(t,x,T_{N+1}),\quad \forall\ T-t\in [T_{N+1},T_N],\ (t,x)\in \mathcal C. \end{align*} $$

Let

(3.9) $$ \begin{align} &E_{N*}=\rho_{N*}^{p}+\rho_{N*}\partial^{\alpha}\phi_{N*}\partial_{\alpha}\phi_{N*}-b\Box\rho_{N*},\quad J_{N*}= 2\partial^{\alpha}\rho_{N*}\partial_{\alpha}\phi_{N*}+\rho_{N*}\Box\phi_{N*}. \end{align} $$

Then by (2.24), for all $N\in \mathbb {Z}_{\geq N_0-1}$ , we have

$$ \begin{align*} &E_{*}(t,x)=E_{N*}(t,x,T_{N+1}),\quad J_{*}(t,x)=J_{N*}(t,x,T_{N+1}),\quad \forall\ T-t\in [T_{N+1},T_N],\ (t,x)\in \mathcal C. \end{align*} $$

Now (3.2) is reduced to the proof of

(3.10) $$ \begin{align} (T-t)^{-2k_N}(D^jE_{N*},D^jJ_{N*})\in L^{\infty}(\mathcal C\times(0,1]),\quad \forall\ j\in\mathbb{Z}\cap[0,k_N],\ N\in \mathbb{Z}_{\geq N_0-1}. \end{align} $$

For $t\in [0, T), x\in \mathbb {R}^d, s\in (0,+\infty )$ , let

$$ \begin{align*} &\rho_{N,n}(t,x,s):=\left\{\begin{array}{ll} \rho_{n}(t,x) & n\in\mathbb{Z}\cap[0,N] \\ \rho_{n}^*(t,x,s) &n=N+1 \\ 0 & n\in\mathbb{Z}_{\geq N+2} \end{array} \right.,\\ &\phi_{N,n}(t,x,s):=\left\{\begin{array}{ll} \phi_{n}(t,x) & n\in\mathbb{Z}\cap[0,N] \\ \phi_{n}^*(t,x,s) &n=N+1 \\ 0 & n\in\mathbb{Z}_{\geq N+2} \end{array} \right.. \end{align*} $$

where

(3.11) $$ \begin{align} &\rho_{n}^*(t,x,s):=\eta\left((T-t)/{s}\right)\rho_{n}(t,x) ,\ \phi_{n}^*(t,x,s):=\eta\left((T-t)/{s}\right)\phi_{n}(t,x),\quad \forall\ n\in\mathbb{Z}_{\geq 0}. \end{align} $$

As $\rho _n\in {\mathscr X}_{\mu _n}$ , $\phi _n\in {\mathscr X}_{\lambda _n}$ for all $n\in \mathbb {Z}_{\geq 0}$ , we have $\rho _n^*,\rho _{N,n}\in {\mathscr X}_{\mu _n}^*$ , $\phi _n^*,\phi _{N,n}\in {\mathscr X}_{\lambda _n}^*$ for all $n,N\in \mathbb {Z}_{\geq 0}$ . For $t\in [0, T), x\in \mathbb {R}^d, s\in (0,+\infty )$ and $N\in \mathbb {Z}_{\geq 0}$ , we have

$$ \begin{align*} \rho_{N*}(t,x,s):=\sum_{n=0}^{N+1}\rho_{N,n}(t,x,s) b^n,\quad \phi_{N*}(t,x,s):=\sum_{n=0}^{N+1}\phi_{N,n}(t,x,s)b^n. \end{align*} $$

Then by (3.9), (2.5) and $\rho _{N,n}(t,x,s)=\rho _{n}(t,x) $ for $n\in \mathbb {Z}\cap [0,N] $ , we have

(3.12) $$ \begin{align} E_{N*}(t,x,s)=\sum_{n=N+1}^{p(N+1)}E_{N,n}^*(t,x,s) b^n,\quad J_{N*}(t,x,s)=\sum_{n=N+1}^{2(N+1)} J_{N,n}^*(t,x,s)b^n, \end{align} $$

with (note that $\rho _{N,n}(t,x,s)=0$ for $n\in \mathbb {Z}_{\geq N+2} $ )

$$ \begin{align*} E_{N,n}^*&=\sum_{n_1+\cdots+n_p=n}\rho_{N,n_1}\rho_{N,n_2}\cdots\rho_{N,n_p}+\sum_{n_1+n_2+n_3=n}\rho_{N,n_1}\partial^\alpha\phi_{N,n_2}\partial_\alpha\phi_{N,n_3}-\Box\rho_{N,n-1},\\ J_{N,n}^*&=2\sum_{n_1+n_2=n}\partial^\alpha\rho_{N,n_1}\partial_\alpha\phi_{N,n_2}+\sum_{n_1+n_2=n}\rho_{N,n_1}\Box\phi_{N,n_2}. \end{align*} $$

By Lemma 3.4 (i), we have $E_{N,n}^*\in \mathscr {X}_{2n(\beta -1)-2p\beta /(p-1)}^*$ , ${ J_{N,n}^*}\in \mathscr {X}_{(2n+1)(\beta -1)-2p\beta /(p-1)}^*$ , where we have used the facts that $\mu _{n_1}+\cdots +\mu _{n_p}=\mu _{n-1}-2$ if $n_1+\cdots +n_p=n$ , $\mu _{n_1}+\lambda _{n_2}+\lambda _{n_3}-2=\mu _{n-1}-2=2n(\beta -1)-2p\beta /(p-1)$ if $n_1+n_2+n_3=n$ and $\mu _{n_1}+\lambda _{n_2}-2=(2n-1)(\beta -1)-2\beta /(p-1)-2=(2n+1)(\beta -1)-2p\beta /(p-1)$ if $n_1+n_2=n$ .

If $n\geq N+1$ and $j\in \mathbb {Z}\cap [0,k_N]$ , then we get by (3.1) that

$$ \begin{align*} &(2n+1)(\beta-1)-2p\beta/(p-1)-j>2n(\beta-1)-2p\beta/(p-1)-j\\ &>2N(\beta-1)-2p\beta/(p-1)-j>3k_N-j\geq 2k_N. \end{align*} $$

Thus, by Lemma 3.4 (ii), we have $(T-t)^{-2k_N}D^jE_{N,n}^*\in L^\infty (\mathcal C\times (0,1])$ for $n\in \mathbb {Z}\cap [N+1,p(N+1)]$ , and $(T-t)^{-2k_N}D^jJ_{N,n}^*\in L^\infty (\mathcal C\times (0,1])$ for $n\in \mathbb {Z}\cap [N+1,2(N+1)]$ , which along with (3.12) implies (3.10).

Proof of Lemma 3.2.

For $(t,x)\in [0,T)\times \mathbb {R}^d$ , and any fixed $\{T_n\}_{n\geq 0}$ satisfying $T_n=T$ for $0\leq n<N_0$ and $0<T_n\le T_{n-1}/4$ for $n\ge N_0$ , we define $\rho _{*}, \phi _*$ by (2.23), then ( $ (\rho _n^*,\phi _n^*)$ is defined in (3.11))

$$ \begin{align*} &\rho_{*}(t,x)=\sum_{n=0}^{\infty}\rho_n^*(t,x,T_n) b^n, \quad \phi_{*}(t,x)=\sum_{n=0}^{\infty} \phi_n^*(t,x,T_n)b^n,\\ &(\rho_{*}-\rho_0)(t,x)=\sum_{n=1}^{\infty}\rho_n^*(t,x,T_n) b^n,\quad (\phi_{*}-\phi_{0})(t,x)=\sum_{n=1}^{\infty} \phi_n^*(t,x,T_n)b^n. \end{align*} $$

Recall that $\rho _n^*\in {\mathscr X}_{\mu _n}^*$ and $\phi _n^*\in {\mathscr X}_{\lambda _n}^*$ for all $n\in \mathbb {Z}_{\geq 0}$ . By (2.22), we have

$$ \begin{align*} \lambda_n-j&=(2n-1)(\beta-1)-j>(n-1)(\beta-1)-j,\quad \forall\ n\in\mathbb{Z}_{+},\\ \mu_n-j&=2n(\beta-1)-\frac{2\beta}{p-1}-j>n(\beta-1)-\frac{2\beta}{p-1}-j,\quad \forall\ n\in\mathbb{Z}_{+}. \end{align*} $$

Thus, by Lemma 3.4 (ii), for any $j,n\in \mathbb {Z}_{\geq 0}$ , there exists a constant $B_{n,j}>0$ , which is independent of the sequence $\{T_n\}_{n\geq 0}$ , such that for all $(t,x)\in \mathcal C$ , we have

$$ \begin{align*} (T-t)^{j-(n-1)(\beta-1)}|D^j\phi_n^*(t,x,T_n)|+(T-t)^{\frac{2\beta}{p-1}+j-n(\beta-1)}|D^j\rho_n^*(t,x,T_n)|\leq B_{n,j}, \end{align*} $$

which gives (recalling that $\eta ((T-t)/T_n))\neq 0$ implies $T-t\leq 2T_n$ )

$$ \begin{align*} (T-t)^{j}|D^j\phi_n^*(t,x,T_n)|+(T-t)^{\frac{2\beta}{p-1}+j-\beta+1}|D^j\rho_n^*(t,x,T_n)|\leq B_{n,j}(2T_n)^{(n-1)(\beta-1)}. \end{align*} $$

Let $\widetilde {T}_n:=T$ for $0\leq n<N_0$ and for $n\geq N_0$ we let

$$\begin{align*}\widetilde{B}_{n}:=2^n\max\limits_{0\leq j\leq n-N_0}B_{n,j}, \qquad\widetilde{T}_n:=\min\left(\widetilde{B}_{n}^{-1/[(n-1)(\beta-1)]}/2,\widetilde{T}_{n-1}/4\right).\end{align*}$$

Now we prove that $\{\widetilde T_n\}_{n\ge 0}$ is a desired sequence for Lemma 3.2.

Let $\{T_n\}_{n\geq 0}$ be such that $T_n=T$ for $0\le n<N_0$ and $0<T_n\leq \min (\widetilde T_n, T_{n-1}/4)$ for $n\geq N_0$ . Then $\widetilde {B}_{n}(2\widetilde {T}_n)^{(n-1)(\beta -1)}\leq 1 $ for $n\geq N_0$ . Fix $j\in \mathbb {Z}_{\geq 0}$ . For any $(t,x)\in \mathcal C$ , we have

$$ \begin{align*} &(T-t)^{j}|D^j(\phi_*-\phi_0)(t,x)|+(T-t)^{\frac{2\beta}{p-1}+j-\beta+1}|D^j(\rho_*-\rho_0)(t,x)|\\&\quad\leq \sum_{n=1}^{\infty}\left((T-t)^{j}|D^j\phi_n^*(t,x,T_n)|+(T-t)^{\frac{2\beta}{p-1}+j-\beta+1}|D^j\rho_n^*(t,x,T_n)|\right)\\&\quad\leq \sum_{n=1}^{\infty}B_{n,j}(2T_n)^{(n-1)(\beta-1)}\leq \sum_{n=1}^{\infty}B_{n,j}(2\widetilde{T}_n)^{(n-1)(\beta-1)}\\&\quad\leq \sum_{n=1}^{N_0+j-1}B_{n,j}(2\widetilde{T}_n)^{(n-1)(\beta-1)}+\sum_{n=N_0+j}^{\infty}2^{-n}\widetilde{B}_{n}(2\widetilde{T}_n)^{(n-1)(\beta-1)}\\&\quad\leq \sum_{n=1}^{N_0+j-1}B_{n,j}(2\widetilde{T}_n)^{(n-1)(\beta-1)}+\sum_{n=N_0+j}^{\infty}2^{-n} \leq \sum_{n=1}^{N_0+j-1}B_{n,j}(2\widetilde{T}_n)^{(n-1)(\beta-1)}+1, \end{align*} $$

which implies (3.3), as the right hand side is a finite constant independent of $(t,x)\in \mathcal C$ .

Proof of Lemma 3.3.

By (2.7), Lemma A.6 and Lemma 2.3 (iii), we obtain (3.4). It suffices to prove (3.5). By (2.7), we have

$$ \begin{align*} \partial_t\phi_0=(T-t)^{-\beta}\left((\beta-1)\widehat\phi_0(Z)+Z\widehat\phi_0'(Z)\right),\quad \partial_r\phi_0=(T-t)^{-\beta}\widehat\phi_0'(Z). \end{align*} $$

It follows from (2.12) that

(3.13) $$ \begin{align} \widehat\phi_0'(Z)=\frac{(\beta-1)\widehat\phi_0(Z)v(Z)}{1-Zv(Z)},\quad Z\widehat\phi_0'(Z)+(\beta-1)\widehat\phi_0(Z)=\frac{(\beta-1)\widehat\phi_0(Z)}{1-Zv(Z)}. \end{align} $$

Hence,

(3.14) $$ \begin{align} (T-t)^\beta\left(\partial_t\phi_0-|D_x\phi_0|\right)=(\beta-1)\frac{\widehat\phi_0(Z)(1-{|v(Z)|})}{1-Zv(Z)}. \end{align} $$

Since $\beta>1$ , $\widehat \phi _0(Z)>0, v(Z)\in (-1,1), Zv(Z)<1$ for all $Z\in [0,+\infty )$ and $\widehat \phi _0, v\in C^\infty ([0,+\infty ))$ , we know that the right hand side of (3.14) is strictly positive and continuous. Thus, there exists $\widetilde c_1\in (0, T)$ such that

$$ \begin{align*} \inf_{Z\in[0,2]}(\beta-1)\frac{\widehat\phi_0(Z)(1-{|v(Z)|})}{1-Zv(Z)}>\widetilde c_1. \end{align*} $$

On the other hand, by (2.7), we have $(T-t)^{2\beta /(p-1)}\rho _0=\widehat \rho _0(Z)$ . As $\widehat \rho (Z)>0$ and $\widehat \rho \in C([0,+\infty ))$ , there exists $\widetilde c_2\in (0, T)$ such that $\inf _{Z\in [0,2]}\widehat \rho _0(Z)>\widetilde c_2$ . As a consequence, letting $\widetilde c:=\min (\widetilde c_1, \widetilde c_2)\in (0, T)$ , we have (3.5).

Proof. This is a brute force computation. If $u=(1+h)w_*\text {e}^{\text {i}\Phi _*}$ , then for any $\alpha \in \mathbb {Z}\cap [0, d]$ , we have

$$ \begin{align*} \partial_\alpha u=\partial_\alpha hw_*\text{e}^{\text{i}\Phi_*}+(1+h)\partial_\alpha w_*\text{e}^{\text{i}\Phi_*}+\text{i}(1+h)w_*\text{e}^{\text{i}\Phi_*}\partial_\alpha\Phi_*. \end{align*} $$

Hence,

By the definition of $\varphi _1$ , we have $1+(p-1)h_{\text {r}}+h+\varphi _1(h)=(1+h)|1+h|^{p-1}$ . Thus,

$$ \begin{align*} &\Box u=|u|^{p-1}u\Longleftrightarrow\Box h+2\text{i}\partial^\alpha\Phi_*\partial_\alpha h+2\frac{\partial^\alpha w_*}{w_*}\partial_\alpha h+(1+h)\left(w_*^{p-1}+\frac{\text{i} J_*-E_*}{w_*}\right)\\ &\qquad\qquad\qquad\qquad\qquad\qquad =\big(1+(p-1)h_{\text{r}}+h+\varphi_1(h)\big)w_*^{p-1}\\ &\Longleftrightarrow\Box h+2\text{i}\partial^\alpha\Phi_*\partial_\alpha h+2\frac{\partial^\alpha w_*}{w_*}\partial_\alpha h-(p-1)w_*^{p-1}h_{\text{r}}=w_*^{p-1}\varphi_1(h)+\frac{E_*-\text{i} J_*}{w_*}(1+h). \end{align*} $$

This completes the proof of Lemma 4.1.

Proof. On $\mathcal C_1$ , we have

(4.9) $$ \begin{align}X=D\Phi_*,\quad Y=Dw_*/w_*,\quad V_*=(p-1)w_*^{p-1}.\end{align} $$

By (2.28), we have $(T-t)^{\beta }X_0=(T-t)^{\beta }\partial _t\Phi _*\in L^\infty (\mathcal C_1)$ and $(T-t)^{2\beta }V_*\in L^\infty (\mathcal C_1)$ . By (2.29), we have $(T-t)^\beta X_0=(T-t)^{\beta }\partial _t\Phi _*\geq c_0, (T-t)^{2\beta }V_*=(p-1)(T-t)^{2\beta }w_*^{p-1}\geq (p-1)c_0^{p-1}$ on $\mathcal C_1$ . This proves (4.6).

It follows from (2.28) that $(T-t)^{\beta +1}|DX|=(T-t)^{\beta +1}|D^2\Phi _*|\in L^\infty (\mathcal C_1)$ , hence by $(T-t)^\beta X_0\geq c_0$ on $\mathcal C_1$ , we have $(T-t)|DX|/X_0\in L^\infty (\mathcal C_1)$ . Similarly, by using (2.28) and (4.6), we get $(T-t)(|Y|+|DV_*|/V_*)\in L^\infty (\mathcal C_1)$ . This proves (4.5).

Next we prove (4.7) and (4.8). Recall the product rule: for smooth $f,g$ and $(\alpha _0, \alpha _1,\cdots , \alpha _d)\in \mathbb {Z}_{\geq 0}^{d+1}$ , we have (see [Reference Hardy24])

$$ \begin{align*} \frac{\partial^{\alpha_0+\alpha_1+\cdots+\alpha_d}}{\partial_t^{\alpha_0}\partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d}}(fg)= &\sum_{j=0}^{\alpha_0}\sum_{j_1=0}^{\alpha_1}\cdots\sum_{j_d=0}^{\alpha_d}\binom{\alpha_0}{j_0}\binom{\alpha_1}{j_1}\cdots\binom{\alpha_d}{j_d}\\ &\qquad\qquad\times\frac{\partial^{j_0+j_1+\cdots+j_d}f}{\partial_t^{j_0}\partial_{x_1}^{j_1}\cdots\partial_{x_d}^{j_d}}\cdot \frac{\partial^{\alpha_0-j_0+\alpha_1-j_1+\cdots+\alpha_d-j_d}g}{\partial_t^{\alpha_0-j_0}\partial_{x_1}^{\alpha_1-j_1}\cdots\partial_{x_d}^{\alpha_d-j_d}}. \end{align*} $$

Hence,

(4.10) $$ \begin{align} |D^n(fg)|&\lesssim_n\sum_{j=0}^n|D^jf||D^{n-j}g|,\quad |D_x^n(fg)|\lesssim_n\sum_{j=0}^n|D_x^jf||D_x^{n-j}g|,\quad \forall\ n\in\mathbb{Z}_{\geq 0}, \end{align} $$

(4.11) $$ \begin{align} |gD^nf|&\lesssim_n |D^n(fg)|+\sum_{j=0}^{n-1}|D^jf||D^{n-j}g|,\quad\forall\ n\in\mathbb{Z}_{+}. \end{align} $$

As $X=D\Phi _*$ on $\mathcal C_1$ , we get by (2.28) that

(4.12) $$ \begin{align} (T-t)^{\beta+j}|D^jX|\in L^\infty(\mathcal C_1),\quad\forall\ j\in\mathbb{Z}_{\geq 0}. \end{align} $$

Now we use the induction argument to prove that

(4.13) $$ \begin{align} (T-t)^{1+j}|D^jY|\in L^\infty(\mathcal C_1),\quad\forall\ j\in\mathbb{Z}_{\geq 0}. \end{align} $$

By (4.5), we know that (4.13) holds for $j=0$ . Assume that (4.13) holds for all $j\in \mathbb {Z}\cap [0, n-1]$ for some $n\in \mathbb {Z}_{+}$ . Note that $Dw_*=w_*Y$ on $\mathcal C_1$ , hence by (4.11) we have

$$\begin{align*}|w_*D^nY|\lesssim_n|D^n(Dw_*)|+\sum_{j=1}^{n}|D^jw_*||D^{n-j}Y|\quad\text{on}\quad \mathcal C_1.\end{align*}$$

Using (2.29), (2.28) and the induction assumption, we obtain

$$ \begin{align*} &(T-t)^{1+n}|D^nY|\leq \frac1{c_0}(T-t)^{1+n+\frac{2\beta}{p-1}}|w_*D^nY|\\& \quad\lesssim_n (T-t)^{1+n+\frac{2\beta}{p-1}}|D^{1+n}w_*|+\sum_{j=1}^{n}(T-t)^{\frac{2\beta}{p-1}+j}|D^jw_*|(T-t)^{1+n-j}|D^{n-j}Y|\in L^\infty(\mathcal C_1). \end{align*} $$

This proves (4.13).

Now we prove that

(4.14) $$ \begin{align} (T-t)^{\frac{2\beta}{p-1}m+j}\left|D^j(w_*^m)\right|\in L^\infty(\mathcal C_1),\quad \forall\ m\in\mathbb{Z}_{+},\ \forall\ j\in\mathbb{Z}_{\geq 0}. \end{align} $$

By (2.28), we know that (4.14) holds for $m=1$ . We assume that (4.14) holds for $m-1$ , where $m\in \mathbb {Z}\cap [2,+\infty )$ . By (4.10), for $j\in \mathbb {Z}_{\geq 0}$ we have

$$ \begin{align*} \left|D^j(w_*^m)\right|=\left|D^j(w_*^{m-1}w_*)\right|\lesssim_j \sum_{i=0}^j\left|D^i(w_*^{m-1})\right|\left|D^{j-i}w_*\right|, \end{align*} $$

which gives

$$ \begin{align*} (T-t)^{\frac{2\beta}{p-1}m+j}\left|D^j(w_*^m)\right|&\lesssim_j\sum_{i=0}^j(T-t)^{\frac{2\beta}{p-1}(m-1)+i}\left|D^i(w_*^{m-1})\right|(T-t)^{\frac{2\beta}{p-1}+j-i}\left|D^{j-i}w_*\right|\\ &\in L^\infty(\mathcal C_1). \end{align*} $$

By the induction argument, we have (4.14). Letting $m=p-1$ in (4.14), we get (using (4.9))

(4.15) $$ \begin{align} (T-t)^{2\beta+j}|D^jV_*|\in L^\infty(\mathcal C_1),\quad\forall\ j\in\mathbb{Z}_{\geq 0}. \end{align} $$

Finally, we estimate $N_*$ . Let $\widetilde \xi (t,x):=\xi (3|x|/(4(T-t)))$ . Then by Lemma 2.3 (iii), we have $(T-t)^{j}D^j\widetilde \xi \in L^\infty (\mathcal C)$ . Let $\widetilde N_*:=(E_*-\text {i} J_*)\widetilde \xi $ , then $N_*=\widetilde N_*/w_*$ . By (4.10), we have

$$ \begin{align*} \left|D^j\widetilde N_*\right|\lesssim_j \sum_{i=0}^j\left|D^i(E_*-\text{i} J_*)\right|\left|D^{j-i}\widetilde \xi\right|, \end{align*} $$

hence by (2.27), for all $\lambda>0$ and $j\in \mathbb {Z}_{\geq 0}$ we have

(4.16) $$ \begin{align} (T-t)^{-\lambda}\left|D^j\widetilde N_*\right|\lesssim_j\sum_{i=0}^j(T-t)^{-(\lambda+j-i)}\left|D^i(E_*-\text{i} J_*)\right|(T-t)^{j-i}\left|D^{j-i}\widetilde \xi\right|\in L^\infty(\mathcal C). \end{align} $$

Now we use the induction argument to prove that

(4.17) $$ \begin{align} (T-t)^{-\lambda}|D^jN_*|\in L^\infty(\mathcal C_1),\quad\forall\ j\in\mathbb{Z}_{\geq 0},\ \forall\ \lambda>0. \end{align} $$

For $j=0$ , (4.17) follows from (4.16) and (2.29). Assume that (4.17) holds for all $j\in \mathbb {Z}\cap [0, n-1]$ for some $n\in \mathbb {Z}_{+}$ . As $\widetilde N_*=w_*N_*$ , we get by (4.11) that

$$\begin{align*}\left|w_*D^nN_*\right|\lesssim_n \left|D^n\widetilde N_*\right|+\sum_{j=1}^{n}|D^jw_*||D^{n-j}N_*|\quad\text{on}\quad \mathcal C_1.\end{align*}$$

Using (2.27), (2.28), (2.29) and the induction assumption, for any $\lambda>0$ we obtain

$$ \begin{align*} &(T-t)^{-\lambda}|D^nN_*|\leq \frac1{c_0}(T-t)^{-\lambda+\frac{2\beta}{p-1}}\left|w_*D^nN_*\right|\\ \lesssim_n& (T-t)^{\frac{2\beta}{p-1}}(T-t)^{-\lambda}\left|D^n\widetilde N_*\right|+\sum_{j=1}^{n}(T-t)^{\frac{2\beta}{p-1}+j}\left|D^jw_*\right|(T-t)^{-(\lambda+j)}\left|D^{n-j}N_*\right| \in L^\infty(\mathcal C_1). \end{align*} $$

This proves (4.17) for $j=n$ . By induction, we have (4.17), which is equivalent to (4.8).

Taking $\lambda =1$ in (4.8), we get

(4.18) $$ \begin{align} (T-t)^{1+j}|D^jN_*|\leq T^{j+2}(T-t)^{-1}|D^jN_*|\in L^\infty(\mathcal C_1),\quad\forall\ j\in\mathbb{Z}_{\geq 0}. \end{align} $$

Therefore, (4.7) follows from (4.12), (4.13), (4.15) and (4.18).

Proof. Let $T_*\in (0, c_0)$ and $h\in C_c^\infty ([T-T_*, T)\times \mathbb {R}^d;\mathbb {C})$ be such that

$$ \begin{align*}\operatorname{supp}_xh(t,\cdot)\subset\{x\in\mathbb{R}^d: |x|\leq 4(T-t)/3\},\quad \forall\ t\in[T-T_*, T).\end{align*} $$

We define the energy momentum tensor $T[h]$ by

(4.23) $$ \begin{align} T[h]_{\mu\nu}:=\operatorname{Re}\left(\partial_\mu h\overline{\partial_\nu h}\right)-\frac12m_{\mu\nu}\left(\partial^\alpha h\overline{\partial_\alpha h}+V_*h_{\text{r}}^2\right), \quad\forall\ \mu,\nu\in\mathbb{Z}\cap[0, d], \end{align} $$

where we have used the Einstein’s convention in $\partial ^\alpha h\overline {\partial _\alpha h}$ . Then we have

(4.24) $$ \begin{align} E_0[h](t)=\int_{\mathbb{R}^d}T[h]_{00}(t,x)\,\mathrm dx,\quad \forall\ t\in[T-T_*, T). \end{align} $$

We define

(4.25) $$ \begin{align} P_\mu^X[h]:=T[h]_{\mu\nu}X^\nu,\quad \forall\ \mu\in\mathbb{Z}\cap[0,d]. \end{align} $$

Let’s first claim that there exists a constant $\widetilde c_0>0$ such that

(4.26) $$ \begin{align} P_0^X[h]\leq \widetilde c_0T[h]_{00}X^0\leq0\ \text{on}\ \mathcal C_*:=\{(t,x)\in[T-T_*, T)\times\mathbb{R}^d: |x|\leq 4(T-t)/3\}; \end{align} $$

and there exists a constant $C_*>0$ such that

(4.27) $$ \begin{align} \left|\partial^\mu P_\mu^X[h]\right|\leq C_*(T-t)^{-\beta}\left((T-t)^{-1}T[h]_{00}+\sqrt{T[h]_{00}}|\mathcal Lh|\right)\quad\text{on}\quad\mathcal C_*, \end{align} $$

and moreover, for all $j\in \mathbb {Z}_{\geq 0}$ ,

(4.28) $$ \begin{align} \left\|D_x^j(\mathcal LD_xh-D_x\mathcal Lh)\right\|_{L_x^2}\lesssim_j \sum_{i=0}^j(T-t)^{-\beta-1+i-j}\sqrt{E_i[h](t)}\quad\forall\ t\in[T-T_*, T), \end{align} $$

where the implicit constants only depend on $X, Y, V_*, N_*$ (and they are independent of h).

Now we prove (4.22) by the induction argument. We first consider $j=0$ . For all $t\in [T-T_*, T)$ , by (4.24), $X^0=-X_0$ , $\operatorname {supp}_xh(t,\cdot )\subset \{x\in \mathbb {R}^d: |x|\leq 4(T-t)/3\}$ , (4.26) and (4.6), we have

(4.29) $$ \begin{align} E_0[h](t)\leq \int_{\mathbb{R}^d}\frac{-P_0^X[h](t,x)}{\widetilde c_0X_0(t,x)}\,\mathrm dx\leq \frac{M}{\widetilde c_0}(T-t)^{\beta}\int_{\mathbb{R}^d}-P_0^X[h](t,x)\,\mathrm dx. \end{align} $$

Let

$$\begin{align*}\widetilde E_0[h](t)=\int_{\mathbb{R}^d}-P_0^X[h](t,x)\,\mathrm dx\geq0, \quad\forall\ t\in[T-T_*, T).\end{align*}$$

By the divergence theorem (recall that $\partial ^0=-\partial _0=-\partial _t$ ), we get

$$ \begin{align*} \frac{\mathrm d}{\mathrm dt}\widetilde E_0[h](t)=\int_{\mathbb{R}^d}\partial^0P_0^X[h](t,x)\,\mathrm dx=\int_{\mathbb{R}^d}\partial^\mu P_\mu^X[h](t,x)\,\mathrm dx, \quad\forall\ t\in[T-T_*, T). \end{align*} $$

Using (4.27), (4.24) and Cauchy’s inequality, we obtain

$$ \begin{align*} &\left|\frac{\mathrm d}{\mathrm dt}\widetilde E_0[h](t)\right|\leq C_*(T-t)^{-\beta}\left((T-t)^{-1}E_0[h](t)+\sqrt{E_0[h](t)}\|\mathcal Lh(t)\|_{L_x^2}\right),\quad\forall\ t\in[T-T_*, T). \end{align*} $$

Hence, by (4.29) and $h\in C_c^\infty ([T-T_*, T)\times \mathbb {R}^d;\mathbb {C})$ , for all $t\in [T-T_*, T)$ we have

$$ \begin{align*} E_0[h](t)&\leq \frac M{\widetilde c_0}(T-t)^\beta\widetilde E_0[h](t)\leq \frac M{\widetilde c_0}(T-t)^\beta\int_t^T\left|\frac{\mathrm d}{\mathrm dt}\widetilde E_0[h](s)\right|\,\mathrm ds\\&\leq \frac{MC_*}{\widetilde c_0}(T-t)^\beta\int_t^T(T-s)^{-\beta}\left((T-s)^{-1}E_0[h](s)+\sqrt{E_0[h](s)}\|\mathcal Lh(s)\|_{L_x^2}\right)\,\mathrm ds. \end{align*} $$

By Grönwall’s lemma, we have

$$ \begin{align*} \sqrt{E_0[h](t)}\leq\frac{MC_*}{2\widetilde c_0}\int_t^T\left(\frac{T-t}{T-s}\right)^{\frac{MC_0/\widetilde c_0+\beta}{2}}\|\mathcal Lh(s)\|_{L_x^2}\,\mathrm ds,\quad\forall\ t\in[T-T_*, T). \end{align*} $$

Letting $M_1:=\frac {MC_*/\widetilde c_0+\beta }{2}>0$ , we know that (4.22) holds for $j=0$ .

Let $n\in \mathbb {Z}_{+}$ . We assume that (4.22) holds for all $j\in \mathbb {Z}\cap [0,n-1]$ . Then by (4.22) for $j=n-1$ and (4.28), for $t\in [T-T_*, T)$ we have (also using (4.20) and (4.21))

$$ \begin{align*} &\sqrt{E_n[h](t)}=\sqrt{E_{n-1}[D_xh](t)}\lesssim_n\int_t^T\left(\frac{T-t}{T-s}\right)^{M_1}\sum_{j=0}^{n-1}\frac{\|D_x^j\mathcal LD_xh(s)\|_{L_x^2}}{(T-s)^{(n-1-j)\beta}}\,\mathrm ds\\&\quad\lesssim_n \int_t^T\left(\frac{T-t}{T-s}\right)^{M_1}\sum_{j=0}^{n-1}\frac{\|D_x^{j+1}\mathcal Lh(s)\|_{L_x^2}+\sum_{i=0}^j(T-s)^{-\beta-1+i-j}\sqrt{E_i[h](s)}}{(T-s)^{(n-1-j)\beta}}\,\mathrm ds\\&\quad\lesssim_n \int_t^T\left(\frac{T-t}{T-s}\right)^{M_1}\sum_{j=1}^{n}\frac{\|D_x^{j}\mathcal Lh(s)\|_{L_x^2}}{(T-s)^{(n-j)\beta}}\,\mathrm ds+I_n(t), \end{align*} $$

where

$$ \begin{align*} I_n(t):=\sum_{j=0}^{n-1}\sum_{i=0}^j\int_t^T\left(\frac{T-t}{T-s}\right)^{M_1}(T-s)^{-1+i-j-(n-j)\beta}\sqrt{E_i[h](s)}\,\mathrm ds. \end{align*} $$

For $T-T_*\leq t<s<T$ , $j\geq i\geq 0$ we have $0<T-s<T_*<c_0<T=1$ and $ (T-s)^{-1+i-j-(n-j)\beta }=(T-s)^{-1-(n-i)\beta +(j-i)(\beta -1)}\leq (T-s)^{-1-(n-i)\beta }$ (as $ \beta>1$ ). Then

$$ \begin{align*} &I_n(t)\leq n\sum_{i=0}^{n-1}\int_t^T\left(\frac{T-t}{T-s}\right)^{M_1}(T-s)^{-1-(n-i)\beta}\sqrt{E_i[h](s)}\,\mathrm ds. \end{align*} $$

Using the induction assumption and Fubini’s theorem, we have

$$ \begin{align*} I_n(t)&\lesssim_n\sum_{i=0}^{n-1}\sum_{j=0}^i \int_t^T\frac{\left((T-t)/(T-s)\right)^{M_1}}{(T-s)^{1+(n-i)\beta}} \int_s^T\left(\frac{T-s}{T-\tau}\right)^{M_1}\frac{\|D_x^j\mathcal Lh(\tau)\|_{L_x^2}}{(T-\tau)^{(i-j)\beta}}\,\mathrm d\tau\,\mathrm ds\\&=\sum_{i=0}^{n-1}\sum_{j=0}^i\int_t^T\left(\frac{T-t}{T-\tau}\right)^{M_1}\frac{\|D_x^j\mathcal Lh(\tau)\|_{L_x^2}}{(T-\tau)^{(i-j)\beta}}\int_t^\tau\frac{\mathrm ds}{(T-s)^{1+(n-i)\beta}}\,\mathrm d\tau\\&\leq\sum_{i=0}^{n-1}\sum_{j=0}^i\int_t^T\left(\frac{T-t}{T-\tau}\right)^{M_1}\frac{\|D_x^j\mathcal Lh(\tau)\|_{L_x^2}}{(T-\tau)^{(i-j)\beta}}\frac{1}{(T-\tau)^{(n-i)\beta}}\,\mathrm d\tau\\&\leq n\sum_{j=0}^{n-1}\int_t^T\left(\frac{T-t}{T-\tau}\right)^{M_1}\frac{\|D_x^j\mathcal Lh(\tau)\|_{L_x^2}}{(T-\tau)^{(n-j)\beta}}\,\mathrm d\tau. \end{align*} $$

Therefore, we obtain (4.22) for $j=n$ . This proves (4.22) for all $j\in \mathbb {Z}_{\geq 0}$ .

Thus, it remains to prove (4.26), (4.27) and (4.28). We start with

$$\begin{align*}P_0^X[h]=T[h]_{0\nu}X^\nu=T[h]_{00}X^0+\sum_{i=1}^dT[h]_{0i}X^i.\end{align*}$$

On $\mathcal C_*\subset \mathcal C_1$ , by (4.2), we have $-X^0=X_0=\partial _t\Phi _*$ and $X^i=X_i=\partial _i\Phi _*$ for $i\in \mathbb {Z}\cap [1,d]$ , hence by (4.23) and Cauchy’s inequality,

$$ \begin{align*} \left|\sum_{i=1}^dT[h]_{0i}X^i\right|&\leq\sum_{i=1}^d|\partial_th||\partial_ih||\partial_i\Phi_*|\leq |\partial_th||D_xh||D_x\Phi_*|\leq \frac{|\partial_th|^2+|D_xh|^2}{2}|D_x\Phi_*|\\ &\leq T[h]_{00}|D_x\Phi_*|. \end{align*} $$

On the other hand, by (2.28) and (2.29), there exists a constant $\widetilde c_0\in (0,1)$ such that

$$\begin{align*}\partial_t\Phi_*-|D_x\Phi_*|\geq c_0(T-t)^{-\beta}\geq \widetilde c_0\partial_t\Phi_*>0\quad\text{on}\quad \mathcal C_1.\end{align*}$$

Thus, we have $|X|\leq |\partial _t\Phi _*|+|D_x\Phi _*|\leq 2\partial _t\Phi _*=2X_0 $ and

$$\begin{align*}\left|\sum_{i=1}^dT[h]_{0i}X^i\right|\leq T[h]_{00}|D_x\Phi_*|\leq T[h]_{00}(1-\widetilde c_0)\partial_t\Phi_*=T[h]_{00}(1-\widetilde c_0)X_0,\end{align*}$$

hence

$$ \begin{align*} P_0^X[h]\leq T[h]_{00}X^0+T[h]_{00}(1-\widetilde c_0)X_0=\widetilde c_0T[h]_{00}X^0\leq0\quad\text{on}\quad \mathcal C_*. \end{align*} $$

This proves (4.26).

As for (4.27), we compute

$$ \begin{align*} \partial^\mu T[h]_{\mu\nu}&=\operatorname{Re}\left(\Box h\overline{\partial_\nu h}\right)+\operatorname{Re}\left(\partial_\mu h\partial^\mu\overline{\partial_\nu h}\right)-\frac12\partial_\nu\left(\partial^\alpha h\overline{\partial_\alpha h}+V_*h_{\text{r}}^2\right)\\ &=\operatorname{Re}\left(\Box h\overline{\partial_\nu h}\right)+\frac12\operatorname{Re}\partial_\nu\left(\partial_\mu h\overline{\partial^\mu h}\right)-\frac12\partial_\nu\left(\partial^\alpha h\overline{\partial_\alpha h}\right)-h_{\text{r}}\partial_\nu h_{\text{r}}V_*-\frac12 h_{\text{r}}^2\partial_\nu V_*\\ &=\operatorname{Re}\left(\Box h\overline{\partial_\nu h}\right)-h_{\text{r}}\partial_\nu h_{\text{r}}V_*-\frac12 h_{\text{r}}^2\partial_\nu V_* \end{align*} $$

for $\nu \in \mathbb {Z}\cap [0,d]$ . Hence,

$$ \begin{align*} \partial^\mu P_\mu^X[h]&=T[h]_{\mu\nu}\partial^\mu X^\nu+(\partial^\mu T[h]_{\mu\nu})X^\nu\\ &=T[h]_{\mu\nu}(\pi^X)^{\mu\nu}+\operatorname{Re}\left(\Box hX^\nu\overline{\partial_\nu h}\right)-h_{\text{r}}X^\nu\partial_\nu h_{\text r}V_*-\frac12h_{\text{r}}^2X^\nu\partial_\nu V_*\\ &=T[h]_{\mu\nu}(\pi^X)^{\mu\nu}+\operatorname{Re}\left(\Box h\overline{Xh}\right)-V_*h_{\text r}Xh_{\text r}-\frac12h_{\text r}^2XV_*, \end{align*} $$

where we have used the fact that $X^\nu $ is real-valued for $\nu \in \mathbb {Z}\cap [0, d]$ , $Xh=X^\nu \partial _\nu h$ and we define

(4.30) $$ \begin{align} (\pi^X)^{\mu\nu}:=\frac{\partial^\mu X^\nu+\partial^\nu X^\mu}2,\quad\forall\ \mu,\nu\in\mathbb{Z}\cap[0,d]. \end{align} $$

Hence, it follows from (4.19) that

(4.31) $$ \begin{align} \partial^\mu P_\mu^X[h]&=T[h]_{\mu\nu}(\pi^X)^{\mu\nu}+\operatorname{Re}\left(\Box h\overline{Xh}\right)-V_*h_{\text r}Xh_{\text r}-\frac12h_{\text r}^2XV_*\nonumber\\ &=T[h]_{\mu\nu}(\pi^X)^{\mu\nu}-\frac12h_{\text r}^2XV_*+\operatorname{Re}\left(\mathcal Lh\overline{Xh}\right)-2\operatorname{Re}\left(Yh\overline{Xh}\right). \end{align} $$

By (4.23), we have $|T[h]_{\mu \nu }|\leq T[h]_{00}$ for all $\mu ,\nu \in \mathbb {Z}\cap [0,d]$ and $|Dh|^2\leq 2T[h]_{00}$ . Thus, by (4.31), $|X|\leq 2 X_0$ , (4.5) and (4.6), on $\mathcal C_*$ we have (note that $|XV_*|\leq |X||DV_*|$ , $|Xh|\leq |X||Dh|$ , $|Yh|\leq |Y||Dh|$ , see footnote 8)

$$ \begin{align*} \left|\partial^\mu P_\mu^X[h]\right|&\lesssim T[h]_{00}|DX|+T[h]_{00}X_0\frac{|DV_*|}{V_*}+|\mathcal Lh||X||Dh|+|Y||Dh||X||Dh|\\ &\lesssim T[h]_{00}(T-t)^{-1}X_0+|\mathcal Lh|X_0\sqrt{T[h]_{00}}\\ &\lesssim (T-t)^{-\beta}\left((T-t)^{-1}T[h]_{00}+\sqrt{T[h]_{00}}|\mathcal Lh|\right), \end{align*} $$

which gives (4.27).

Finally, we prove (4.28). By (4.19), we have

$$ \begin{align*} \mathcal LD_xh-D_x\mathcal Lh=-2\text{i} D_xX^{\alpha}\partial_{\alpha}h-2D_xY^{\alpha}\partial_{\alpha}h+D_xV_*\cdot h_{\text r}. \end{align*} $$

Let $j\in \mathbb {Z}_{\geq 0}$ , by (4.10) and (4.7), for any $t\in [T-T_*, T)$ we have

$$ \begin{align*} \|D_x^j(D_xX^{\alpha}\partial_{\alpha}h)(t)\|_{L_x^2}&\lesssim_j\sum_{i=0}^j\|D_x^{j-i}D_xX(t)\|_{L_x^\infty}\|DD_x^{i}h(t)\|_{L_x^2}\\ &\lesssim_j\sum_{i=0}^j(T-t)^{-\beta-1+i-j}\sqrt{E_i[h](t)}. \end{align*} $$

Similarly, we have (recalling $\beta>1$ )

$$ \begin{align*} \|D_x^j(D_xY^{\alpha}\partial_{\alpha}h)(t)\|_{L_x^2}\lesssim_j \sum_{i=0}^j(T-t)^{-2+i-j}\sqrt{E_i[h](t)}\lesssim_j\sum_{i=0}^j(T-t)^{-\beta-1+i-j}\sqrt{E_i[h](t)}. \end{align*} $$

By (4.6) and (4.7), we have

$$ \begin{align*} \|D_x^j(D_xV_*\cdot h_r)\|_{L_x^2}&\lesssim_j\sum_{i=0}^j\|D_x^{j-i+1}V_*/\sqrt{V_*}\|_{L_x^\infty}\|\sqrt{V_*}D_x^ih_{\text{r}}\|_{L_x^2}\\ &\lesssim_j\sum_{i=0}^j(T-t)^{-\beta-1+i-j}\sqrt{E_i[h](t)}. \end{align*} $$

Hence, we get (4.28).

Proof of Lemma 4.4.

Let $f\in C_c^\infty ([T-c_0, T)\times \mathbb {R}^d;\mathbb {C})$ be such that $|D_x^jf|\leq |D_x^jN_*|$ for all $j\in \mathbb {Z}_{\geq 0}$ . We assume that $\varepsilon \in (0, c_0)$ satisfies $f(t,x)=0$ for all $(t,x)\in (T-\varepsilon , T)\times \mathbb {R}^d$ . By the standard local well-posedness theory (Theorem 6.4.11 in [Reference Hörmander25]), there is a unique local solution $h\in C^\infty ((T-T_+, T)\times \mathbb {R}^d;\mathbb {C})$ to (4.32) with $(h, \partial _th)|_{t=T-\varepsilon /2}=(0, 0)$ , where $\varepsilon <T_+\leq c_0$ corresponds to the left life span of h; moreover, if $T_+<c_0$ , then

(4.38) $$ \begin{align} \limsup_{t\downarrow T-T_+}\|h(t,\cdot)\|_{L^\infty(\mathbb{R}^d)}=+\infty. \end{align} $$

By the uniqueness and $f(t,x)=0$ for all $(t,x)\in (T-\varepsilon , T)\times \mathbb {R}^d$ , we have $h(t,x)=0$ for all $(t,x)\in (T-\varepsilon , T)\times \mathbb {R}^d$ . Moreover, by $|f|\leq |N_*|$ , (4.4) and the finite speed of propagation, we have

$$\begin{align*}\operatorname{supp}_xh(t,\cdot)\subset \{x\in\mathbb{R}^d: |x|\leq 4(T-t)/3\},\quad\forall\ t\in[T-T_+, T).\end{align*}$$

Let $c_2\in (0, c_0)$ be such that $C_{0,2\beta }\cdot c_2<1/2$ , where $C_{0,2\beta }>0$ is given by (4.37). Note that $c_2$ is independent of f and $T_*$ . We claim that $T_+\geq c_2$ . We assume in contrary that $\varepsilon <T_+<c_2$ . Let

(4.39) $$ \begin{align} \mathscr{E}:=\left\{T_0\in(0, T_+): \|h(t,\cdot)\|_{L^\infty(\mathbb{R}^d)}\leq (T-t)^{2\beta-1} \text{ for all }t\in[T-T_0, T)\right\}. \end{align} $$

Then $(0,\varepsilon )\subset {\mathscr {E}}$ . Let $T_s:=\sup {\mathscr {E}}\in [\varepsilon , T_+]$ . By (4.38), we have $T_s<T_+$ , hence $T_s\in \mathscr {E}$ and $T_s<T_+<c_2$ . By (4.37), we have

$$ \begin{align*} |h(t,x)|\leq C_{0,2\beta}(T-t)^{2\beta}=C_{0,2\beta}(T-t)(T-t)^{2\beta-1}\leq C_{0,2\beta}c_2(T-t)^{2\beta-1}<\frac12(T-t)^{2\beta-1} \end{align*} $$

for all $t\in [T-T_s, T)\subset [T-c_2, T)$ . Thus, by the continuity we have $T_s+\delta \in \mathscr {E}$ for some $\delta>0$ . This contradicts with $T_s=\sup \mathscr {E}$ . Therefore, $T_+\geq c_2$ and $\|h(t,\cdot )\|_{L^\infty (\mathbb {R}^d)}\leq (T-t)^{2\beta -1}$ for all $t\in [T-c_2, T)$ . Now Lemma 4.5 (letting $T_*=c_2$ ) implies Lemma 4.4.

Proof of Lemma 4.5.

Assume that h solves (4.32). Then ( $\mathcal L$ is defined in (4.19))

$$\begin{align*}\mathcal Lh=\frac1{p-1}V_*\varphi_1(h)+N_*h+f.\end{align*}$$

We claim that for each $j\in \mathbb {Z}_{\geq 0}$ , there exists a constant $\widetilde C_j>0$ such that

(4.40) $$ \begin{align} \sqrt{E_j[h](t)}\leq \widetilde C_j\int_t^T\left(\frac{T-t}{T-s}\right)^{M_1}\sum_{i=0}^j\frac{\left\|D_x^i f(s)\right\|_{L_x^2}}{(T-s)^{(j-i)\beta}}\mathrm ds,\quad\forall\ t\in[T-T_*, T). \end{align} $$

By the definition of $\varphi _1$ , we know that $\varphi _1$ is a polynomial on $(h, \overline {h})$ of the form $\varphi _1=\sum _{2\leq i+j\leq p}c_{i,j}h^i\overline {h}^j$ , with $c_{i,j}\in \mathbb {R}$ , thus

(4.41) $$ \begin{align} |\varphi_1(h)|\lesssim |h|^{2}+|h|^p,\quad\forall\ h\in\mathbb{C}. \end{align} $$

Hence, $\operatorname {supp}_x\varphi _1(h)(t,\cdot )\subset \{x\in \mathbb {R}^d: |x|\leq 4(T-t)/3\}$ for all $t\in [T-T_*, T)$ . For $j\in \mathbb {Z}_{\geq 0}$ and $t\in [T-T_*, T)$ , by (4.10), (4.7) and Poincaré’s inequality, we have

$$ \begin{align*} \left\|D_x^j(V_*\varphi_1(h))(t)\right\|_{L_x^2}&\lesssim_j \sum_{i=0}^j\|D_x^{j-i}V_*(t)\|_{L_x^\infty}\|D_x^i\varphi_1(h)(t)\|_{L_x^2}\\ &\lesssim_j \sum_{i=0}^j(T-t)^{-(2\beta+j-i)}(T-t)^{j-i}\|D_x^j\varphi_1(h)(t)\|_{L_x^2}\\&\lesssim_j (T-t)^{-2\beta}\|D_x^j\varphi_1(h)(t)\|_{L_x^2}\end{align*} $$

Using the classical product estimate,

(4.42) $$ \begin{align} \|D_x^n(fg)\|_{L_x^2}\lesssim_n \|f\|_{L_x^\infty}\|D_x^ng\|_{L_x^2}+\|g\|_{L_x^\infty}\|D_x^nf\|_{L_x^2},\quad\forall\ n\in\mathbb{Z}_{\geq0}, \end{align} $$

and (4.35), we infer

$$ \begin{align*} &\|D_x^n(h^i\overline{h}^j)\|_{L_x^2}\lesssim_{n,i,j} \|h\|_{L_x^\infty}^{i+j-1}\|D_x^nh\|_{L_x^2},\quad\forall\ n,i,j\in\mathbb{Z}_{\geq0},\ i+j\geq2,\\ &\|D_x^j\varphi_1(h)(t)\|_{L_x^2}\lesssim_j\left(\|h(t)\|_{L_x^\infty}+\|h(t)\|_{L_x^\infty}^{p-1}\right)\|D_x^jh(t)\|_{L_x^2}\lesssim_j (T-t)^{2\beta-1}\|D_x^jh(t)\|_{L_x^2}, \\ &\left\|D_x^j(V_*\varphi_1(h))(t)\right\|_{L_x^2}\lesssim_j(T-t)^{-2\beta} (T-t)^{2\beta-1}\|D_x^jh(t)\|_{L_x^2}=(T-t)^{-1}\|D_x^jh(t)\|_{L_x^2}; \end{align*} $$

Similarly, by (4.7) and Poincaré’s inequality, we have

$$ \begin{align*} \left\|D_x^j(N_*h)(t)\right\|_{L_x^2}\lesssim_j(T-t)^{-1}\|D_x^jh(t)\|_{L_x^2}. \end{align*} $$

Therefore, for each $j\in \mathbb {Z}_{\geq 0}$ , there holds

(4.43) $$ \begin{align} \left\|D_x^j\mathcal Lh(t)\right\|_{L_x^2}\lesssim_j (T-t)^{-1}\|D_x^jh(t)\|_{L_x^2}+\|D_x^jf(t)\|_{L_x^2}. \end{align} $$

By (4.43), (4.22), for any $j\in \mathbb {Z}_{\geq 0}$ and $t\in [T-T_*, T)$ we have

(4.44) $$ \begin{align} \sqrt{E_j[h](t)}&\lesssim_j\int_t^T\left(\frac{T-t}{T-s}\right)^{M_1}\sum_{i=0}^j\frac{(T-s)^{-1}\|D_x^ih(s)\|_{L_x^2}+\|D_x^if(s)\|_{L_x^2}}{(T-s)^{(j-i)\beta}}\mathrm ds. \end{align} $$

It follows from Poincaré’s inequality and $\operatorname {supp}_xh(t,\cdot )\subset \{x\in \mathbb {R}^d: |x|\leq 4(T-t)/3\}$ that

(4.45) $$ \begin{align}(T-t)^{-1}\|D_x^jh(t)\|_{L_x^2}\lesssim\|D_x^{j+1}h(t)\|_{L_x^2}\leq\sqrt{2E_j[h](t)},\quad \forall\ t\in [T-T_*, T),\ j\in\mathbb{Z}_{\geq0}.\end{align} $$

Here we also used the definitions of $E_0$ and $E_j$ in (4.20) and (4.21). Next we use the induction argument to prove (4.40).

For $j=0$ , by (4.44) and (4.45), there exists a constant $C_0'>0$ satisfying

$$ \begin{align*} \sqrt{E_0[h](t)}\leq C_0'\int_t^T\left(\frac{T-t}{T-s}\right)^{M_1}\left(\sqrt{E_0[h](s)}+\|f(s)\|_{L_x^2}\right)\,\mathrm ds,\quad\forall\ t\in[T-T_*, T). \end{align*} $$

By Grönwall’s lemma, we get

$$ \begin{align*} (T-t)^{-M_1}\sqrt{E_0[h](t)}&\leq C_0'\int_t^T(T-s)^{-M_1}\text{e}^{C_0'(s-t)}\|f(s)\|_{L_x^2}\,\mathrm ds\\ &\leq C_0'\text{e}^{C_0'T}\int_t^T(T-s)^{-M_1}\|f(s)\|_{L_x^2}\,\mathrm ds \end{align*} $$

for all $t\in [T-T_*, T)$ . This proves (4.40) for $j=0$ . Let $n\in \mathbb {Z}_{+}$ , assume that (4.40) holds for $j\in \mathbb {Z}\cap [0, n-1]$ . By (4.20), (4.21) and $ \beta>1$ , we have

$$ \begin{align*} (T-s)^{-1}\|D_x^nh(s)\|_{L_x^2}\leq (T-s)^{-1}\sqrt{2E_{n-1}[h](s)}\leq (T-s)^{-\beta}\sqrt{2E_{n-1}[h](s)}, \end{align*} $$

for $ s\in [T-T_*, T)$ . Then by (4.44) for $j=n$ , (4.45) for $j=i<n$ , and the induction assumption, we have (as $0<T-t\leq T_*<c_0<1$ )

$$ \begin{align*} \sqrt{E_n[h](t)}&\lesssim_n\int_t^T\left(\frac{T-t}{T-s}\right)^{M_1}\left(\|D_x^nf(s)\|_{L_x^2}+\sum_{j=0}^{n-1}\frac{\sqrt{E_j[h](s)}+\|D_x^jf(s)\|_{L_x^2}}{(T-s)^{(n-j)\beta}}\right)\,\mathrm ds\\ &\lesssim_n \int_t^T\left(\frac{T-t}{T-s}\right)^{M_1}\sum_{j=0}^n\frac{\|D_x^jf(s)\|_{L_x^2}}{(T-s)^{(n-j)\beta}}\,\mathrm ds+I_n(t), \end{align*} $$

where

$$ \begin{align*} I_n(t)&:=\sum_{j=0}^{n-1}\int_t^T\frac{\big((T-t)/(T-s)\big)^{M_1}}{(T-s)^{(n-j)\beta}}\int_s^T\left(\frac{T-s}{T-\tau}\right)^{M_1}\sum_{i=0}^j\frac{\|D_x^if(\tau)\|_{L_x^2}}{(T-\tau)^{(j-i)\beta}}\,\mathrm d\tau\,\mathrm ds\\ &=\sum_{j=0}^{n-1}\sum_{i=0}^j\int_t^T\left(\frac{T-t}{T-\tau}\right)^{M_1}\frac{\|D_x^if(\tau)\|_{L_x^2}}{(T-\tau)^{(j-i)\beta}}\int_t^{\tau}\frac{\mathrm ds}{(T-s)^{(n-j)\beta}}\,\mathrm d\tau\\ &\leq \sum_{j=0}^{n-1}\sum_{i=0}^j\int_t^T\left(\frac{T-t}{T-\tau}\right)^{M_1}\frac{\|D_x^if(\tau)\|_{L_x^2}}{(T-\tau)^{(j-i)\beta}}\frac{1}{(T-\tau)^{(n-j)\beta}}\,\mathrm d\tau\\ &\leq n\sum_{i=0}^{n-1}\int_t^T\left(\frac{T-t}{T-\tau}\right)^{M_1}\frac{\|D_x^if(\tau)\|_{L_x^2}}{(T-\tau)^{(n-i)\beta}}\,\mathrm d\tau.\end{align*} $$

Thus, (4.40) holds for $j=n$ . Therefore, by the the induction, (4.40) holds for all $j\in \mathbb {Z}_{\geq 0}$ .

As $|D_x^jf|\leq |D_x^jN_*|$ for all $j\in \mathbb {Z}_{\geq 0}$ , by (4.4) and (4.8), for all $j\in \mathbb {Z}_{\geq 0}$ and $\lambda>0$ there exists a constant $\widetilde M_{j,\lambda }>0$ which is independent of f and $T_*$ such that $\|D_x^jf(t)\|_{L^2}\leq \widetilde M_{j,\lambda }(T-t)^\lambda $ for all $t\in [T-T_*, T)$ . Using (4.40), (4.21) and (4.45), for all $j\in \mathbb {Z}_{\geq 0}$ and $\lambda>0$ there exists a constant $M_{j,\lambda }'>0$ independent of f and $T_*$ (depending on $\widetilde M_{j,\lambda '} $ for some $\lambda '>\lambda $ ) such that

$$ \begin{align*} \|h(t)\|_{H_x^j}+\|\partial_th(t)\|_{H_x^j}\leq M_{j,\lambda}'(T-t)^\lambda,\quad\forall\ t\in[T-T_*, T). \end{align*} $$

By Sobolev’s embedding theorem ( $H_x^d(\mathbb {R}^d)\hookrightarrow L_x^\infty (\mathbb {R}^d)$ ), we have (4.37). It remains to prove (4.36). By (4.41), (4.35), (4.32), (4.37), (4.7) ( $j=0$ ) and (4.8) ( $j=0, \lambda =1$ ), we know that there there exist constants $ C_2'>0, C_3'>0$ such that

$$ \begin{align*} |\Box h|&\leq C_2'(T-t)^{-2\beta}|D^{\leq 1}h|+|f|\leq C_2'(T-t)^{-2\beta}(C_{0, 2\beta}+C_{1,2\beta})(T-t)^{2\beta}+|N_*| \leq C_3' \end{align*} $$

on $[T-T_*, T)\times \mathbb {R}^d$ , which implies (4.36).

Proof of Proposition 2.8.

Let $\xi _1=1-\xi $ , then $\xi _1|_{[0,4/5]}=0, \xi _1|_{[1,+\infty )}=1$ . Let

$$\begin{align*}\varepsilon_n:=c_2/2^n,\quad f_n(t,x):=N_*(t,x)\xi_1((T-t)/\varepsilon_n), \quad \forall\ [T-c_0, T)\times\mathbb{R}^d,\ \forall\ n\in\mathbb{Z}_{+}.\end{align*}$$

Then for each $n\in \mathbb {Z}_{+}$ , we have $f_n\in C^\infty ([T-c_0, T)\times \mathbb {R}^d;\mathbb {C})$ and $\operatorname {supp}f_n\subset \{(t,x)\in [T-c_0, T-4\varepsilon _n/5]\times \mathbb {R}^d: |x|\leq 4(T-t)/3\}$ , hence $f_n\in C_c^\infty ([T-c_0, T)\times \mathbb {R}^d;\mathbb {C})$ ; as $f_n$ equals to $N_*$ multiplied by a function in t that takes values in $[0, 1]$ , we have $|D_x^jf_n|\leq |D_x^jN_*|$ for all $j\in \mathbb {Z}_{+}$ . By Lemma 4.4, for each $n\in \mathbb {Z}_{+}$ , there exists $h_n\in C_c^\infty ([T-c_2, T)\times \mathbb {R}^d;\mathbb {C})$ satisfying

(4.46) $$ \begin{align} \Box h_n+2\text{i} Xh_n+2Yh_n-V_*(h_n)_{\text{r}}-\frac1{p-1}V_*\varphi_1(h_n)-N_*h_n=f_n=N_*\xi_1\left(\frac{T-t}{\varepsilon_n}\right) \end{align} $$

on $[T-c_2, T)\times \mathbb {R}^d$ . Moreover, for $n\in \mathbb {Z}_{+}$ , we have

(4.47) $$ \begin{align} \operatorname{supp}_xh_n(t,\cdot)\subset\{x\in\mathbb{R}^d: |x|\leq4(T-t)/3\},\quad\forall\ t\in [T-c_2, T), \end{align} $$

and for $j\in \mathbb {Z}_{\geq 0},\lambda>0$ we have

(4.48) $$ \begin{align} |\Box h_n(t,x)|\leq C_\Box, \quad\forall\ t\in [T-c_2, T),\ \forall\ x\in\mathbb{R}^d, \end{align} $$

(4.49) $$ \begin{align} |D_x^jh_n(t,x)|+|\partial_tD_x^jh_n(t,x)|\leq C_{j,\lambda}(T-t)^\lambda,\quad \forall\ t\in [T-c_2, T),\ \forall\ x\in\mathbb{R}^d, \end{align} $$

where $C_\Box $ and $C_{j,\lambda }$ are given by Lemma 4.4. By $h_n\in C_c^\infty ([T-c_2, T)\times \mathbb {R}^d;\mathbb {C})$ , (4.47), (4.48), (4.49) and the Arzelà–Ascoli theorem, there exists a subsequence of $\{h_n\}_{n=1}^\infty $ , which is still denoted by $\{h_n\}_{n=1}^\infty $ Footnote ⁹, such that $h_n\to h$ in $C^1([T-c_0, T)\times \mathbb {R}^d)$ for some $h\in C^1([T-c_0, T)\times \mathbb {R}^d;\mathbb {C})$ and (here $D^{\leq 1}f:=(f,Df)$ )

(4.50) $$ \begin{align} \left\|D^{\leq 1}(h_n-h_{n+1})\right\|_{L^\infty([T-c_0, T)\times \mathbb{R}^d)}\leq 2^{-n},\quad \forall\ n\in\mathbb{Z}_{+}. \end{align} $$

Letting $j=2$ , $\lambda =1$ in (4.49), by (4.50) and the Gagliardo–Nirenberg inequality, we have

(4.51) $$ \begin{align} \left\|D_xD(h_n-h_{n+1})\right\|_{L^\infty([T-c_0, T)\times \mathbb{R}^d)}\leq \widetilde C_0 2^{-n/2},\quad \forall\ n\in\mathbb{Z}_{+} \end{align} $$

for some constant $\widetilde C_0>0$ which is independent of $n\in \mathbb {Z}_{+}$ . Letting $j=0$ and $\lambda =1$ in (4.49), by the definition of $\varphi _1$ , there exists a constant $\widetilde C_1>0$ such that for all $n\in \mathbb {Z}_+$ and $(t,x)\in [T-c_2, T)\times \mathbb {R}^d$ , we have

$$\begin{align*}|\varphi_1(h_n)-\varphi_1(h_{n+1})|\leq \widetilde C_1|h_n-h_{n+1}|.\end{align*}$$

Combining this with (4.46), (4.49), (4.7) ( $j=0$ ), and (4.8) ( $j=0, \lambda =1$ ), we know that there exist constants $\widetilde C_2>0, \widetilde C_3>0$ such that for all $n\in \mathbb {Z}_{+}$ , we haveFootnote ¹⁰

$$ \begin{align*} |\Box h_{n}-\Box h_{n+1}|&\leq \widetilde C_2(T-t)^{-2\beta}\left|D^{\leq 1}(h_n-h_{n+1})\right|+\widetilde C_2\varepsilon_n\\ &\overset{(4.49)}{\underset{(4.50)}{\leq}}\widetilde C_2(T-t)^{-2\beta}\min\left(2^{-n}, 2(C_{0, 4\beta}+C_{1,4\beta})(T-t)^{4\beta}\right)+\widetilde C_2\varepsilon_n\\ &\leq \widetilde C_3(2^{-n/2}+\varepsilon_n) \end{align*} $$

on $[T-c_2, T)\times \mathbb {R}^d$ . Hence, by (4.50) and (4.51) we know that $\{h_n\}_{n=1}^\infty $ is Cauchy in $C^2([T-c_2, T)\times \mathbb {R}^d;\mathbb {C})$ , hence $h\in C^2([T-c_2, T)\times \mathbb {R}^d;\mathbb {C})$ and $h_n\to h$ in $C^2$ . Moreover, by (4.49), (4.50) and the Gagliardo–Nirenberg inequality we know that $\{D_x^jD^{\leq 1}h_n\}$ is Cauchy in $L^\infty ([T-c_2, T)\times \mathbb {R}^d)$ for all $j\in \mathbb {Z}_{\geq 0}$ . Hence (also using (4.47)) $h(t,\cdot ), \partial _th(t,\cdot )\in C_c^\infty (\mathbb {R}^d)$ with $\operatorname {supp}_xh(t,\cdot )\subset \{x\in \mathbb {R}^d: |x|\leq 4(T-t)/3\}$ . Moreover, h solves the equation (as $f_n\to N_*$ )

(4.52) $$ \begin{align} \Box h+2\text{i} Xh+2Yh-V_*h_{\text{r}}-\frac1{p-1}V_*\varphi_1(h)-N_*h=N_*\quad\text{on}\quad [T-c_2, T)\times\mathbb{R}^d. \end{align} $$

By (4.49), we have

$$\begin{align*}|h(t,x)|\leq C_{0,1}(T-t),\quad \forall\ (t,x)\in[T-c_2, T)\times\mathbb{R}^d.\end{align*}$$

Let $c_1\in (0,c_2)\subset (0, T)$ be such that $C_{0, 1}c_1<1/2$ , hence $|h(t,x)|<1/2$ for all $(t,x)\in [T-c_1, T)\times \mathbb {R}^d$ . Let

$$ \begin{align*}u(t,x):=(1+h(t,x))w_*(t,x)\xi\left(\frac{3|x|}{5(T-t)}\right)\exp\left(\text{i}\Phi_*(t,x)\xi\left(\frac{3|x|}{5(T-t)}\right)\right)\end{align*} $$

for $(t,x)\in [T-c_1, T)\times \mathbb {R}^d$ . Then $u\in C^2([T-c_1, T)\times \mathbb {R}^d;\mathbb {C})$ with $u(t,\cdot ), \partial _tu(t,\cdot )\in C_c^\infty (\mathbb {R}^d)$ and $\operatorname {supp}_xu(t,\cdot ), \operatorname {supp}_x\partial _tu(t,\cdot )\subset \{x\in \mathbb {R}^d: |x|\leq 5(T-t)/3\}$ for $t\in [T-c_1, T)$ . Moreover, on $\mathcal C_0:=\{(t,x)\in (T-c_1, T)\times \mathbb {R}^d: |x|\leq T-t\}$ we have $u=(1+h)w_*\text {e}^{\text {i}\Phi _*}$ , and by (4.2), (4.3) and (4.52), we know that h satisfies (4.1) on $\mathcal C_0$ , hence by Lemma 4.1 we know that $\Box u=|u|^{p-1}u$ on $\mathcal C_0$ . Finally, by $|h|<1/2$ , (2.28) (for $j=0$ ) and (2.29) we have (2.31) on $\mathcal C_0$ .

This completes the proof of Proposition 2.8.

Proof. As $f\in C_{\text {e}}^\infty ([0,+\infty ))$ , there exists a function $\widetilde f\in C^\infty ([0,+\infty ))$ such that $f(Z)=\widetilde f(Z^2)$ for all $Z\in [0,+\infty )$ , hence $F(x)=\widetilde f(|x|^2)$ for all $x\in \mathbb {R}^d$ . The smoothness of F follows from the smoothness of $\widetilde f$ and $x\mapsto |x|^2$ .

Proof. By the definition of ${\mathscr X}_0$ in (2.20), it suffices to prove the result for $f=\widehat f(Z)\tau ^n$ for some $\widehat f\in C^\infty _{\text {e}}([0,+\infty ))$ and some $n\in \mathbb {Z}_{\geq 0}$ .

As $\partial _\tau f=n\widehat f(Z)\tau ^{n-1}$ , we have $\partial _\tau f\in {\mathscr X}_0$ ( $\partial _\tau f=0$ for $n=0$ ).

Note that $Z\partial _Z f=Z\widehat f'(Z)\tau ^n, \partial _Z^2 f=\widehat f"(Z)\tau ^n, \partial _Z f/Z=\frac {1}{Z}\widehat f'(Z)\tau ^n$ . As $\widehat f\in C^\infty _{\text {e}}([0,+\infty ))$ , by (5.3) we have $\widehat f'\in C^\infty _{\text {o}}([0,+\infty ))$ , then by the definitions of $C^\infty _{\text {e}}([0,+\infty ))$ and $C^\infty _{\text {o}}([0,+\infty ))$ we have $Z\widehat f'(Z),\frac {1}{Z}\widehat f'(Z)\in C^\infty _{\text {e}}([0,+\infty ))$ , by (5.4) we have $\widehat f"\in C^\infty _{\text {e}}([0,+\infty ))$ . Thus, $Z\partial _Z f$ , $\partial _Z^2 f$ , $\partial _Z f/Z\in {\mathscr X}_0$ .

Proof.

1. (i) Let $f(t,x)=(T-t)^\lambda \widehat f(\tau , Z)$ for some $\widehat f\in {\mathscr X}_0$ and $g(t,x)=(T-t)^\mu \widehat g(\tau , Z)$ for some $\widehat g\in {\mathscr X}_0$ . Then $(fg)(t,x)=(T-t)^{\lambda +\mu }(\widehat f\widehat g)(\tau , Z)$ . As ${\mathscr X}_0 $ is a ring, we have $\widehat f\widehat g\in {\mathscr X}_0$ and then $fg\in {\mathscr X}_{\lambda +\mu }$ . It is direct to check that

   $$ \begin{align*} \partial_t f=(T-t)^{\lambda-1}\left(\partial_\tau+Z\partial_Z-\lambda\right)\widehat f,\quad \Delta f=(T-t)^{\lambda-2}\left(\partial_Z^2+(k/Z)\partial_Z\right)\widehat f. \end{align*} $$
   Since $\widehat f\in {\mathscr X}_0$ , by Lemma 5.2 we have $\partial _\tau \widehat f, Z\partial _Z\widehat f, \partial _Z^2\widehat f, \partial _Z\widehat f/Z\in {\mathscr X}_0$ , which gives
   $$\begin{align*}\left(\partial_\tau+Z\partial_Z-\lambda\right)\widehat f\in{\mathscr X}_0,\quad \left(\partial_Z^2+(k/Z)\partial_Z\right)\widehat f\in{\mathscr X}_0.\end{align*}$$
   Thus, $\partial _t f\in {\mathscr X}_{\lambda -1}$ , $\Delta f\in {\mathscr X}_{\lambda -2}$ . As a consequence, we have $\partial _t^2 f\in {\mathscr X}_{\lambda -2}$ , $\Box f=-\partial _t^2 f+\Delta f\in {\mathscr X}_{\lambda -2}$ and (also using $fg\in {\mathscr X}_{\lambda +\mu } $ for $ f\in {\mathscr X}_{\lambda }$ , $ g\in {\mathscr X}_{\mu }$ )
   $$\begin{align*}\Box(fg)\in{\mathscr X}_{(\lambda+\mu)-2}, \quad (\Box f)g\in{\mathscr X}_{(\lambda-2)+\mu}, \quad f\Box g\in{\mathscr X}_{\lambda+(\mu-2)},\end{align*}$$
   hence,
   $$ \begin{align*} &\partial^\alpha f\partial_\alpha g=[\Box(fg)-(\Box f)g-f\Box g]/2\in{\mathscr X}_{\lambda+\mu-2},\\ & \partial^\alpha(f\partial_\alpha g)=\partial^\alpha f\partial_\alpha g+f\Box g\in{\mathscr X}_{\lambda+\mu-2}. \end{align*} $$

2. (iii) Let $\lambda ,\mu \in \mathbb {R}$ and $j\in \mathbb {Z}_{\geq 0}$ be such that $\lambda \geq j+\mu $ . Let $f(t,x)=(T-t)^\lambda \widehat f(Z)$ for some $\widehat f\in C^\infty _{\text {e}}([0,+\infty ))$ . Assume that $\alpha _0,\alpha _1,\cdots ,\alpha _d\in \mathbb {Z}_{\geq 0}$ are such that $\alpha _0+\cdots +\alpha _d=j$ . We only need to prove that

   (5.11) $$ \begin{align} (T-t)^{-\mu}\partial_t^{\alpha_0}\partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d}f\in L^\infty(\mathcal C). \end{align} $$
   Let $\widetilde f(x):=\widehat f(|x|)$ for $x\in \mathbb {R}^d$ , then by Lemma 5.1 we have $\widetilde f\in C^\infty (\mathbb {R}^d)$ . Let $j'=\alpha _1+\cdots +\alpha _d\in \mathbb {Z}\cap [0,j]$ , and we let
   $$\begin{align*}f_\alpha:=(-(\lambda-j')+x\cdot\nabla_x)(-(\lambda-j'-1)+x\cdot\nabla_x)\cdots(-(\lambda-j+1)+x\cdot\nabla_x)\partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d}\widetilde f.\end{align*}$$
   Then $f_\alpha \in C^\infty (\mathbb {R}^d)$ and one can check by direct computation that
   $$\begin{align*}\partial_t^{\alpha_0}\partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d}f(t,x)=(T-t)^{\lambda-j}f_\alpha(x/(T-t)),\quad \forall\ (t,x)\in[0, T)\times\mathbb{R}^d.\end{align*}$$
   As $|x/(T-t)|<2$ for $(t,x)\in \mathcal C$ and $\lambda -j-\mu \geq 0$ , we have (5.11).

3. (ii) Let $\lambda ,\mu \in \mathbb {R}$ and $j\in \mathbb {Z}_{\geq 0}$ be such that $\lambda> j+\mu $ . By the definitions of ${\mathscr X}_0$ and ${\mathscr X}_\lambda $ , it suffices to prove $(T-t)^{-\mu }D^jf\in L^\infty (\mathcal C)$ for $f(t,x)=(T-t)^\lambda \widehat f(Z)\tau ^n$ for some $\widehat f\in C^\infty _{\text {e}}([0,+\infty ))$ and some $n\in \mathbb {Z}_{\geq 0}$ . Let $P(\tau ):=\tau ^n$ and $\widetilde P(t):=P(\tau )=P(-\ln (T-t))$ . Then by the induction, for any $i\in \mathbb {Z}_{\geq 0}$ , there is a polynomial $P_i(\tau )$ such that $\widetilde P^{(i)}(t)=(T-t)^{-i}P_i(\tau )$ . Hence,

   (5.12) $$ \begin{align} (T-t)^{i+\varepsilon}\widetilde P^{(i)}(t)\in L^\infty([0,T)),\quad\forall\ i\in\mathbb{Z}_{\geq 0},\ \forall\ \varepsilon>0. \end{align} $$
   Let $\widetilde f(t,x):=(T-t)^\lambda \widehat f(Z)$ for $(t,x)\in [0,T)\times \mathbb {R}^d$ , then $f(t,x)=\widetilde f(t,x)\widetilde P(t)$ for $(t,x)\in [0,T)\times \mathbb {R}^d$ , and by (iii) we have
   (5.13) $$ \begin{align} (T-t)^{i-\lambda}D^i\widetilde f\in L^\infty(\mathcal C),\quad \forall\ i\in\mathbb{Z}_{\geq 0}. \end{align} $$
   Assume that $\alpha _0,\alpha _1,\cdots ,\alpha _d\in \mathbb {Z}_{\geq 0}$ are such that $\alpha _0+\cdots +\alpha _d=j$ . It suffices to prove
   (5.14) $$ \begin{align} (T-t)^{-\mu}\partial_t^{\alpha_0}\partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d}(\widetilde f(t,x)\widetilde P(t))\in L^\infty(\mathcal C). \end{align} $$
   By Leibnitz’s product rule, we have
   $$ \begin{align*} &(T-t)^{-\mu}\partial_t^{\alpha_0}\partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d}(\widetilde f(t,x)\widetilde P(t))\\&\quad =(T-t)^{-\mu}\sum_{i=0}^{\alpha_0}\binom{\alpha_0}{i}\widetilde P^{(i)}(t)\partial_t^{\alpha_0-i}\partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d}\widetilde f(t,x)\\&\quad =\sum_{i=0}^{\alpha_0}\binom{\alpha_0}{i}(T-t)^{\lambda-j-\mu+i}\widetilde P^{(i)}(t)\cdot(T-t)^{j-i-\lambda}\partial_t^{\alpha_0-i}\partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d}\widetilde f(t,x). \end{align*} $$
   Then (5.14) follows from $\lambda -j-\mu>0$ , (5.12) and (5.13).