Proof. Without loss of generality, we can assume $f \not \equiv 0$ . Clearly,

$$ \begin{align*} f(x)-f(x/2)=\int^x_{x/2}f^{\prime}(s) {\: \rm d} s \leq \tfrac{1}{2} x f^{\prime}(x). \end{align*} $$

Since f is completely monotone, it is a positive function and

$$ \begin{align*} f(x/2) \geq \tfrac{1}{2} x|f^{\prime}(x)|, \end{align*} $$

which, together with the doubling property, gives

$$ \begin{align*} f(x)\geq C x|f^{\prime}(x)| \end{align*} $$

for $x> 2 x_0$ . Hence, we obtain our assertion in the case $x_0 = 0$ . If $x_0> 0$ , we observe that the function

$$ \begin{align*} [x_0, 2 x_0] \ni x \mapsto \frac{x |f^{\prime}(x)|}{f(x)} \end{align*} $$

is continuous and positive and thus bounded. This completes the proof.

Proof. Let $\lambda>1$ . For $y> x > x_0$ , we have

$$ \begin{align*} f(\lambda x)-f(\lambda y) = -\int^{\lambda y}_{\lambda x} f'(s) {\: \rm d} s &= -\lambda \int^y_x f'(\lambda s) {\: \rm d} s \\ &\geq - c \lambda^{1+\tau} \int^y_x f'(s) {\: \rm d} s = c \lambda^{1+\tau} (f(x)-f(y)); \end{align*} $$

thus,

$$ \begin{align*} f(\lambda x) \geq c \lambda^{1+\tau} f(x) + f(\lambda y) - c \lambda^{1+\tau} f(y). \end{align*} $$

Since f is nonnegative and nonincreasing, we can take y approaching infinity to get

$$ \begin{align*} f(\lambda x) &\geq c \lambda^{1+\tau} f(x) + \big(1 - c \lambda^{1+\tau}\big) \lim_{y \to \infty} f(y) \\ &\geq c \lambda^{1+\tau} f(x), \end{align*} $$

where in the last inequality we have also used that $1 \geq c \lambda ^{1+\tau }$ . The second part of the proposition can be proved in much the same way.

Proof. Assume first that $\phi ^{\prime }\in \textrm {WLSC}(\alpha -1,c,x_0)$ . Without loss of generality, we can assume $\phi ^{\prime } \not \equiv 0$ . We claim that (2.12) holds true. In view of (2.9), it is enough to show that there is $C \geq 1$ such that for all $x> x_0$ ,

$$ \begin{align*} \phi(x) \leq C x \phi^{\prime}(x). \end{align*} $$

First, let us observe that, by the weak lower scaling property of $\phi ^{\prime }$ ,

(2.13) $$ \begin{align} \nonumber \phi(x) -\phi(x_0) &= \int_{x_0}^{x} \phi^{\prime}(s) {\: \rm d} s \\ \nonumber &\leq c^{-1} \phi^{\prime}(x) \int_{x_0}^x \big(s/x\big)^{-1+\alpha} {\: \rm d} s \\ &\leq \frac{1}{c \alpha} x \phi^{\prime}(x). \end{align} $$

Thus, we get the assertion in the case $x_0 = 0$ . If $x_0> 0$ , it is enough to show that there is $C> 0$ such that for all $x> x_0$ ,

(2.14) $$ \begin{align} x \phi^{\prime}(x) \geq C. \end{align} $$

Since $\phi ' \in \textrm {WLSC}(\alpha -1,c,x_0)$ , the function

$$ \begin{align*} (x_0, \infty) \ni x \mapsto x \phi'(x) \end{align*} $$

is almost increasing. Hence, for $x \geq 2 x_0$ we have

$$ \begin{align*} x \phi'(x) \geq c 2 x_0 \phi'(2 x_0). \end{align*} $$

To conclude (2.14), we notice that $\phi '(x)$ is positive and continuous in $[x_0, 2x_0]$ . Now, by (2.14) we get

$$ \begin{align*} x \phi'(x) \geq C \phi(x_0) \end{align*} $$

for all $x> x_0$ , which, together with (2.13), implies (2.12) and the scaling property of $\phi $ follows.

Now assume that $\phi \in \textrm {WLSC}(\alpha ,c,x_0)$ . By monotonicity of $\phi '$ , for $0 < s < t$ ,

$$ \begin{align*} \frac{\phi(tx)-\phi(sx)}{\phi(x)} \leq \frac{x(t-s)\phi'(sx)}{\phi(x)}. \end{align*} $$

For $s=1$ , by the lower scaling we get

$$ \begin{align*} \frac{x(t-1)\phi'(x)}{\phi(x)} \geq \frac{\phi(tx)}{\phi(x)}-1 \geq ct^{\alpha}-1, \end{align*} $$

for all $x>x_0$ . Thus, for $t = 2^{1/\alpha } c^{-1/\alpha }$ , we obtain that $x\phi '(x) \gtrsim \phi (x)$ for all $x>x_0$ . Invoking (2.9), we conclude (2.12). In particular, $\phi '$ has the weak lower scaling property. This completes the proof.

Proof. Without loss of generality, we can assume $\phi ^{\prime \prime } \not \equiv 0$ . By the scaling property, for $x> x_0$ we have

$$ \begin{align*} \frac{\phi'(x)-b}{x(-\phi^{\prime\prime}(x))} = \int_x^{\infty} \frac{t}{x} \frac{(-\phi^{\prime\prime}(t))}{(-\phi^{\prime\prime}(x))} \frac{{\textrm{d}} t}{t} \leq C \int_x^{\infty} \bigg( \frac{t}{x} \bigg)^{-1+\beta} \frac{{\textrm{d}} t}{t} = C \frac{1}{1-\beta}, \end{align*} $$

which concludes the proof.

Proof. By monotonicity of $-f'$ , for $0 < s < t$ ,

(2.15) $$ \begin{align} \frac{-x(t-s)f'(tx)}{f(x)} \leq \frac{f(sx)-f(tx)}{f(x)} \leq \frac{-x(t-s)f'(sx)}{f(x)}. \end{align} $$

Taking $s=1$ in the second inequality, the weak upper scaling property yields

$$ \begin{align*} \frac{-x(t-1)f'(x)}{f(x)} \geq 1- \frac{f(tx)}{f(x)} \geq 1-ct^{\beta-1}, \end{align*} $$

for all $x>x_0$ . By selecting $t> 1$ such that $ct^{\beta -1} \leq \frac 12$ , we obtain $x\big (-f'(x)\big ) \gtrsim f(x)$ for $x>x_0$ . Similarly, taking $t=1$ in the first inequality in (2.15), by the weak lower scaling property we get

$$ \begin{align*} \frac{-x(1-s)f'(x)}{f(x)} \leq \frac{f(sx)}{f(x)}-1 \leq c^{-1}s^{\alpha-1}-1, \end{align*} $$

for all $x>x_0/s$ . By selecting $0 < s < 1$ such that $s^{\alpha -1}\geq 2c$ , we obtain $x\big ( -f'(x)\big ) \lesssim f(x)$ for $x>x_0/s$ . Hence,

(2.16) $$ \begin{align} f(x) \approx x\big( -f'(x) \big), \end{align} $$

for all $x>x_0/s$ . Therefore, lower and upper scaling properties follow from (2.16) and the scaling properties of f. This finishes the proof for the case $x_0 = 0$ . If $x_0> 0$ , we notice that both f ad $-f'$ are positive and continuous; thus, at the possible expense of worsening the constants, we get (2.16) for all $x>x_0$ .

Proof. Let $f\colon (0, \infty ) \rightarrow \mathbb {R}$ be a function defined as

$$ \begin{align*} f(t)=\int_{(0, t)} s^2 \: \mu ({\textrm{d}} s). \end{align*} $$

We observe that, by the Fubini–Tonelli theorem, for $x> 0$ we have

$$ \begin{align*} \mathcal{L} f(x) &= \int_0^\infty e^{-x t} \int_{(0, t)} s^2 \: \mu({\textrm{d}} s) {\: \rm d} t \\ &= \int_{(0, \infty)} s^2 \int_s^\infty e^{-x t} {\: \rm d} t \: \mu({\textrm{d}} s) = x^{-1} (-\phi^{\prime\prime}(x)). \end{align*} $$

Since f is nondecreasing, for any $s> 0$ ,

$$ \begin{align*} -\phi^{\prime\prime}(x)=x \mathcal{L} f(x) &\geq \int_s^{\infty} e^{-t} f\big( t/x \big) {\: \rm d} t\\ &\geq e^{-s} f\big(s /x \big). \end{align*} $$

Hence, for any $u> 2$ ,

$$ \begin{align*} -\phi^{\prime\prime}(x) &=\int_0^u e^{-s} f\big( s/x \big) {\: \rm d} s + \int_u^{\infty} e^{-s} f\big(s/x \big) {\: \rm d} s \\ &\leq f\big(u / x \big) + \int_u^{\infty} e^{-s/2} (-\phi^{\prime\prime}(x/2) ) {\: \rm d} s. \end{align*} $$

Therefore, setting $x = \lambda u> 2 x_0$ , by the weak lower scaling property of $-\phi ^{\prime \prime }$ ,

$$ \begin{align*} f(1/\lambda) &\geq -\phi^{\prime\prime}(u \lambda) - 2 e^{-u/2} (-\phi^{\prime\prime}(u \lambda/2)) \\ &\geq \big(2^{\alpha-2}c - 2 e^{-u/2}\big) (-\phi^{\prime\prime}(u \lambda/2)). \end{align*} $$

At this stage, we select $u> 2$ such that

$$ \begin{align*} 2^{\alpha-2}c - 2 e^{-u/2} \geq 2^{-2} c. \end{align*} $$

Then again, by the weak lower scaling property of $-\phi ^{\prime \prime }$ , for $\lambda>x_0$ ,

$$ \begin{align*} f(1/\lambda) &\geq c 2^{-2} (-\phi^{\prime\prime}(u \lambda/2))\geq c^2 2^{-\alpha} u^{\alpha-2} (-\phi^{\prime\prime}(\lambda)), \end{align*} $$

which ends the proof.

Proof. Let us define

$$ \begin{align*} f(\lambda)=a+b\lambda+\int^\infty_0\frac{\lambda u}{\lambda u +1} \mu({\textrm{d}} u). \end{align*} $$

By [Reference Schilling, Song and Vondraček48, Theorem 6.2 (vi)] the function f is a complete Bernstein function. Since for $y> 0$ ,

$$ \begin{align*} \frac{y}{y +1} \approx \big(1-e^{-y}\big), \end{align*} $$

we get $f(\lambda ) \approx \phi (\lambda )$ . Moreover,

$$ \begin{align*} f^{\prime\prime}(\lambda)=-2\int^\infty_0\frac{u^2}{{(\lambda u+1)^3}}\mu({\textrm{d}}u). \end{align*} $$

Hence, by Lemma 2.8 we obtain $-f^{\prime \prime }(\lambda )\approx -\phi ^{\prime \prime }(\lambda )$ for $\lambda>x_0$ .

Proof. Without loss of generality, we can assume $b = 0$ . Indeed, otherwise it is enough to consider a shifted process $\widetilde {T_t} = T_t - tb$ . Next, let us observe that for any Borel set $B \subset \mathbb {R}$ , we have

(4.19) $$ \begin{align} \nonumber \nu(B) & \lesssim \int_{(\delta(B), \infty)} \big(1 - e^{-s/\delta(B)}\big) \: \nu({\textrm{d}} s) \\ &\leq \phi\big(1/\delta(B)\big). \end{align} $$

Furthermore, for $\delta (B)<1/x_0$ , by Proposition 4.1 and Remark 3.2,

$$ \begin{align*} \nu(B) &\leq h(\delta(B)) \\ &\lesssim \varphi^*(1/\delta(B)). \end{align*} $$

Thus, $\nu (B) \lesssim \zeta (\delta (B))$ . We claim that $\zeta $ has doubling property on $(0, \infty )$ . Indeed, since $-\phi ^{\prime \prime }$ is nonincreasing function with the weak lower scaling property, it has doubling property on $(x_0, \infty )$ ; thus, for $0 < s < x_0^{-1}$ ,

$$ \begin{align*} \zeta\big(\tfrac{1}{2} s\big) \approx 4 s^{-2} (-\phi^{\prime\prime}(2/s)) \lesssim s^{-2} (-\phi^{\prime\prime}(1/s)) \lesssim \zeta(s). \end{align*} $$

This completes the argument in the case $x_0 = 0$ . If $x_0> 0$ , then by (2.10), for $s> 2 x_0^{-1}$ we have

$$ \begin{align*} \zeta\big(\tfrac{1}{2} s\big) = A \phi(2/s) \leq 2 A \phi(1/s) \leq 2 \zeta(s). \end{align*} $$

Lastly, the function

$$ \begin{align*} \big[\tfrac{1}{2} x_0, x_0] \ni x \mapsto \frac{\varphi^*(2 x)}{\phi(x)} \end{align*} $$

is continuous and thus it is bounded.

Next, for $s> 0$ and $x \in \mathbb {R}$ ,

$$ \begin{align*} s \vee {\lvert {x} \rvert} - \tfrac{1}{2} {\lvert {x} \rvert} \geq \tfrac{1}{2} s; \end{align*} $$

thus, by motonicity and the doubling property of $\zeta $ , we get

$$ \begin{align*} \zeta\big(s \vee {\lvert {x} \rvert} - \tfrac{1}{2} {\lvert {x} \rvert} \big) \lesssim \zeta(s). \end{align*} $$

Hence, by (2.4) and (4.2), for $r> 0$ ,

(4.20) $$ \begin{align} \nonumber \int_{(r, \infty)} \zeta\big(s \vee x - \tfrac{1}{2} x \big) \: \nu({\textrm{d}} x) &\lesssim \zeta(s) h(r) \\ &\lesssim \zeta(s) \psi^*(1/r). \end{align} $$

Since $\psi ^*$ has the weak lower scaling property and satisfies (4.3), by [Reference Grzywny and Szczypkowski23, Proposition 3.4] together with Proposition 4.2, there are $C> 0$ and $t_1 \in (0, \infty ]$ such that for all $t \in (0, t_1)$ ,

(4.21) $$ \begin{align} \int_{\mathbb{R}} e^{-t \Re \psi(\xi)} {\lvert {\xi} \rvert} {\: \rm d} \xi \leq C \big(\psi^{-1}(1/t)\big)^2. \end{align} $$

If $x_0 = 0$ , then $t_1 = \infty $ . If $t_1 < 1/\varphi (x_0)$ , we can expand the above estimate for $t_1\leq t< 1/\varphi (x_0)$ using positivity of the right-hand side and monotonicity of the left-hand side.

In view of (4.19), (4.20) and (4.21), by [Reference Kaleta and Sztonyk32, Theorem 1] with $\gamma =\,0$ , there are $C_1,C_2,C_3> 0$ such that for all $t \in (0 , 1/\varphi (x_0))$ and $x \in \mathbb {R}$ ,

$$ \begin{align*} & p\Big(t, x+t b_{1/\psi^{-1}(1/t)} \Big) \\ &\quad\leq C_1 \psi^{-1}(1/t) \cdot \min{ \bigg\{ 1, t \zeta\big(\tfrac{1}{4} {\lvert {x} \rvert} \big) + \exp\Big\{-C_2 {\lvert {x} \rvert} \psi^{-1}(1/t) \log \big(1 + C_3 {\lvert {x} \rvert} \psi^{-1}(1/t)\big) \Big\} \bigg\}}. \end{align*} $$

Let us consider $x> 0$ and $t \in (0, 1/\varphi (x_0))$ such that $t \zeta (x) \leq 1$ . We claim that

(4.22) $$ \begin{align} \exp\Big\{-C_2 x \psi^{-1}(1/t) \log\big(1 + C_3 x \psi^{-1}(1/t)\big)\Big\} \lesssim t \zeta(x). \end{align} $$

First suppose that $x>x_0^{-1}$ . Let us observe that the function

$$ \begin{align*} [0, \infty) \ni u \mapsto u \exp\Big\{-C_2 u \log \big(1+C_3 u\big) \Big\} \end{align*} $$

is bounded. Therefore,

(4.23) $$ \begin{align} \exp\Big\{-C_2 x \psi^{-1}(1/t) \log\big(1 + C_3 x \psi^{-1}(1/t)\big)\Big\} \lesssim \frac{1}{x \psi^{-1}(1/t)}. \end{align} $$

Since $x \phi ^{-1}(1/t) \geq 1$ , by (2.10), we have

(4.24) $$ \begin{align} t \phi(1/x) = \frac{\phi(1/x)}{\phi\big(x \phi^{-1}(1/t) \cdot 1/x\big)} \geq \frac{1}{x \phi^{-1}(1/t)}. \end{align} $$

Next, in light of (2.9), for all $y> 0$ ,

$$ \begin{align*} \tfrac{1}{2} \varphi^*(y) \leq \phi(y); \end{align*} $$

hence, by the monotonicity of $\phi ^{-1}$ ,

(4.25) $$ \begin{align} \nonumber \phi^{-1}(1/t) &= \phi^{-1}\big(\tfrac{1}{2} \varphi^*(\varphi^{-1}(2/t))\big) \\ \nonumber &\leq \phi^{-1}\big(\phi(\varphi^{-1}(2/t))\big) \\ \nonumber &= \varphi^{-1}(2/t) \\ &\leq C \psi^{-1}(1/t), \end{align} $$

where in the last step we have used Proposition 4.3. Putting (4.23), (4.24) and (4.25) together, we obtain (4.22) as claimed.

Now let $0<x \leq x_0^{-1}$ . Observe that the function

$$ \begin{align*} [0, \infty) \ni u \mapsto u^2 \exp\Big\{-C_2 u \log \big(1+C_3 u\big) \Big\} \end{align*} $$

is also bounded. Hence,

(4.26) $$ \begin{align} \exp\Big\{-C_2 x \psi^{-1}(1/t) \log\big(1 + C_3 x \psi^{-1}(1/t)\big)\Big\} \lesssim \frac{1}{\big(x \psi^{-1}(1/t)\big)^2}. \end{align} $$

Since $x \varphi ^{-1}(1/t) \geq 1$ , using (4.5) we get

(4.27) $$ \begin{align} t\varphi^*(1/x) = \frac{\varphi^*(1/x)}{\varphi^*\big( x\varphi^{-1}(1/t) \cdot 1/x \big)} \geq \frac{1}{\big( x\varphi^{-1}(1/t) \big)^2}. \end{align} $$

Hence, putting together (4.26) and (4.27) and invoking Proposition 4.2, we again obtain (4.22).

Finally, using doubling property of $\zeta $ we get

$$ \begin{align*} \zeta\big(\tfrac{1}{4} x\big) \lesssim \zeta(x); \end{align*} $$

thus, another application of Proposition 4.3 leads to (4.17).

For the proof of (4.18), we observe that

$$ \begin{align*} \phi'(\lambda) = \int_{(0, \infty)} x e^{-\lambda x} \: \nu({\textrm{d}} x) \geq e^{-1} \int_{(0, 1/\lambda)} x \: \nu({\textrm{d}} x). \end{align*} $$

Thus,

$$ \begin{align*} b_{1/\psi^{-1}(1/t)} = \int_{(0, 1/\psi^{-1}(1/t))} x \: \nu({\textrm{d}}x) \leq e \phi'(\psi^{-1}(1/t)). \end{align*} $$

Hence, by monotonicity and the doubling property of $\zeta $ , for $x> 2e t \phi '(\psi ^{-1}(1/t))$ , we obtain

$$ \begin{align*} \zeta\Big(x - t b_{1/\psi^{-1}(1/t)}\Big) \leq \zeta\bigg(\frac{x}{2}\bigg) \lesssim \zeta (x), \end{align*} $$

and the theorem follows.

Proof. Without loss of generality, we can assume $b = 0$ . Let us observe that for any $\lambda> 0$ ,

$$ \begin{align*} \phi(\lambda) \geq \int_0^{1/\lambda} \big(1 - e^{-\lambda s} \big) \nu(s) {\: \rm d} s \gtrsim \nu(1/\lambda) \lambda^{-1} \end{align*} $$

and

$$ \begin{align*} -\phi^{\prime\prime}(\lambda) \geq \int_0^{1/\lambda} s^2 e^{-\lambda s} \nu(s) {\: \rm d} s \gtrsim \nu(1/\lambda) \lambda^{-3}. \end{align*} $$

Hence,

(4.30) $$ \begin{align} \nu(x) \lesssim \eta(x) \qquad \text{for all } x> 0. \end{align} $$

Since $\eta $ is nonincreasing, for any Borel subset $B \subset \mathbb {R}$ ,

(4.31) $$ \begin{align} \nu(B) &\lesssim \int_{B \cap (0,\infty)} \eta(x) {\: \rm d} x \lesssim \eta\big(\delta(B)\big) \operatorname{\mathrm{diam}}(B). \end{align} $$

Arguing as in the proof of Theorem 4.7, we conclude that $\eta $ has a doubling property on $(0,\infty )$ . Using that and monotonicity of $\eta $ , for $s> 0$ and $x \in \mathbb {R}$ ,

$$ \begin{align*} \eta\big(s \vee x - \tfrac{1}{2} x\big) \leq \eta\big(\tfrac{1}{2} s \big) \lesssim \eta(s). \end{align*} $$

Therefore, by (4.2), for $r> 0$ ,

(4.32) $$ \begin{align} \int_r^\infty \eta\big(s \vee x - \tfrac{1}{2} x\big) \nu(x) {\: \rm d} x \lesssim \eta(s) \psi^*(1/r). \end{align} $$

Since $\psi ^*$ has the weak lower scaling property and satisfies (4.3), by [Reference Grzywny and Szczypkowski23, Theorem 3.1] and Proposition 4.2, there are $C> 0$ and $t_1 \in (0, \infty ]$ such that for all $t \in (0, t_1)$ ,

(4.33) $$ \begin{align} \int_{\mathbb{R}} e^{-t\Re \psi(\xi)} {\: \rm d} \xi \leq C \psi^{-1}(1/t). \end{align} $$

If $x_0 = 0$ , then $t_1 = \infty $ . If $t_1 < 48/\varphi (x_0)$ , we can expand the above estimate for $t_1\leq t< 48/\varphi (x_0)$ using positivity of the right-hand side and monotonicity of the left-hand side.

In view of (4.31), (4.32) and (4.33), by [Reference Grzywny and Szczypkowski22, Theorem 2.1], there is $C> 0$ such that for all $t \in (0, 1/\varphi (x_0))$ and $x \in \mathbb {R}$ ,

$$ \begin{align*} p\Big(t,x+tb_{1/\psi^{-1}(1/t)}\Big) \leq C \psi^{-1}(1/t) \cdot \min{ \Big\{ 1, t \big(\psi^{-1}(1/t)\big)^{-1} \eta({\lvert {x} \rvert})+\big(1+{\lvert {x} \rvert}\psi^{-1}(1/t)\big)^{-3} \Big\}}. \end{align*} $$

We claim that

(4.34) $$ \begin{align} \frac{\psi^{-1}(1/t)}{\big(1+{\lvert {x} \rvert}\psi^{-1}(1/t)\big)^3} \lesssim t \eta({\lvert {x} \rvert}) \end{align} $$

whenever $t \eta ({\lvert {x} \rvert }) \leq \tfrac {A}{2} \varphi ^{-1}(1/t)$ .

First, let us show that for any $\epsilon \in (0, 1]$ , the condition $t \eta ({\lvert {x} \rvert }) \leq \tfrac {A \epsilon }{2} \varphi ^{-1}(1/t)$ implies that

(4.35) $$ \begin{align} t \varphi^*\bigg(\frac{1}{{\lvert {x} \rvert}}\bigg) \leq \epsilon {\lvert {x} \rvert} \varphi^{-1}\bigg(\frac{1}{t}\bigg). \end{align} $$

Indeed, by (2.9), we have ${\lvert {x} \rvert } \eta ({\lvert {x} \rvert }) \geq \tfrac {A}{2} \varphi ^*(1/{\lvert {x} \rvert })$ ; thus,

$$ \begin{align*} \epsilon {\lvert {x} \rvert} \varphi^{-1}\left(\frac{1}{t}\right) \geq \frac{2}{A} t {\lvert {x} \rvert} \eta({\lvert {x} \rvert}) \geq t \varphi^*\left(\frac{1}{{\lvert {x} \rvert}}\right). \end{align*} $$

Notice also that $\epsilon ^{1/3} {\lvert {x} \rvert } \varphi ^{-1}(1/t) \geq 1$ since otherwise, by (4.5),

$$ \begin{align*} 1 < t \varphi^*\bigg(\frac{1}{\epsilon^{1/3} {\lvert {x} \rvert}}\bigg) < \frac{1}{\epsilon^{2/3}} t \varphi^*\bigg(\frac{1}{{\lvert {x} \rvert}}\bigg), \end{align*} $$

which entails that $\epsilon ^{2/3} < t \varphi ^*(1/{\lvert {x} \rvert })$ ; that is, $\epsilon ^{1/3} {\lvert {x} \rvert } \varphi ^{-1}(1/t) < \epsilon ^{-2/3} t \varphi ^*(1/{\lvert {x} \rvert })$ , contrary to (4.35).

To show (4.34), let us suppose that $t \eta ({\lvert {x} \rvert }) \leq \tfrac {A}{2} \varphi ^{-1}(1/t)$ ; thus, ${\lvert {x} \rvert } \varphi ^{-1}(1/t) \geq 1$ . By (4.5), we have

$$ \begin{align*} t \varphi^*(1/{\lvert {x} \rvert}) = \frac{\varphi^*(1/{\lvert {x} \rvert})}{\varphi^*\big({\lvert {x} \rvert} \varphi^{-1}(1/t) \cdot 1/{\lvert {x} \rvert}\big)} \geq \frac{1}{({\lvert {x} \rvert} \varphi^{-1}(1/t))^2}, \end{align*} $$

which, by Proposition 4.3, gives

$$ \begin{align*} t {\lvert {x} \rvert} \eta({\lvert {x} \rvert}) \geq \tfrac{A}{2} t \varphi^*(1/{\lvert {x} \rvert}) \gtrsim \frac{|x|\psi^{-1}(1/t)}{(1 + {\lvert {x} \rvert} \psi^{-1}(1/t))^3}, \end{align*} $$

proving (4.34), and (4.28) follows. The inequality (4.29) holds by the same argument as in the proof of Theorem 4.7.

Proof. We are going to apply the estimates [Reference Pruitt45, (3.2)]. To do so, we need to express the Lévy–Khintchine exponent of $T_s - s b_\lambda $ in the form used in [Reference Pruitt45, Section 3], namely,

$$ \begin{align*} \tilde{\psi}(\xi) &= \psi(\xi) + i \xi b_\lambda \\ &= -i\xi \bigg(b - b_\lambda + \int_{(0, \infty)} \frac{y}{1 + |y|^2} \: \nu({\textrm{d}} y) \bigg) - \int_{(0, \infty)} \bigg(e^{i\xi y} - 1 - \frac{i y \xi}{1 + |y|^2} \bigg) \: \nu({\textrm{d}} y). \end{align*} $$

Since

$$ \begin{align*} \int_{(0, \lambda]} \frac{y |y|^2}{1+|y|^2} \: \nu({\textrm{d}} y) - \int_{(\lambda, \infty)} \frac{y}{1+|y|^2} \: \nu({\textrm{d}} y) = \int_{(0, \lambda]} y \: \nu({\textrm{d}} y) - \int_{(0, \infty)} \frac{y}{1+|y|^2} \: \nu({\textrm{d}} y), \end{align*} $$

we have

$$ \begin{align*} M(\lambda) &= \frac{1}{\lambda} \bigg| b - b_\lambda + \int_{(0, \infty)} \frac{y}{1 + |y|^2} \: \nu({\textrm{d}} y) + \int_{(0, \lambda]} \frac{y |y|^2}{1+|y|^2} \: \nu({\textrm{d}} y) - \int_{(\lambda, \infty)} \frac{y}{1+|y|^2} \: \nu({\textrm{d}} y) \bigg| \\ &= \frac{1}{\lambda} \bigg| b - b_\lambda + \int_{(0, \lambda]} y \: \nu({\textrm{d}} y) \bigg| = 0. \end{align*} $$

Hence, by [Reference Pruitt45, (3.2)],

$$ \begin{align*} \mathbb{P} \Big( \mathop{\mathrm{sup}}\limits_{0 \leq s \leq t} \big|T_s- s b_\lambda \big|\geq \lambda \Big) \leq c s h(\lambda), \end{align*} $$

as desired.

Proof. First let us observe that it is enough to prove that (4.40) holds true for all $t \in (0,1/\varphi (x_0))$ and all $x> 0$ satisfying

$$ \begin{align*} -\frac{\rho_1}{\varphi^{-1}(M/t)} \leq x - t \phi'\big(\varphi^{-1}({M}/t)\big) \leq \frac{\rho_2}{\varphi^{-1}(M/t)}. \end{align*} $$

Indeed, since $\varphi ^{-1}$ is nondecreasing and has upper scaling property (see Proposition 4.3), it has a doubling property. Hence, the lemma will follow immediately with possibly modified $\rho _0$ .

Without loss of generality, we can assume that $b = 0$ . Let $\lambda> 0$ , whose value will be specified later. We decompose the Lévy measure $\nu ({\mathrm {d}} x)$ as follows: Let $\nu _1({\mathrm {d}} x)$ be the restriction of $\tfrac 12 \nu ({\mathrm {d}} x)$ to the interval $(0,\lambda ]$ and

$$ \begin{align*} \nu_2({\textrm{d}} x)=\nu({\textrm{d}} x)-\nu_1({\textrm{d}} x). \end{align*} $$

We set

$$ \begin{align*} \phi_1(u) = \int_{(0, \infty)} \big(1 - e^{-u s}\big) \: \nu_1({\textrm{d}} s), \qquad \phi_2(u) = \int_{(0, \infty)} \big(1 - e^{-u s}\big) \: \nu_2({\textrm{d}} s). \end{align*} $$

Let us denote by $\mathbf {T}^{(j)}$ the subordinator having the Laplace exponent $\phi _j$ , for $j \in \{ 1, 2 \}$ . Let $\psi _j(\xi ) = \phi _j(-i\xi )$ . Notice that $\tfrac 12 \nu \leq \nu _2 \leq \nu $ ; thus,

$$ \begin{align*} \tfrac12 \phi \leq \phi_2 \leq \phi, \end{align*} $$

and for every $n \in \mathbb {N}$ ,

(4.41) $$ \begin{align} \tfrac12 (-1)^{n+1} \phi^{(n)} \leq (-1)^{n+1} \phi_2^{(n)} \leq (-1)^{n+1} \phi^{(n)}. \end{align} $$

Therefore, for all $u> 0$ ,

(4.42) $$ \begin{align} \tfrac12 \varphi(u) \leq \varphi_2(u) \leq \varphi(u). \end{align} $$

Next, by Theorem 3.3, the random variables $T_t^{(2)}$ and $T_t$ are absolutely continuous. Let us denote by $p^{(2)}(t, \:\cdot \:)$ and $p(t, \:\cdot \:)$ the densities of $T^{(2)}_t$ and $T_t$ , respectively.

Let $M \geq 2 M_0+1$ , where $M_0$ is determined in Corollary 3.5 for the process $\mathbf {T}^{(2)}$ . For $0 < t < 1/\varphi (x_0)$ , we set

$$ \begin{align*} x_t = t \phi_2'\big(\varphi^{-1}(M/t)\big). \end{align*} $$

Since $\varphi ^{-1}(M/t)> x_0$ , we have

$$ \begin{align*} \frac{x_t}{t} = \phi_2'\big(\varphi^{-1}(M/t)\big) \leq \phi'_2(x_0). \end{align*} $$

Let

$$ \begin{align*} w_2 = (\phi_2')^{-1}(x_t/t) =\varphi^{-1}(M/t). \end{align*} $$

Then, by (4.42) we get

$$ \begin{align*} \varphi_2(w_2) \geq \frac12 \varphi \big( \varphi^{-1}(M/t) \big) = \frac{M}{2t} \geq \frac{M_0}{t}. \end{align*} $$

Moreover, by Corollary 4.5 together with (4.42) we get

$$ \begin{align*} t\big( \phi_2(w_2) - w_2 \phi_2'(w_2) \big) \lesssim t \varphi_2\big(w_2\big) \lesssim 1. \end{align*} $$

Hence, by Corollary 3.5,

(4.43) $$ \begin{align} p^{(2)}(t, x_t) \gtrsim \frac{1}{\sqrt{t (-\phi_2^{\prime\prime})(w_2)}}. \end{align} $$

Notice that, by (4.41) and Remark 3.4, the implied constant in (4.43) is independent of t and $\lambda $ . Since

$$ \begin{align*} (-\phi_2^{\prime\prime})(w_2) \leq (-\phi^{\prime\prime})\big( \varphi^{-1}(M/t)\big) =\frac{M}{t \big(\varphi^{-1}(M/t)\big)^{2}}, \end{align*} $$

by (4.43) and monotonicity of $\varphi ^{-1}$ , we get

(4.44) $$ \begin{align} p^{(2)}(t, x_t) \geq C_1 \varphi^{-1} (1/t), \end{align} $$

for some constant $C_1> 0$ .

Next, by the Fourier inversion formula

$$ \begin{align*} \mathop{\mathrm{sup}}\limits_{x \in \mathbb{R}} \big| \partial_x p^{(2)}(t,x) \big| &\lesssim \int_{\mathbb{R}} e^{-t \Re \psi_2(\xi)} {\lvert {\xi} \rvert} {\: \rm d} \xi \\ &\lesssim \int_{\mathbb{R}} e^{-\frac{t}{2} \Re \psi(\xi)} {\lvert {\xi} \rvert} {\: \rm d} \xi; \end{align*} $$

thus, by [Reference Grzywny and Szczypkowski23, Proposition 3.4] and Propositions 4.2 and 4.3 we see that there is $C_2> 0$ such that for all $t \in (0, 1/\varphi (x_0))$ ,

$$ \begin{align*} \mathop{\mathrm{sup}}\limits_{x \in \mathbb{R}} \big| \partial_x p^{(2)}(t,x) \big| \leq C_2 \big(\varphi^{-1}(1/t)\big)^2. \end{align*} $$

By the mean value theorem, for $y \in \mathbb {R}$ , we get

$$ \begin{align*} \big| p^{(2)}(t, y + x_t) - p^{(2)}(t, x_t) \big| \leq C_2 {\lvert {y} \rvert} \big(\varphi^{-1}(1/t)\big)^2. \end{align*} $$

Hence, for $y \in \mathbb {R}$ satisfying

$$ \begin{align*} {\lvert {y} \rvert} \leq \frac{C_1} {2 C_2 \varphi^{-1}(1/t)}, \end{align*} $$

by (4.44), we get

$$ \begin{align*} p^{(2)}(t, y + x_t) &\geq p^{(2)}(t, x_t) - C_2 {\lvert {y} \rvert} \big(\varphi^{-1}(1/t)\big)^2 \\ &\geq \frac{C_1}{2} \varphi^{-1}(1/t). \end{align*} $$

Therefore,

$$ \begin{align*} p(t,x) &= \int_{\mathbb{R}} p^{(2)}(t,x-y) \mathbb{P} \big( T_t^{(1)}\in {\: \rm d} y \big) \\ &\geq \frac{C_1}{2} \varphi^{-1}(1/t) \cdot \mathbb{P} \bigg( \big|x - x_t - T_t^{(1)} \big| \leq \frac{C_0}{\varphi^{-1}(1/t)} \bigg) \\ &\geq \frac{C_1}{2} \varphi^{-1}(1/t) \cdot \mathbb{P} \bigg( \big|x - x_t - T_t^{(1)} \big| \leq \frac{C_0}{\varphi^{-1}(M/t)} \bigg) \\ &= \frac{C_1}{2} \varphi^{-1}(1/t) \cdot \mathbb{P} \bigg( \Big|x - \tilde{x}_t - \big( \tfrac12 tb_{\lambda} - (\tilde{x}_t - x_t) \big) - \big(T_t^{(1)}-\tfrac12 tb_{\lambda}\big) \Big| \leq \frac{C_0}{\varphi^{-1}(M/t)} \bigg), \end{align*} $$

where we have set $C_0=C_1(2C_2)^{-1}$ and

$$ \begin{align*} \tilde{x}_t = t\phi' \big(\varphi^{-1}(M/t)\big). \end{align*} $$

Let $\rho _0 = \tfrac {1}{2} C_0$ and

(4.45) $$ \begin{align} \lambda = \frac{1}{\varphi^{-1}(M/t)}. \end{align} $$

We have

$$ \begin{align*} \frac{1}{2} tb_{\lambda} - (\tilde{x}_t - x_t) &= \frac12tb_{\lambda} - t\phi_1'(1/\lambda) \\ &= \frac{t}{2} \int_{(0,\lambda]} s\big(1-e^{-s/\lambda} \big) \: \nu({\textrm{d}} s). \end{align*} $$

Thus, $\tfrac 12 tb_{\lambda } - (\tilde {x}_t - x_t)$ is nonnegative, and in view of (4.2) and (4.45),

(4.46) $$ \begin{align} \nonumber \tfrac12 tb_{\lambda} - (\tilde{x}_t - x_t) &\leq C_3 t \lambda \varphi(1/\lambda) \\ &= \frac{C_3M}{\varphi^{-1}(M/t)}, \end{align} $$

for some constant $C_3>0$ . Next, setting

$$ \begin{align*} \rho(t) = \lambda^{-1}\Big( \tfrac12 tb_{\lambda} - (\tilde{x}_t - x_t)\Big), \end{align*} $$

we get

(4.47) $$ \begin{align} \nonumber & \inf_{t \in (0, 1/\varphi(x_0))}{\bigg\{ \mathbb{P} \bigg( \Big| x -\tilde{x}_t - \lambda\rho(t) - \big( T_t^{(1)} - \tfrac12 tb_{\lambda}\big) \Big| \leq C_0 \lambda \bigg) \colon x \geq 0, -\rho_1 \lambda \leq x- \tilde{x}_t \leq \rho_2\lambda \bigg\}} \\ &\qquad\qquad\geq \inf_{t \in (0, 1/\varphi(x_0))}{ \Big\{ \mathbb{P}\left( \Big| y - \lambda^{-1} \big( T_t^{(1)} -\tfrac12 tb_{\lambda} \big) \Big| \leq C_0 \right) \colon -\rho_1 -\rho(t) \leq y \leq \rho_2 \Big\}}. \end{align} $$

Hence, the problem is reduced to showing that the infimum above is positive. Let us consider a collection $\{Y_t \colon t \in (0, 1/\varphi (x_0)) \}$ of infinitely divisible nonnegative random variables $Y_t = \lambda ^{-1}\big ( T^{(1)}_t-\tfrac 12 tb_{\lambda }\big )$ . The Lévy measure corresponding to $Y_t$ is

(4.48) $$ \begin{align} \mu_t(B) = t \nu_1\big(\lambda B\big) \end{align} $$

for any Borel subset $B \subset \mathbb {R}$ . Since for each $R> 1$ ,

$$ \begin{align*} b_{R \lambda}^{(1)} &= \int_{(0, R \lambda]} y \: \nu_1({\textrm{d}} y) \\ &= \frac{1}{2} \int_{(0, \lambda]} y \: \nu({\textrm{d}} y) = \frac{1}{2} b_\lambda, \end{align*} $$

by Proposition 4.10,

$$ \begin{align*} \mathbb{P}\big(|Y_t| \geq R\big) &= \mathbb{P}\left(\Big| T_t^{(1)}-\tfrac12 tb_{\lambda} \Big| \geq R \lambda \right) \\ &\lesssim t \int_{(0, \infty)} \min{\big\{1, R^{-2} \lambda^{-2} s^2\big\}} \: \nu_1({\textrm{d}} s); \end{align*} $$

thus,

$$ \begin{align*} \mathbb{P}\big(|Y_t| \geq R\big) &\lesssim t \lambda^{-2} R^{-2} \int_{(0, \lambda]} s^2 \: \nu({\textrm{d}} s) \\ &\lesssim t R^{-2} h(\lambda) \\ &\lesssim t R^{-2} \varphi(1/\lambda), \end{align*} $$

where in the last estimate we have used (4.2). Therefore, recalling (4.45), we conclude that the collection is tight. Next, let $\big ((Y_{t_n}, y_n) \colon n \in \mathbb {N}\big )$ be a sequence realising the infimum in (4.47). By the Prokhorov theorem, we can assume that $(Y_{t_n} \colon n \in \mathbb {N})$ is weakly convergent to the random variable $Y_0$ . We note that $Y_{t_n}$ has the probability distribution supported in $\big [-\tfrac 12t_n\lambda _n^{-1}b_{\lambda _n}, \infty \big )$ where $\lambda _n$ is defined as $\lambda $ corresponding to $t_n$ .

Suppose that $(t_n : n \in \mathbb {N})$ contains a subsequence convergent to $t_0> 0$ . Then $Y_0 = Y_{t_0}$ and the support of its probability distribution equals $\big [-\tfrac 12t_0\lambda _0^{-1}b_{\lambda _0}, \infty \big )$ . Since $\rho (t_0) \leq \tfrac 12t_0\lambda _0^{-1}b_{\lambda _0}$ , we easily conclude that the infimum in (4.47) is positive.

Hence, it remains to investigate the case when $(t_n \colon n \in \mathbb {N})$ has no positive accumulation points. If zero is the only accumulation point, then $(\lambda _n \colon n \in \mathbb {N})$ has a subsequence convergent to zero. Otherwise, $(t_n)$ diverges to infinity; thus, $x_0 = 0$ and $(\lambda _n)$ contains a subsequence diverging to infinity. In view of (4.46), $\rho (t)$ is uniformly bounded in t. Thus, after taking a subsequence, we may and do assume that there exists a limit

$$ \begin{align*} \tilde{\rho} = \lim_{n \to \infty} \rho(t_n). \end{align*} $$

By compactness we can also assume that $(y_n \colon n \in \mathbb {N})$ converges to $y_0 \in [-\rho _1-\tilde{\rho},\, \rho _2]$ . Consequently, to prove that the infimum in (4.47) is positive, it is sufficient to show that

(4.49) $$ \begin{align} \mathbb{P}\big(|y_0 - Y_0| \leq \tfrac{1}{2} C_0 \big)> 0. \end{align} $$

Observe that (4.49) is trivially satisfied if the support of the probability distribution of $Y_0$ is the whole real line. Therefore, we can assume that $Y_0$ is purely non-Gaussian. In view of [Reference Sato47, Theorem 8.7], it is also infinitely divisible.

Given $w \colon \mathbb {R} \rightarrow \mathbb {R}$ a continuous function satisfying

(4.50) $$ \begin{align} \big| w(x) - 1 \big| \leq C' {\lvert {x} \rvert}, \qquad\text{and}\qquad \big| w(x)\big| \leq C' {\lvert {x} \rvert}^{-1}, \end{align} $$

we write the Lévy–Khintchine exponent of $Y_{t_n}$ in the form

$$ \begin{align*} \psi_n(\xi) = -i \xi \gamma_n - \int_{(0, \infty)} \big(e^{i\xi s} -1 -i\xi s w(s) \big) \: \mu_{t_n}({\textrm{d}} s), \end{align*} $$

where

$$ \begin{align*} \gamma_n = \int_{(0, \infty)} s w(s) \: \mu_{t_n}({\textrm{d}} s) - \tfrac12 \lambda_n^{-1}t_n b_{\lambda_n}. \end{align*} $$

Since $(Y_{t_n} : n \in \mathbb {N})$ converges weakly to $Y_0$ , there are $\gamma _0 \in \mathbb {R}$ and $\sigma $ -finite measure $\mu _0$ on $(0, \infty )$ satisfying

$$ \begin{align*} \int_{(0, \infty)} \min{\big\{1, s^2 \big\}} \: \mu_0({\textrm{d}} s) < \infty, \end{align*} $$

such that the Lévy–Khintchine exponent of $Y_0$ is

$$ \begin{align*} \psi_0(\xi) = -i\xi \gamma_0 - \int_{(0, \infty)} \big(e^{i\xi s} -1 -i\xi s w(s) \big) \: \mu_0({\textrm{d}} s), \end{align*} $$

where

(4.51) $$ \begin{align} \gamma_0 = \lim_{n \to \infty} \gamma_n. \end{align} $$

Moreover, for any bounded continuous function $f\colon \mathbb {R} \rightarrow \mathbb {R}$ vanishing in a neighbourhood of zero, we have

(4.52) $$ \begin{align} \lim_{n \to \infty} \int_{(0, \infty)} f(s) \: \mu_{t_n}({\textrm{d}} s) = \int_{(0, \infty)} f(s) \: \mu_{0}({\textrm{d}} s). \end{align} $$

Next, let us fix w satisfying (4.50) which equals $1$ on $[0,1]$ . In view of (4.48) and the definition of $\nu _1$ , the support of $\mu _{t_n}$ is contained in $[0, 1]$ . Hence, $\gamma _n=0$ for every $n \in \mathbb {N}$ and, consequently, $\gamma _0=0$ . We also conclude that $\operatorname {\mathrm {supp}} \mu _0 \subset [0,1]$ .

At this stage we consider the cases (i) and (ii) separately. In (ii) we need to distinguish two possibilities: if $(t_n)$ tends toward zero, then $(\lambda _n)$ also approaches zero, and we impose that $-\phi ^{\prime \prime }$ is a function regularly varying at infinity with index $-1$ ; otherwise, $(t_n)$ tends toward infinity as well as $(\lambda _n)$ , and thus $x_0 = 0$ , and we additionally assume that $-\phi ^{\prime \prime }$ is a function regularly varying at zero with index $-1$ . For the sake of clarity of presentation, we restrict attention to the first possibility only. In the second one the reasoning is analogous. We show that the support of the probability distribution of $Y_0$ is the whole real line. By [Reference Sato47, Theorem 24.10], the latter can be deduced from

(4.53) $$ \begin{align} \int_{(0, \infty)} \min\{1, s\} \: \mu_0({\textrm{d}} s) = \infty. \end{align} $$

Since $\operatorname {\mathrm {supp}} \mu _0 \subset [0, 1]$ , for each $\epsilon \in (0,1)$ we can write

$$ \begin{align*} \int_{(0, \infty)} \min\{1, s\} \: \mu_0({\textrm{d}} s) \geq \int_{(\epsilon/2, 1]} s \: \mu_0({\textrm{d}} s); \end{align*} $$

thus, to conclude (4.53), it is enough to show that

(4.54) $$ \begin{align} \int_{(\epsilon/2, 1]} s \: \mu_0({\textrm{d}} s) \gtrsim \log \epsilon^{-1}. \end{align} $$

For the proof, for any $\epsilon \in (0,1)$ we define the following bounded continuous function:

(4.55) $$ \begin{align} f_{\epsilon}(s) = \begin{cases} 0 & \text{if } s<\epsilon/2,\\ 2s-\epsilon, & \text{if } \epsilon/2 \leq s < \epsilon,\\ s & \text{if } \epsilon \leq s < 1,\\ 1 & \text{if }s \geq 1. \end{cases} \end{align} $$

We have, in view of (4.52),

(4.56) $$ \begin{align} \int_{(\epsilon/2,1]} s \: \mu_0({\textrm{d}}s) \geq \int_{(0,1]}f_{\epsilon}(s) \: \mu_0({\textrm{d}}s) = \lim_{n \to \infty} \int_{(0,1]}f_{\epsilon}(s) \: \mu_{t_n}({\textrm{d}}s) \geq \liminf_{n \to \infty} \int_{(\epsilon,1]} s\:\mu_{t_n}({\textrm{d}}s). \end{align} $$

Let us estimate the last integral. We write

$$ \begin{align*} \int_{(\epsilon, 1]} s \: \mu_t (\rm{d} s) \nonumber &= t\lambda^{-1} \int_{(\lambda \epsilon, \lambda]}s \: \nu_1({\textrm{d}}s) \\ &= \tfrac{1}{2} t \lambda^{-1}\int_{(\lambda \epsilon, \lambda ]}s \: \nu({\textrm{d}}s). \end{align*} $$

By the Fubini–Tonelli theorem, we get

$$ \begin{align*} \int_{{(\lambda\epsilon,\lambda]}} s \:\nu({\textrm{d}}s) = \int_{\lambda \epsilon}^{\lambda}u^{-2} \int_{(0,u]} s^2\:\nu({\textrm{d}}s) \:{\textrm{d}}u + \lambda K(\lambda) - \lambda \epsilon K(\lambda \epsilon). \end{align*} $$

Thus,

(4.57) $$ \begin{align} 2 \int_{{(\epsilon,1]}} s \:\mu_t({\textrm{d}}s) = t\lambda^{-1} \int_{\lambda \epsilon}^{\lambda} K(u)\:{\textrm{d}}u + tK(\lambda) - t\epsilon K(\lambda \epsilon). \end{align} $$

Setting $z = 1/\lambda $ , by (4.2) and (4.45), we obtain

$$ \begin{align*} t K(\lambda) \approx t \varphi(z) \approx 1. \end{align*} $$

Moreover, since $\varphi $ is a $1$ -regularly varying function at infinity, we have

$$ \begin{align*} t\epsilon K(\lambda \epsilon) \approx t \epsilon \varphi(z/\epsilon) = M \epsilon \frac{\varphi(z/\epsilon)}{\varphi(z)}\to M, \end{align*} $$

as z tends to infinity. Therefore, it remains to estimate the integral in (4.57). Using (4.2) we get

$$ \begin{align*} t\lambda^{-1} \int_{\lambda \epsilon}^{\lambda} K(u)\:{\textrm{d}}u &\approx \frac{z}{\varphi(z)} \int_{\epsilon z^{-1}}^{z^{-1}}\varphi \big( u^{-1} \big)\:{\textrm{d}}u \\ &=\frac{z}{\varphi(z)}\int_z^{\epsilon^{-1}z}u^{-2}\varphi(u)\:{\textrm{d}}u \\ &= \frac{\phi'(z)-\phi' \big(\epsilon^{-1}z\big)}{z \big(-\phi^{\prime\prime}(z)\big)}. \end{align*} $$

Since $-\phi ^{\prime \prime }(s)=s^{-1} \ell (s)$ for a certain function $\ell $ slowly varying at infinity, by [Reference Bingham, Goldie and Teugels3, Theorem 1.5.6],

$$ \begin{align*} \frac{\phi'(z)-\phi' \big(\epsilon^{-1}z\big)}{z \big(-\phi^{\prime\prime}(z)\big)} = \int_1^{\epsilon^{-1}} \frac{\ell(zt)}{\ell(z)} \frac{{\textrm{d}}t}{t} \to \log \epsilon^{-1}, \end{align*} $$

as z tends to infinity. Hence,

$$ \begin{align*} \liminf_{n \to \infty} \int_{(\epsilon, 1]} s \: \mu_{t_n}({\textrm{d}} s) \gtrsim \log \epsilon^{-1}, \end{align*} $$

which by (4.56) implies (4.54).

Next, let us consider the case (i); that is, when $-\phi ^{\prime \prime } \in \mathrm {WUSC}(\beta -2,C,x_0)$ with $C \geq 1$ and $\alpha \leq \beta <1$ . We claim that for all $\epsilon \in (0, 1)$ ,

(4.58) $$ \begin{align} \int_{(0, \epsilon)} s^2 \: \mu_0({\textrm{d}} s)>0. \end{align} $$

To see this, it is enough to show that there is $C> 0$ such that for all $\epsilon \in (0, 1]$ and $t \in (0, 1/\varphi (x_0))$ ,

(4.59) $$ \begin{align} \int_{(0, \epsilon)} s^2 \: \mu_t({\textrm{d}} s) \geq C \epsilon^{2-\alpha}. \end{align} $$

For the proof, we select a continuous function on $\mathbb {R}$ such that

and for each $\tau> 0$ set

$$ \begin{align*} \eta_\tau(x) = \eta(\tau^{-1} x). \end{align*} $$

Since for $0 < 2 \tau < \epsilon $ ,

$$ \begin{align*} \int_{(0, \infty)} s^2 \big(\eta_\epsilon(s) - \eta_\tau(s)\big) \: \mu_t({\textrm{d}} s) + \int_{(0, 2\tau)} s^2 \eta_\tau(s) \: \mu_t({\textrm{d}} s) &\geq \int_{(0, \epsilon)} s^2 \: \mu_t({\textrm{d}} s), \end{align*} $$

by (4.59) and (4.52),

$$ \begin{align*} \int_{(0, \infty)} s^2 \big(\eta_\epsilon(s) - \eta_\tau(s)\big) \: \mu_0({\textrm{d}} s) + \limsup_{n \to \infty} \int_{(0, \infty)} s^2 \eta_\tau(s) \: \mu_{t_n}({\textrm{d}} s) \geq C \epsilon^{2-\alpha}. \end{align*} $$

Since $Y_{t_n}$ and $Y_0$ are purely non-Gaussian, by [Reference Sato47, Theorem 8.7(2)],

$$ \begin{align*} \lim_{\tau \to 0^+} \limsup_{n \to \infty} \int_{(-\tau, \tau)} s^2 \: \mu_{t_n} ({\textrm{d}} s) = 0; \end{align*} $$

thus,

$$ \begin{align*} \int_{(0, \epsilon)} s^2 \:\mu_0({\textrm{d}} s) \geq C \epsilon^{2-\alpha}, \end{align*} $$

which entails (4.58).

We now turn to showing (4.59). We have

$$ \begin{align*} \int_{(0, \epsilon)} s^2 \: \mu_t({\textrm{d}} s) &= t \lambda^{-2} \int_{(0, \lambda \epsilon)} s^2 \: \nu_1({\textrm{d}} s) \\ &= \tfrac{1}{2} t \lambda^{-2} \int_{(0, \lambda \epsilon)} s^2 \: \nu({\textrm{d}} s)\\ &= \tfrac{1}{2} t \epsilon^2 K(\lambda \epsilon); \end{align*} $$

thus, by (4.2) and the weak lower scaling property of $\varphi $ ,

$$ \begin{align*} \int_{(0, \epsilon)} s^2 \:\mu_t({\textrm{d}} s) &\gtrsim t \epsilon^2 \varphi\big(\epsilon^{-1} \lambda^{-1}\big) \\ &\gtrsim t \epsilon^{2-\alpha} \varphi(1/\lambda), \end{align*} $$

which, together with the definition of $\lambda $ , implies (4.59).

Since the support of the probability distribution of $Y_0$ is not the whole real line, by [Reference Picard42, Lemma 2.5], the inequality (4.58) implies that

(4.60) $$ \begin{align} \int_{(0, \infty)} \min\{1, s\} \: \mu_0({\textrm{d}} s) < \infty \end{align} $$

and the support of $Y_0$ equals $[\chi , \infty )$ where

(4.61) $$ \begin{align} \chi = \gamma_0 - \int_{(0, \infty)} sw(s) \: \mu_0({\textrm{d}} s) = -\int_{(0, 1]} s \: \mu_0({\textrm{d}} s). \end{align} $$

To conclude (4.49), it is enough to show that $\chi \leq -\tilde {\rho }$ . Since $\rho (t_n) \leq \frac {1}{2}t_n \lambda _n^{-1} b_{\lambda _n}$ , the latter can be deduced from

(4.62) $$ \begin{align} \begin{aligned} \chi &= -\lim_{n \to \infty} \frac{1}{2} t_n \lambda_n^{-1} b_{\lambda_n} \\ &= -\lim_{n \to \infty} \int_{(0, 1]} s \: \mu_{t_n}({\textrm{d}} s), \end{aligned} \end{align} $$

where the last equality is a consequence of (4.48) since

(4.63) $$ \begin{align} \int_{(0, 1]} s \: \mu_t (\rm{d} s) &= t\lambda^{-1} \int_{(0, \lambda]} s \: \nu_1({\textrm{d}}s) \nonumber\\ &= \tfrac{1}{2} t \lambda^{-1}\int_{(0, \lambda] }s \: \nu({\textrm{d}}s). \end{align} $$

Therefore, the problem is reduced to showing (4.62). By the monotone convergence theorem and (4.52), we have

(4.64) $$ \begin{align} \begin{aligned} \chi &= -\lim_{\epsilon \to 0^+} \int_{(0,1]} f_{\epsilon}(s) \: \mu_0({\textrm{d}}s) \\ &= -\lim_{\epsilon \to 0^+} \lim_{n \to \infty}\int_{(0,1]} f_{\epsilon}(s) \: \mu_{t_n}({\textrm{d}}s) \end{aligned} \end{align} $$

and

(4.65) $$ \begin{align} \lim_{\epsilon \to 0^+} \int_{(0,1]} f_{\epsilon}(s) \: \mu_{t_n}({\textrm{d}}s) = \int_{(0, 1]} s \: \mu_{t_n}({\textrm{d}} s), \end{align} $$

where $f_{\epsilon }$ is as in (4.55). Hence, we just need to justify the change in the order of limits. In view of the Moore–Osgood theorem [Reference Graves19, Chapter VII], it is enough to show that the limit in (4.65) is uniform with respect to $n \in \mathbb {N}$ .

We write

$$ \begin{align*} \bigg| \int_{(0, 1]} s \: \mu_t({\textrm{d}} s) - \int_{(0, 1]} f_\epsilon(s) \: \mu_t({\textrm{d}} s) \bigg| &\leq \int_{(0, \epsilon/2]} s \: \mu_t({\textrm{d}} s) + \int_{(\epsilon/2, \epsilon]} (\epsilon-s) \: \mu_t({\textrm{d}} s) \\ &\leq \int_{(0, \epsilon]} s \: \mu_t({\textrm{d}} s). \end{align*} $$

By (4.63) and the Fubini–Tonelli theorem, we have

$$ \begin{align*} 2 t^{-1} \lambda \int_{(0, \epsilon]} s \: \mu_t({\textrm{d}} s) = \int_{(0, \lambda \epsilon]}s \: \nu({\textrm{d}}s) &= \int_0^{\lambda \epsilon} u^{-2} \int_{(0,u]} s^2 \: \nu({\textrm{d}} s) {\: \rm d} u + \lambda \epsilon K (\lambda \epsilon) \\ &\approx \int_0^{\lambda \epsilon} \varphi\big(u^{-1}\big) {\: \rm d} u + \lambda \epsilon \varphi\big(\lambda^{-1}\epsilon^{-1}\big). \end{align*} $$

By almost monotonicity of $\varphi $ ,

(4.66) $$ \begin{align} \nonumber \int_{(0, \epsilon]} s \: \mu_t ({\textrm{d}} s) & \approx t \lambda^{-1} \int_0^{\lambda \epsilon} \varphi\big(u^{-1}\big) {\: \rm d} u + t \epsilon \varphi\big(\lambda^{-1}\epsilon^{-1}\big) \\ &\approx t \lambda^{-1} \int_0^{\lambda \epsilon} \varphi\big(u^{-1}\big) {\: \rm d} u. \end{align} $$

Now, setting $z = \varphi ^{-1}(M/t)$ , by (4.45), we get

(4.67) $$ \begin{align} \nonumber t \lambda^{-1} \int_0^{\lambda \epsilon} \varphi\big(u^{-1}\big) {\: \rm d} u &= t\varphi^{-1}(M/t) \int_0^{\epsilon/\varphi^{-1}(M/t)} \varphi \big(u^{-1}\big) {\: \rm d} u \\ \nonumber &\approx \frac{z}{\varphi(z)}\int_0^{\epsilon z^{-1}} \varphi \big( u^{-1} \big) {\: \rm d} u \\ \nonumber &= \frac{z}{\varphi(z)}\int_{\epsilon^{-1} z}^{\infty} u^{-2}\varphi(u) {\: \rm d} u \nonumber \\ &= \frac{\phi'\big(\epsilon^{-1} z\big)}{z \big(-\phi^{\prime\prime}(z)\big)}. \end{align} $$

In view of Proposition 2.4, by the upper scaling of $-\phi ^{\prime \prime }$ , there is $c> 0$ such that for all $z> x_0$ ,

$$ \begin{align*} \frac{\phi'\big(\epsilon^{-1} z\big)}{z \big(-\phi^{\prime\prime}(z)\big)} \leq c \epsilon^{1-\beta}. \end{align*} $$

Hence, the limit in (4.65) is uniform with respect to $n \in \mathbb {N}$ , which justifies (4.62). This completes the proof of (4.49) and the theorem follows.

Proof. First let us notice that Corollary 2.7 implies that $-\phi ^{\prime \prime } \in \mathrm {WLSC}(\alpha -2,c,x_0) \cap \mathrm {WUSC}(\beta -2,C,x_0)$ . Therefore, the hypothesis of Theorem 4.11 is satisfied.

It is enough to show the first inequality in (4.68) since the latter is an easy consequence of (4.28) and Proposition 4.6. For $t \in (0, 1/\varphi (x_0))$ and $M \geq 1$ , we set

$$ \begin{align*} x_t = t \phi' \big( \varphi^{-1}(M/t) \big). \end{align*} $$

By Proposition 4.6, the function $\varphi ^{-1}$ possesses the weak lower scaling property. Moreover, there is $C_1 \geq 1$ such that for all $r> \varphi (x_0)$ ,

(4.69) $$ \begin{align} C_1^{-1} \varphi^{-1}(r) \leq \phi^{-1}(r) \leq C_1 \varphi^{-1}(r). \end{align} $$

Hence, by Proposition 2.4, there is $C_2 \geq 1$ , such that

(4.70) $$ \begin{align} x_t \leq C_2 M^{1-1/\beta} \frac{1}{\varphi^{-1}(1/t)}. \end{align} $$

We select $M \geq 1$ satisfying

$$ \begin{align*} C_1 C_2 M^{1-1/\beta} < \chi_1. \end{align*} $$

Let $\rho _1 = \rho _0/2$ where $\rho _0$ is determined in Theorem 4.11. Then, by (4.69) and (4.70), we have

(4.71) $$ \begin{align} \nonumber x_t - \frac{\rho_1}{\varphi^{-1}(1/t)} &\leq C_1 C_2 M^{1-1/\beta} \frac{1}{\phi^{-1}(1/t)} \\ &< \frac{\chi_1}{\phi^{-1}(1/t)}. \end{align} $$

Now set $\rho _2 = C_1 \chi _2$ . Then, by (4.69), we have

(4.72) $$ \begin{align} x_t + \frac{\rho_2}{\varphi^{-1}(1/t)}> \frac{\rho_2}{C_1 \phi^{-1}(1/t)} = \frac{\chi_2}{\phi^{-1}(1/t)}. \end{align} $$

Putting (4.72) and (4.71) together, we conclude that

$$ \begin{align*} \bigg[\frac{\chi_1}{\phi^{-1}(1/t)}, \frac{\chi_2}{\phi^{-1}(1/t)} \bigg] \subseteq \left( x_t - \frac{\rho_1}{\varphi^{-1}(1/t)}, x_t + \frac{\rho_2}{\varphi^{-1}(1/t)} \right). \end{align*} $$

Therefore, by Theorem 4.11, for all $t \in (0, 1/\varphi (x_0))$ and $x> 0$ satisfying

$$ \begin{align*} \chi_1 \leq x \phi^{-1}(1/t) \leq \chi_2, \end{align*} $$

we have

$$ \begin{align*} p(t,x) \gtrsim \varphi^{-1}(1/t). \end{align*} $$

In view of (4.69), this completes the proof of the theorem.

Proof. Let $\lambda> 0$ . We begin by decomposing the Lévy measure $\nu ({\mathrm {d}} x)$ . Let $\nu _1({\mathrm {d}} x) = \nu _1(x) {\: \rm d} x$ and $\nu _2({\mathrm {d}}x) = \nu _2(x) {\: \rm d} x$ where

For $u> 0$ , we set

$$ \begin{align*} \phi_1(u) = b u + \int_{(0, \infty)} \big(1-e^{-u s}\big) \: \nu_1({\textrm{d}} s) \qquad\text{and}\qquad \phi_2(u) = \int_{(0, \infty)} \big(1 - e^{-u s}\big) \: \nu_2({\textrm{d}} s). \end{align*} $$

Let $\mathbf {T}^{(j)}$ be the Lévy process having the Laplace exponent $\phi _j$ , for $j \in \{1, 2\}$ . Since $\tfrac {1}{2} \nu \leq \nu _1 \leq \nu $ , we have

$$ \begin{align*} \tfrac{1}{2} \phi \leq \phi_1 \leq \phi, \end{align*} $$

and for all $n \in \mathbb {N}$ ,

(4.73) $$ \begin{align} \tfrac{1}{2} (-1)^{n+1} \phi^{(n)} \leq (-1)^{n+1} \phi_1^{(n)} \leq (-1)^{n+1} \phi^{(n)}. \end{align} $$

Thus,

$$ \begin{align*} \tfrac{1}{2} \varphi \leq \varphi_1 \leq \varphi, \end{align*} $$

and so for all $u> 0$ ,

(4.74) $$ \begin{align} \varphi^{-1}_1(u/2) \leq \varphi^{-1}(u) \leq \varphi^{-1}_1(u). \end{align} $$

In particular, $-\phi _1^{\prime \prime }$ has the weak lower scaling property. Therefore, by Theorem 3.3, $T_t^{(1)}$ and $T_t$ are absolutely continuous. Let us denote by $p(t, \:\cdot \:)$ and $p^{(1)}(t, \:\cdot \:)$ the densities of $T_t$ and $T_t^{(1)}$ , respectively. Observe that $\mathbf {T}^{(2)}$ is a compound Poisson process with the probability distribution denoted by $P_t({\mathrm {d}} x)$ . By [Reference Sato47, Remark 27.3],

(4.75) $$ \begin{align} P_t({\textrm{d}} x) \geq t e^{-t\nu_2(\mathbb{R})} \nu_2(x) {\: \rm d} x. \end{align} $$

We apply Theorem 4.11 to the process $\mathbf {T}^{(1)}$ . For $t> 0$ , we set

$$ \begin{align*} x_t = t \phi'_1\big(\varphi_1^{-1}({M_0}/t)\big). \end{align*} $$

Then there are $C> 0$ and $\rho _0> 0$ , such that for all $t \in (0, 1/\varphi (x_0))$ and $x \geq 0$ satisfying

$$ \begin{align*} x_t - \frac{\rho_0}{\varphi_1^{-1}(1/t)} \leq x \leq x_t + \frac{\rho_0}{\varphi_1^{-1}(1/t)}, \end{align*} $$

we have

$$ \begin{align*} p^{(1)}(t, x) \geq C \varphi_1^{-1}(1/t). \end{align*} $$

Therefore, if

$$ \begin{align*} \lambda = x_t + \frac{\rho_0}{\varphi_1^{-1}(1/t)}, \end{align*} $$

then

(4.76) $$ \begin{align} \int_0^\lambda p^{(1)}(t, x) {\: \rm d} x \gtrsim 1. \end{align} $$

Next, if $\lambda \geq \rho _0/\varphi ^{-1}(1/t)$ , then, by (4.2),

(4.77) $$ \begin{align} \nonumber t \nu_2(\mathbb{R}) &= \tfrac{1}{2} t \int_\lambda^\infty \: \nu(x) {\: \rm d} x \\ \nonumber &\leq \tfrac{1}{2} t h\big(\rho_0/\varphi^{-1}(1/t)\big) \\ \nonumber &\lesssim t h\big(1/\varphi^{-1}(1/t)\big)\\ &\lesssim 1, \end{align} $$

where the penultimate inequality follows either by monotonicity of h or by [Reference Grzywny and Szczypkowski23, Lemma 2.1 (4)]. Finally, by (4.75) and (4.77), for $x \geq 2 \lambda $ we can compute

$$ \begin{align*} p(t, x) &= \int_{\mathbb{R}} p^{(1)}(t, x - y) P_t({\textrm{d}} y) \\ &\gtrsim t \int_{\mathbb{R}} p^{(1)}(t, x-y) \nu_2(y) {\: \rm d} y \\ &=\tfrac{1}{2} t \int_\lambda^x p^{(1)}(t, x-y) \nu(y) {\: \rm d} y. \end{align*} $$

Hence, by the monotonicity of $\nu $ , we get

$$ \begin{align*} p(t, x) &\gtrsim t \nu(x) \int_0^{x - \lambda} p^{(1)}(t, y) {\: \rm d} y\\ &\geq t \nu(x) \int_0^\lambda p^{(1)}(t, y) {\: \rm d} y \\ &\gtrsim t \nu(x), \end{align*} $$

where in the last estimate we have used (4.76). Using (4.73) and (4.74), we can easily show that

$$ \begin{align*} \lambda = x_t + \frac{\rho_0}{\varphi_1^{-1}(1/t)} \leq t \phi'\big(\varphi^{-1}({M_0}/t)\big) +\frac{\rho_0}{\varphi^{-1}(1/t)}, \end{align*} $$

and the proposition follows.

Proof. Let $a \in (0,1]$ . Recall that by (4.30) we have $\nu (s) \leq C_1 s^{-3} \big ( -\phi ^{\prime \prime }(1/s) \big )$ for any $s>0$ . Hence, for any $u>0$ ,

(4.78) $$ \begin{align} \nonumber -\phi^{\prime\prime}(u) &= \int_0^{au^{-1}} s^2e^{-us} \nu(s) {\: \rm d} s + \int_{au^{-1}}^{\infty} s^2e^{-us} \nu(s) {\: \rm d} s \\ &\leq C_1 \int_0^{au^{-1}} s^{-1}e^{-us} \big( -\phi^{\prime\prime}(1/s) \big) {\: \rm d} s + C_2 \nu(au^{-1}) \int_{au^{-1}}^{\infty} s^2e^{-us} {\: \rm d} s, \end{align} $$

where $C_2$ is a constant from the almost monotonicity of $\nu $ . If $u> x_0$ , then by the scaling property of $-\phi ^{\prime \prime }$ we obtain

$$ \begin{align*} C_1 \int_0^{au^{-1}} s^{-1}e^{-us} \big( -\phi^{\prime\prime}(1/s) \big) {\: \rm d} s &\leq C\int_0^{au^{-1}} s^{-1}e^{-us} (su)^{-\gamma} \big( -\phi^{\prime\prime}(u) \big) {\: \rm d} s \\ &\leq C \big( -\phi^{\prime\prime}(u) \big) \int_0^as^{-1-\gamma} e^{-s} {\: \rm d} s. \end{align*} $$

By selecting $a \in (0, 1]$ such that

$$ \begin{align*} 2C \int_0^a s^{-1-\gamma}e^{-s} {\: \rm d} s \leq 1, \end{align*} $$

we get

$$ \begin{align*} \int_0^{au^{-1}} s^{-1}e^{-us} \big( -\phi^{\prime\prime}(1/s) \big) {\: \rm d} s \leq \tfrac{1}{2} \big( -\phi^{\prime\prime}(u) \big). \end{align*} $$

Since

$$ \begin{align*} \int_{au^{-1}}^{\infty} s^2 e^{-us} {\: \rm d} s = u^{-3} e^{-a}(a^2+2a+2), \end{align*} $$

by (4.78) we obtain

$$ \begin{align*} \nu(au^{-1}) \geq \frac{e^a}{2(a^2+2a+2)}u^3 \big( -\phi^{\prime\prime}(u) \big), \end{align*} $$

provided that $u> x_0$ . Now, by the monotonicity of $-\phi ^{\prime \prime }$ we conclude the proof.

Proof. First let us note that by Corollary 2.7, $-\phi ^{\prime \prime } \in \mathrm {WLSC}(\alpha -2,c,x_0) \cap \mathrm {WUSC}(\beta -2,C,x_0)$ . Therefore, we are in position to apply Proposition 4.14. By Corollary 3.7, for $\chi _1 = \min {\{1, \delta \}}$ , we have

$$ \begin{align*} p(t, x) \approx \big(t (-\phi^{\prime\prime}(w)) \big)^{-\frac{1}{2}} \exp\big\{-t \big(\phi(w) - w \phi'(w)\big) \big\}, \end{align*} $$

whenever $0 < x \phi ^{-1}(1/t) \leq \chi _1$ . Next, let $M_0'$ be $M_0$ determined by Proposition 4.14. By Proposition 2.4, (4.2) and monotonicity of $\varphi ^{-1}$ , for $t \in (0, 1/\varphi (x_0))$ , we get

$$ \begin{align*} t \phi'\big(\psi^{-1}(1/t)\big) \lesssim \frac{1}{\psi^{-1}(1/t)} {\qquad \text{and} \qquad t\phi'\big(\varphi^{-1}(M_0'/t)\big) \lesssim \frac{1}{\varphi^{-1}(1/t)};} \end{align*} $$

thus, by Propositions 4.3 and 4.6, there is $C_1> 0$ such that

$$ \begin{align*} 2 e t \phi'\big(\psi^{-1}(1/t)\big) \leq C_1 \frac{1}{\phi^{-1}(1/t)} \end{align*} $$

and

$$ \begin{align*} 2t\phi' \big(\varphi^{-1}(M_0'/t)\big)+ \frac{2 \rho'_0}{\varphi^{-1}(1/t)} \leq C_1 \frac{1}{\phi^{-1}(1/t)} \end{align*} $$

where $\rho _0'$ is the value of $\rho _0$ determined in Proposition 4.14. Let $\chi _2 = \max {\{1, C_1, \chi _1\}}$ . By Proposition 4.14 and Corollary 4.16, there is $a \in (0,1]$ such that if $x \phi ^{-1}(1/t)> \chi _2$ and $0 < x< a/x_0$ , then