2.1 Greedy approximation schemes

The following definitions and notation are taken from [Reference Darken, Donahue, Gurvits and Sontag6]. Let X be a real Banach space, and suppose that $S\subseteq X$ is some arbitrary set. Let $\mathrm {co}({S})$ denote the set of all convex combinations of elements of S, that is, objects of the form

$$ \begin{align*} \lambda_1 y_1+\cdots +\lambda_n y_n \end{align*} $$

for $n\geq 1$ , $y_1,\ldots ,y_n\in S,$ and $\lambda _1,\ldots ,\lambda _n\in [0,1]$ with $\lambda _1+\cdots +\lambda _n=1$ . Now suppose that $x^\ast \in \overline {\mathrm {co}({S})}$ , where $\overline {\mathrm {co}({S})}$ denotes the closure of $\mathrm {co}({S})$ . A natural question arises:

In other words, we consider incremental sequences $(x_n)$ of the form

$$ \begin{align*} x_{n+1}=(1-\lambda_n)x_n+\lambda_ny_n \end{align*} $$

for $x_0\in S$ , $y_n\in S,$ and $\lambda _n\in [0,1)$ (we assume that $\lambda _n<1$ else we would have $x_{n+1}=y_n\in S$ and we are back where we started). The most natural way of choosing $y_n$ and $\lambda _n$ at each step is to require them to be optimal, which leads naturally to the following slightly more general notion.

Definition 2.1. A sequence $(x_n)$ of elements of X is called greedy with respect to $x^\ast \in X$ and S if for all $n\in \mathbb {N}$ :

$$ \begin{align*} \|{x_{n+1}-x^\ast}\|\leq \inf\{ {\|{(1-\lambda)x_n+\lambda y-x^\ast}\|} \ | \ {y\in S, \lambda\in [0,1)}\}. \end{align*} $$

To account for the fact that S may not be compact, and thus the infimum in the above definition not attained, we loosen the notion in order to incorporate error terms $(\epsilon _n)$ , leading to our main definition of greedy approximation scheme.

Definition 2.2. Let $(\epsilon _n)$ be a sequence of positive real numbers. A sequence $(x_n)$ of elements of X is called $(\epsilon _n)$ -greedy with respect to $x^\ast \in X$ and S if for all $n\in \mathbb {N}$ :

$$ \begin{align*} \|{x_{n+1}-x^\ast}\|\leq \inf\{ {\|{(1-\lambda)x_n+\lambda y-x^\ast}\|} \ | \ {y\in S, \lambda\in [0,1)}\}+\epsilon_n. \end{align*} $$

Greedy approximation schemes in Hilbert spaces were first studied by Jones [Reference Jones13], where convergence to $x^\ast \in \overline {\mathrm {co}({S})}$ with rate $\mathcal {O}(1/\sqrt {n})$ is proven, and the connection of such schemes to artificial neural networks is highlighted where, roughly speaking, improving an approximant by combining with a new element of S can be seen as a generalisation of improving the accuracy of a neural network by adding an additional neuron (see [Reference Jones13, Section 4] or [Reference Darken, Donahue, Gurvits and Sontag6, Section 1.3] for further details of this connection).

When X is a general Banach space, on the other hand, convergence is no longer guaranteed. The example given in [Reference Darken, Donahue, Gurvits and Sontag6] is $\mathbb {R}^2$ under the $L^1$ norm: Here, if we take $S:=\{(0,-1),(2,1/2),(-2,1/2)\}$ so that $(0,0)\in \mathrm {co}({S})$ , the only greedy incremental scheme starting from $x_0:=(0,-1)$ is $x_0=x_1= x_2=\cdots $ , since there is no way to strictly decrease the distance to $(0,0)$ through convex combinations with $(2,1/2)$ or $(-2,1/2)$ . Geometrically speaking, the problem here is that the unit ball in this space is a diamond, and the line segment between $(0,-1)$ and $(2,1/2)$ only intersects this unit ball at $(0,-1)$ , which would not be the case with the Euclidean metric.

The critical issue is connected to the notion of smoothness (more informally, spaces where “balls don’t have corners”). It turns out that one can establish convergence of greedy algorithms in general Banach spaces by assuming additional smoothness properties. However, the proofs are more difficult, and do not always come with corresponding rates. Establishing rates in such cases is one of the goals of this article.

2.2 Geometric properties of Banach spaces

In what follows we present some basic facts about Banach spaces: Further details for this section can be found in, e.g., [Reference Chidume3, Chapter 2] and [Reference Temlyakov47, Chapter 6]. Let $X^\ast $ denote the dual of X, and $J:X\to 2^{X^\ast }$ the so-called normalized duality mapping function defined by

$$ \begin{align*} J(x):=\{ {y\in X^\ast} \ | \ {y(x)=\|{x}\|^2=\|{y}\|^2}\}. \end{align*} $$

A space X is defined to be smooth if J is single-valued, in which case we let $j:X\to X^\ast $ denote the corresponding unique duality map.

In the special case that X is a Hilbert space the duality map is just the inner product $j(x)=\langle {x,-}\rangle $ , and as such the duality map often plays the role of mimicking an inner product in Banach spaces. Crudely speaking, the nicer the duality map, the more X behaves like a Hilbert space. In this respect, an important notion is the modulus of smoothness

$$ \begin{align*} \rho_X(t):=\sup\left\{\left.\frac{\|{x+ty}\|+\|{x-ty}\|}{2}-1 \, \right| \, \|{x}\|=\|{y}\|=1 \right\} \end{align*} $$

for $t\in (0,\infty )$ , which in a certain sense gives a quantitative measure of ‘niceness’ in this context. A Banach space X is uniformly smooth if

$$ \begin{align*} \lim_{t\to 0}\frac{\rho_X(t)}{t}=0. \end{align*} $$

The following standard lemma (cf. [Reference Temlyakov47, Lemma 6.1]) connects the duality map with the modulus of smoothness. We give its proof in full as this forms part of the overall proof that will be analysed.

Lemma 2.3. Suppose that X is smooth. Take $x\neq 0$ and let $F_x:=j(x)/\|{x}\|$ . Then for any $y\in X$ and $t\in (0,\infty )$ :

$$ \begin{align*} \|{x+ty}\|\leq \|{x}\|\left(1+2\rho_X\left(t\|{y}\|/\|{x}\|\right)\right)+tF_x(y). \end{align*} $$

Proof From the definition of $\rho _X$ we obtain

$$ \begin{align*} \|{x+ty}\|+\|{x-ty}\|\leq 2\|{x}\|\left(1+\rho_X\left(t\|{y}\|/\|{x}\|\right)\right) \end{align*} $$

and using in addition that from $\|{F_x}\|=1$ we have

$$ \begin{align*} \|{x-ty}\|\geq F_x(x-ty)=F_x(x)-tF_x(y)=\|{x}\|-tF_x(y) \end{align*} $$

the result follows.

Remark 2.4. Using Lemma 2.3 in along with $\|{x+ty}\|\geq F_x(x+ty)=\|{x}\|+tF_x(y)$ proves

$$ \begin{align*} 0\leq \|{x+ty}\|-\|{x}\|-tF_x(y)\leq 2\rho_X\left(t\|{y}\|/\|{x}\|\right). \end{align*} $$

For $\|{x}\|=\|{y}\|$ it follows that

$$ \begin{align*} 0\leq \frac{\|{x+ty}\|-\|{x}\|}{t}-F_x(y)\leq \frac{2\rho_X(t)}{t} \end{align*} $$

and thus uniformly smooth spaces have the nice geometric property that

$$ \begin{align*} \lim_{t\to 0}\left(\frac{\|{x+ty}\|-\|{x}\|}{t}\right)=F_x(y) \end{align*} $$

and moreover the limit is attained uniformly in x and y with $\|{y}\|=\|{x}\|$ . In other words, X has a uniformly Gâteuax differentiable norm, and in fact this is an equivalent characterisation of being uniformly smooth.

2.3 Convergence of greedy incremental sequences in Banach spaces

We now state and prove the main result that we will analyse: Roughly speaking, this says that if X is uniformly smooth, if $(x_n)$ is an $(\epsilon _n)$ -greedy approximation schemes with respect to $x^\ast $ , then $x_n\to x^\ast $ provided that $\sum _{i=0}^\infty \epsilon _i<\infty $ . The material in this section is taken entirely from [Reference Darken, Donahue, Gurvits and Sontag6], with the proof only very slightly reformulated from its original presentation.

We first require a lemma that applies smoothness to greedy algorithms, essentially showing that sequences that are $(\epsilon _n)$ -greedy with respect to $x^\ast $ are, in particular, quasi-Fejér monotone with respect to $x^\ast $ in a certain sense (see [Reference Combettes4, Reference Combettes5] for surveys of this fundamental concept and its role in convex optimization). The result below incorporates Lemma 3.3 of [Reference Darken, Donahue, Gurvits and Sontag6], along with part of the main proof of Theorem 3.4 from the same paper, and we have deliberately re-structured things in this way in order to separate out those parts of the proof that use functional analysis of some kind. This is convenient because it allows us, in the proof of Theorem 2.6, to focus on the main combinatorial structure of the overall proof, making it easier to organise the quantitative analysis that follows.

Lemma 2.5 (Cf. Lemma 3.3 of [Reference Darken, Donahue, Gurvits and Sontag6])

Let X be a Banach space with modulus of smoothness $\rho _X$ . Suppose that $S\subseteq X$ and $x^\ast \in \overline {\mathrm {co}({S})}$ , that $(x_n)$ is $(\epsilon _n)$ -greedy with respect to $x^\ast $ and S, and $K>0$ is such that $\sup \{ {\|{y-x^\ast }\|} \ | \ {y\in S}\}\leq K$ . Then for any $n\in \mathbb {N}$ and $b>0$ , we have

$$ \begin{align*} \|{x_{n+1}-x^\ast}\|\leq (1-\alpha(b))\|{x_n-x^\ast}\|+\epsilon_n \end{align*} $$

whenever $\|{x_n-x^\ast }\|\geq b$ , where

$$ \begin{align*} \alpha(b):=\sup\left\{\left.\lambda\left(1-\frac{2K}{b}\cdot\frac{\rho_X(u(b,\lambda))}{u(b,\lambda)}\right)\ \right| \ \lambda\in [0,1)\right\} \end{align*} $$

for $u(b,\lambda ):=\lambda K/(1-\lambda )b$ .

Proof Fix $n\in \mathbb {N}$ and assume that $\|{x_n-x^\ast }\|\geq b>0$ . For $y\in S$ and $\lambda \in [0,1)$ , writing

$$ \begin{align*} \|{(1-\lambda)x_n+\lambda y-x^\ast}\|=(1-\lambda)\|{(x_n-x^\ast)+\lambda(y-x^\ast)/(1-\lambda)}\| \end{align*} $$

and applying Lemma 2.3 to the right hand side we have

$$ \begin{align*} \begin{aligned} &\|{(1-\lambda)x_n+\lambda y-x^\ast}\|\\&\quad \leq (1-\lambda)\left(1+2\rho_X\left(\frac{\lambda \|{y-x^\ast}\|}{(1-\lambda)\|{x_n-x^\ast}\|}\right)\right)\|{x_n-x^\ast}\|+\lambda F_n(y-x^\ast) \end{aligned} \end{align*} $$

where we write $F_n:=j(x_n-x^\ast )/\|{x_n-x^\ast }\|$ . Using the standard fact that the modulus of smoothness is monotone, it follows from $b\leq \|{x_n-x^\ast }\|$ and $\|{y-x^\ast }\|\leq K$ for $y\in S$ that

$$ \begin{align*} \begin{aligned} (1-\lambda)\rho_X\left(\frac{\lambda \|{y-x^\ast}\|}{(1-\lambda)\|{x_n-x^\ast}\|}\right)&\leq (1-\lambda)\rho_X\left(\frac{\lambda K}{(1-\lambda)b}\right)\\ &=\frac{\lambda K}{b}\cdot\frac{\rho_X(u(b,\lambda))}{u(b,\lambda)}=:\beta(\lambda,n) \end{aligned} \end{align*} $$

for $u(b,\lambda )$ as defined in the statement of the result. Putting these together and defining $\beta (\lambda ,n)$ as indicated in the previous equation, we obtain

(*) $$ \begin{align} &\|{(1-\lambda)x_n+\lambda y-x^\ast}\|\nonumber\\ &\leq \left[1-\lambda\left(1-2\beta(\lambda,n)\right)\right]\|{x_n-x^\ast}\|+\lambda F_n(y-x^\ast). \end{align} $$

Now, since $x^\ast \in \overline {\mathrm {co}({S})}$ , we can make the positive part of $F_n(y-x^\ast )$ arbitrary small, in the sense that for all $\varepsilon>0$ there exists $y\in S$ such that ${F_n(y-x^\ast )\leq \varepsilon }$ . To see this, pick some $z\in \mathrm {co}({S})$ such that $\|{z-x^\ast }\|\leq \varepsilon $ . Writing $z=\sum _{i=1}^k\lambda _iy_i$ for $y_i\in S$ , we must have $F_n(y_i-x^\ast )\leq \varepsilon $ for some $i=1,\ldots ,k$ , else, using that $F_n$ is linear and $\|{F_n}\|\leq 1$ we have

$$ \begin{align*} \varepsilon=\varepsilon\sum_{i=1}^k\lambda_i<\sum_{i=1}^k \lambda_i F_n(y_i-x^\ast)=F_n(z-x^\ast)\leq \|{z-x^\ast}\|\leq \varepsilon \end{align*} $$

Therefore $\inf \{F_n(y-x^\ast )\, | \, y\in S\}\leq 0$ , and more generally the infimum over $y\in S$ and $\lambda \in [0,1)$ of the quantity

$$\begin{align*}\left(1-\lambda\left(1-2\beta(\lambda,n)\right)\right)\|{x_n-x^\ast}\|+\lambda F_n(y-x^\ast) \end{align*}$$

is bounded by

$$\begin{align*}\left(1-\sup\left\{\left.\lambda\left(1-2\beta(\lambda,n)\right)\ \right| \ \lambda\in [0,1)\right\}\right)\|{x_n-x^\ast}\|, \end{align*}$$

which is just $(1-\alpha (b))\|{x_n-x^\ast }\|$ . Thus combining the above with (*) and the definition of being $(\epsilon _n)$ -greedy we have

$$ \begin{align*} \|{x_{n+1}-x^\ast}\|&\leq \inf\{\|{(1-\lambda)x_n+\lambda y-x^\ast}\| \, | \, y\in S,\, \lambda \in [0,1)\}+\epsilon_n\\ &\leq (1-\alpha(b))\|{x_n-x^\ast}\|+\epsilon_n \end{align*} $$

and the lemma is proven.

The main result can now be stated and proved using little more than elementary analysis.

Proof Define $a_n:=\|{x_n-x^\ast }\|$ and let $a_\infty :=\liminf _{n\to \infty } a_n$ . Using the fact that $(x_n)$ is $(\epsilon _n)$ -greedy, we have $a_{n+1}\leq a_n+\epsilon _n$ by definition (by simply setting $\lambda =0$ ), and thus more generally

$$ \begin{align*} a_{n+m}\leq a_n+\sum_{i=n}^{n+m-1}\epsilon_i\leq a_n+\sum_{i=n}^\infty \epsilon_i \end{align*} $$

for any $m,n\in \mathbb {N}$ . But since $\sum _{i=n}^\infty \epsilon _i\to 0$ as $n\to \infty $ , we must in fact have $a_n\to a_\infty $ as $n\to \infty $ . It therefore suffices to show that $a_\infty =0$ .

To this end, suppose for contradiction that $a_\infty>0$ . Since S is bounded there exists $K>0$ is such that $\sup \{ {\|{y-x^\ast }\|} \ | \ {y\in S}\}\leq K$ , and since ${a_\infty /2\leq a_n}$ for n sufficiently large, we can apply Lemma 2.5 to show that

$$ \begin{align*} a_{n+1}\leq (1-\alpha(a_\infty/2))a_n+\epsilon_n \end{align*} $$

for n sufficiently large. Taking the limit as $n\to \infty $ and using that $a_n\to a_\infty $ and $\epsilon _n\to 0$ yields

$$ \begin{align*} a_\infty\leq (1-\alpha(a_\infty/2))a_\infty \end{align*} $$

and therefore $\alpha (a_\infty /2)\leq 0$ . But we have $u(a_\infty /2,\lambda )\to 0$ as $\lambda \to 0$ , and therefore by uniform smoothness of X it follows that

$$ \begin{align*} \frac{\rho_X(u(\lambda,a_\infty/2))}{u(\lambda,a_\infty/2)}\to 0 \ \ \ \text{as } \lambda \to 0 \end{align*} $$

from which we see that $\alpha (a_\infty /2)>0$ , a contradiction. Thus $a_\infty =0$ and we are done.

3.1 The overall structure of the proof of Theorem 2.6

For the remainder of this section, we will fix several things: X will be a Banach space with $\rho $ its modulus of smoothness, $S\subseteq X$ and $x^\ast \in X$ . We suppose that $K>0$ is such that $\|{y-x^\ast }\|\leq K$ for all $y\in S$ , which in particular exists whenever S is bounded. Finally, for now we let $(x_n)$ be an arbitrary sequence in X and $(\epsilon _n)$ an arbitrary sequence of nonnegative real numbers. We treat these throughout as global parameters, and also for convenience fix the notation $a_n:=\|{x_n-x^\ast }\|$ . The goal is therefore to prove that $a_n\to 0$ .

The main technical lemma we required—Lemma 2.5—can then be represented as the single inference

where “ $(\epsilon _n)\text {-}x^\ast \text {-greedy}$ ” is shorthand for the statement “ $(x_n)$ is $(\epsilon _n)$ -greedy with respect to $x^\ast $ ” and the predicate P is defined by

$$ \begin{align*} P(b,a,a',\epsilon):=a'\leq (1-\alpha(b))a+\epsilon \end{align*} $$

for $\alpha (b)$ defined as in Lemma 2.5. Rather than immediately analysing the somewhat intricate proof of that lemma, we will leave it for now and consider how it fits into the main proof of Theorem 2.6. Letting $a_\infty :=\liminf _{n\to \infty }a_n$ , the first step in the main proof has the overall form

(Γ₁)

Of course, there are a number of additional (elementary) steps involved in inferring $a_n\to a_\infty $ from the two premises, but we note that the property of being $(\epsilon _n)$ -greedy is used in a weak way, in that we only require $a_{n+1}\leq a_n+\epsilon _n$ . Moving on to the second main step of the proof, we now include an open assumption $\{{a_\infty>0}\}$ with the aim of reaching a contradiction:

(Γ₂)

and we label this derivation (Γ₂ ). Here, $N_\exists $ is simply some natural number that we know to exist by definition of $a_\infty $ (where the notation is chosen to indicate that this is essentially an epsilon term for which we do not have an explicit value). The conclusion of (Γ₂ ) is simply the statement that

$$ \begin{align*}a_{n+1}\leq (1-\alpha(a_\infty/2))a_n+\epsilon_n\end{align*} $$

for n sufficiently large. Continuing, we have the following crucial step:

(Γ₃)

where we use the shorthand

to represent the proof tree (Γ₁ ), and similarly for (Γ₂ ). The final inference in (Γ₃ ) represents taking the limit as $n\to \infty $ to establish

$$ \begin{align*}a_\infty\leq (1-\alpha(a_\infty/2))a_\infty.\end{align*} $$

We reach our contradiction by using uniform smoothness of the space, with an inference we mark as $(\star )$ below, and from this can therefore derive $a_\infty =0$ using classical logic, eliminating the open assumption $\{{a_\infty>0}\}$ :

(Γ₄)

Formally, there is now one final step in the proof, namely:

(Γ₅)

and thus expanding the definition of (Γ₅ ) in full would give us a complete (high-level) representation of the proof of Theorem 2.6. We now set out to analyse this proof with the three main questions posed at the beginning of the section in mind. We start at the bottom and work back up.

3.2 Using $a_\infty =0$ in the final step

We start by considering precisely how the final step is proven, and asking whether we can extract any computational information at this stage. We first note that in the special case that $(a_n)$ is a sequence of nonnegative reals, the otherwise more complex statement $a_\infty =0$ can be expressed as

(*) $$ \begin{align} \forall b>0,m\in\mathbb{N}\,\exists n\geq m(a_n< b). \end{align} $$

So assuming that we have proven (*), how exactly do we derive $a_n\to 0$ from $a_n\to a_\infty $ ? For a start, we observe that we do not require the full statement $a_n\to a_\infty $ at all: It is sufficient to notice that two premises used to prove this in (Γ₁ )—which is in general a more complex argument—can be combined with (*) in a simple way to establish $a_n\to 0$ . To be more specific, suppose that $f:(0,\infty )\to \mathbb {N}$ is a rate of convergence for $\sum _{i=0}^\infty \epsilon _i<\infty $ in the sense that

$$ \begin{align*} \forall b>0\left(\sum_{i=f(b)}^\infty \epsilon_i<b\right). \end{align*} $$

Then for any $n\geq f(b)$ , using that $\forall n(a_{n+1}\leq a_n+\epsilon _n)$ we must have

$$ \begin{align*} a_{n+k}\leq a_n+\sum_{i=n}^{n+k-1}\epsilon_i\leq a_n+\sum_{i=f(b)}^\infty \epsilon_i< a_n+b \end{align*} $$

for any $n,k\in \mathbb {N}$ and $b>0$ . Now suppose that we have a computable bound for n in (*) for $m:=f(b),$ i.e., a function $\Phi :(0,\infty )\to \mathbb {N}$ such that

$$ \begin{align*} \forall b>0\,\exists n\in [f(b);\Phi(b)](a_n< b), \end{align*} $$

where we use the notation $[f(b);\Phi (b)]:=[f(b),\Phi (b)]\cap \mathbb {N}$ . Then it is not hard to see that $\Phi (b/2)$ must be a rate of convergence for $a_n\to 0$ , and thus we have not only proved that $a_n\to 0$ but shown exactly what quantitative information we need from the assumption $a_\infty =0$ to obtain a rate. In terms of our proof trees, we have essentially transformed (Γ₅ ) into the following computational derivation, eliminating the original (Γ₁ ), where $b>0$ is now some global free variable:

(Γ^c ₅)

where now ( $\Gamma _4^c$ ) is some (currently hypothetical) computation transformation of (Γ₄ ) that allows us to replace $a_\infty =0$ with a derivation of

$$\begin{align*}\exists n\in [f(b); \Phi(b)](a_n< n) \end{align*}$$

for some explicit function $\Phi $ . The challenge has now shifted to transforming the original derivation (Γ₄ ) to a computational one ( $\Gamma _4^c$ ).

3.3 Simplifying (Γ₄)

A crucial observation at this stage is that (Γ₄ ) simplifies: Because we have only used $a_\infty =0$ in the weakened form,

(**) $$ \begin{align} \exists n\geq f(b)(a_n< b), \end{align} $$

where $b>0$ is now some ambient free variable, we can try to substitute this in the conclusion of (Γ₄ ) and then replace the open assumption $\{{a_\infty>0}\}$ with the stronger negation of (**), i.e., the more concrete assumption

$$\begin{align*}\{{\forall n\geq f(b)(a_n\geq b)}\}, \end{align*}$$

and simplify the proof tree accordingly. This process involves a series of straightforward heuristic steps. Starting with (Γ₂ ), we observe that here $\{{a_\infty>0}\}$ is crucially used to establish that $\forall n\geq N_\exists (a_\infty /2\leq a_n)$ for sufficiently large $N_\exists $ , where $a_\infty /2>0$ . For now we see if we can just replace $a_\infty /2$ with b, and set $N_\exists :=f(b)$ as follows:

(Γ^s ₂)

By a simple substitution, (Γ₃ ) then becomes

(Γ^s ₃)

and then the entire modified version of (Γ₄ ) would become

(Γ^s ₄)

where here replacing $a_\infty /2$ with b makes no difference to the way in which we derive a contradiction. Though not yet fully computational, (Γ^s ₄ ) represents a simplified and more explicit version of (Γ₄ ), reflecting throughout our weakened use of $a_\infty =0$ .

3.4 Analysing $(\Gamma _4^s)$

We have demonstrated that in order to obtain a computable rate of convergence for $a_n\to 0$ it suffices to find a bound $\Phi (b)$ for the existential quantifier in $\exists n\geq f(b)(a_n<b)$ , where f is a rate of convergence for $\sum _{i=0}^\infty \epsilon _i<\infty $ . We propose to do this by analysing the simplified version (Γ^s ₄ ) of (Γ₄ ) arrived at above.

Here a key observation is that if we can weaken the open assumption with a bound on how many $n\geq f(b)$ are needed in order to be able to reach a contradiction as before, then this will be exactly the bound we are looking for. In other words, we want to produce $\Phi (b)$ satisfying

Focusing first on the final derivation above, one immediate observation is that $\alpha (b)$ is not necessarily computable, which might pose a problem in the likely case that we want to use property $\alpha (b)>0$ in a computational way. Here we make another assumption that we hope to verify later on, namely, the existence of a computable function $\xi :(0,\infty )\to (0,1)$ witnessing that $\alpha (b)>0,$ i.e., such that $\alpha (b)\geq \xi (b)>0$ for any $b>0$ .

We now shift up the proof tree to the final inference of (Γ^s ₃ ), where we take the limit as $n\to \infty $ to establish $P(b,a_\infty ,a_\infty ,0),$ i.e.,

(†) $$ \begin{align} a_\infty\leq (1-\alpha(b))a_\infty. \end{align} $$

Given the infinitary nature of (†) it is not clear at first glance how we could weaken our open assumption $\{{\forall n\geq f(b)(a_n\geq b)}\}$ to being true only in a finite range. However, a natural question to ask here is the following: If $P(b,a_n,a_{n+1},\epsilon _n)$ fails to be true in the limit—in the sense of (†)—can we show that it also fails to hold for n sufficiently large? In particular, here the only property of $a_n\to a_{\infty }$ that is important is that $a_n$ and $a_{n+1}$ converge to the same value, so could we replace this with $a_n$ and $a_{n+1}$ being sufficiently close together?

To this end, let us take some $\delta>0$ . Then by $a_n\to a_\infty $ and $\epsilon _n\to 0$ there exists, more concretely, some $k\geq f(b)$ such that $a_k-a_{k+1}<\delta $ and $\epsilon _k<\delta $ . Then from $P(b,a_k,a_{k+1},\epsilon _k)$ and our assumption $\alpha (b)\geq \xi (b)$ it follows that

$$ \begin{align*} a_{k+1}\leq (1-\xi(b))(a_{k+1}+\delta)+\delta, \end{align*} $$

which can be rearranged as

$$ \begin{align*} \frac{\xi(b)\cdot a_{k+1}}{2-\xi(b)}\leq \delta. \end{align*} $$

But using also that $b\leq a_{k+1}$ , the above fails for

$$ \begin{align*} \delta_{\xi,b}:=\tfrac{1}{2}\xi(b)\cdot b, \end{align*} $$

so we have reached a contradiction, and we can replace the final inference of (Γ^s ₃ ) with a finitary version which does not use the whole limit as $n\to \infty $ . We now formulate this finitisation of (Γ^s ₃ ) as the following proof tree $(\Gamma _3^f)$ , where $\delta _{\xi ,b}$ is defined as above:

(Γ^f ₃)

Then we have obtained our contradiction as follows:

We now ask: How much of the assumption $\{{\forall n\geq f(b)(a_n\geq b)}\}$ do we need to obtain this contradiction? Inspecting (Γ^f ₃ ) it is readily apparent that if $\Psi (b)$ is a bound on a witness for $\exists k\geq f(b)(a_k-a_{k+1},\epsilon _k<\delta _{\xi ,b})$ , then we can replace the open assumption with

$$ \begin{align*} \{{\forall n\in [f(b);\Psi(b)+1](a_n\geq b)}\} \end{align*} $$

(here the $+1$ is necessary from our additional use of the assumption for $b\leq a_{k+1}$ ), and thus $\Psi (b/2)+1$ is a rate of convergence for $a_n\to 0$ .

3.5 The final step

All that remains in order to obtain our rate of convergence—aside from some assumptions involving uniform smoothness that we have postponed to later—is to analyse the following fragment of our modified proof tree:

We will actually provide a bound $\Psi (N,\delta )$ for the more general statement

$$ \begin{align*} \forall N,\delta\exists k\geq N(a_k-a_{k+1},\epsilon_k<\delta) \end{align*} $$

and then

(+) $$ \begin{align} \Phi(b):=\Psi(f(b/2), \delta_{\xi,b/2})+1 \end{align} $$

would be our rate of convergence for $a_n\to 0$ . Witnessing the above turns out to be simpler that it might look: From $\sum _{i=0}^\infty \epsilon _i<\infty $ we clearly have $\epsilon _k<\delta $ for all $k\geq f(\delta )$ sufficiently large, and then finding k such that we also have $a_k-a_{k+1}<\delta $ does not in any way require the assumption $a_n\to a_\infty $ , rather we simply need a bound on a single element of $(a_n)$ . We now present this construction in general terms as follows.

Lemma 3.1. Let $(a_n)$ , $(K_n),$ and $(\epsilon _n)$ be sequences of nonnegative reals with $K_n\geq a_n$ for all $n\in \mathbb {N}$ and $\epsilon _n\to 0$ as $n\to \infty $ with rate f. Let

$$ \begin{align*} \Psi(N,\delta):=M_{N,\delta}+\Big\lceil \frac{K_{M_{N,\delta}}}{\delta}\Big\rceil \ \ \ \text{for} \ \ \ M_{N,\delta}:=\max\{N,f(\delta)\}. \end{align*} $$

Then for any $N\in \mathbb {N}$ and $\delta>0$ there exists some $k\in [N;\Psi (N,\delta )]$ such that $a_k-a_{k+1}<\delta $ and $\epsilon _k<\delta $ .

Proof Suppose that $a_k-a_{k+1}\geq \delta $ for all $k\in [M_{N,\delta };\Psi (N,\delta )]$ . Then we have

$$ \begin{align*} K_{M_{N,\delta}}\geq a_{M_{N,\delta}}&\geq a_{M_{N,\delta}+1}+\delta\\ &\geq \dots\\ &\geq a_{M_{N,\delta}+\lceil K_{M_{N,\delta}}/\delta\rceil+1}+(\lceil K_{M_{N,\delta}}/\delta\rceil+1)\delta\\ &\geq (\lceil K_{M_{N,\delta}}/\delta\rceil+1)\delta\\ &>K_{M_{N,\delta,}} \end{align*} $$

which is a contradiction. Therefore there exists some

$$\begin{align*}k\in [M_{N,\delta};\Theta(N,\delta)]\subseteq [N;\Theta(N,\delta)] \end{align*}$$

such that $a_k-a_{k+1}<\delta $ , and since $k\geq M_{N,\delta }\geq f(\delta )$ it also follows that $\epsilon _k<\delta $ by definition of f.

Assuming for simplicity that f is monotone, so that we can simplify $\max \{f(b/2),f(\delta _{\xi ,b/2})\}=f(\delta _{\xi ,b/2})$ , and supposing that $\sum _{i=0}^\infty \epsilon _i\leq L$ , by which we have $a_n\leq a_0+\sum _{i=0}^\infty \epsilon _i\leq K+L$ for all $n\in \mathbb {N}$ , we then obtain a concrete rate of convergence for $a_n\to 0$ from (+), namely,

(R) $$ \begin{align} \Phi(b):=f(\mu_{\xi,b})+\Big\lceil \frac{K+L}{\mu_{\xi,b}}\Big\rceil+1 \ \ \ \text{for} \ \ \ \mu_{\xi,b}:=\frac{b}{4}\cdot \xi\left(\frac{b}{2}\right). \end{align} $$

Equipped with this high-level quantitative understanding of the main combinatorial structure of the proof, we now start to put things together in an ordinary mathematical style.

4.1 Abstract convergence lemmas

The result of Section 3 was a preliminary rate of convergence for $\|{x_n-x^\ast }\|\to 0$ in some (as yet hypothetical) computable function $\xi $ satisfying $0<\xi (b)\leq \alpha (b)$ for all $b>0$ , with $\alpha (b)$ defined in terms of the modulus of smoothness of the space as in Lemma 2.5. While there is nothing preventing us from just using this rate directly, it is sensible to first ask whether we can optimize the construction—in particular, whether there are local modifications that either improve our initial convergence rate or weaken any assumptions used. Here we now depart from our rigid analysis of the original proof of Theorem 2.6 and inject into it new insights.

Let us consider our analysis of (Γ^s ₄ ) in Section 3.4. The core of this part of the proof, which allows us to derive our contradiction, is the limiting step in (Γ^s ₃ ), which uses $\sum _{i=0}^\infty \epsilon _i<\infty $ in the weaker form of $\epsilon _n\to 0$ as $n\to \infty $ . A natural question arises: Can we either exploit the strong property $\sum _{i=0}^\infty \epsilon _i<\infty $ in this step to gain a better bound $\Phi (b)$ , or replace $\sum _{i=0}^\infty \epsilon _i<\infty $ with the weaker property $\epsilon _n\to 0$ and $n\to \infty $ throughout the proof? We now show that both are possible.

Our first result captures a refined analysis of (Γ^s ₄ ) that makes full use of the property $\sum _{i=0}^\infty \epsilon _i<\infty $ . We no longer spell out the details using proof trees, now instead adopting a standard presentation.

Lemma 4.1. Let $(a_n)$ and $(\epsilon _n)$ be sequences of nonnegative reals, and ${\xi :[0,\infty )\to [0,1)}$ be a function with $\xi (0)=0$ and $\xi (b)>0$ for $b>0$ , such that for all $n\in \mathbb {N}$ and $b\geq 0,$

$$\begin{align*}a_{n+1}\leq (1-\xi(b))a_n+\epsilon_n \end{align*}$$

whenever $a_n\geq b$ . Suppose that $a_0\leq K$ and $\sum _{i=0}^\infty \epsilon _i\leq L$ with rate f. Then $a_n\to 0$ as $n\to \infty $ with rate

$$\begin{align*}\Phi(b):=f\left(\frac{b}{2}\right)+\Big\lceil \frac{K+L}{\mu_{\xi,b}}\Big\rceil \ \ \ \text{for} \ \ \ \mu_{\xi,b}:=\frac{b}{2}\cdot\xi\left(\frac{b}{2}\right). \end{align*}$$

Proof Fix $b>0$ . We show that there exists some $n\in [f(b/2);\Phi (b)]$ such that $a_n<b/2$ . If this were not the case, then for all n in that range we would have

$$\begin{align*}\mu_{\xi,b}\leq a_n\cdot\xi\left(\frac{b}{2}\right)\leq a_n-a_{n+1}+\epsilon_n, \end{align*}$$

and therefore summing both sides

$$\begin{align*}K+L<\sum_{i=f(b/2)}^{\Phi(b)}\mu_{\xi,b}\leq \sum_{i=f(b/2)}^{\Phi(b)}(a_i-a_{i+1}+\epsilon_i)\leq a_{f(b/2)}+\sum_{i=f(b/2)}^{\Phi(b)}\epsilon_i \end{align*}$$

and now also using that $a_{n+1}\leq a_n+\epsilon _n$ for all $n\in \mathbb {N}$ , we have

$$\begin{align*}a_{f(b/2)}+\sum_{i=f(b/2)}^{\Phi(b)}\epsilon_i\leq a_0+\sum_{i=0}^{f(b/2)-1} \epsilon_i+\sum_{i=f(b/2)}^{\Phi(b)}\epsilon_i\leq a_0+\sum_{i=0}^\infty \epsilon_i \end{align*}$$

and thus we have shown that

$$\begin{align*}K+L<a_0+\sum_{i=0}^\infty \epsilon_i\leq K+L \end{align*}$$

a contradiction. Therefore let $n\in [f(b/2);\Phi (b)]$ be such that $a_n<b/2$ . But then just as in Section 3.2 we have

$$\begin{align*}a_{n+k}\leq a_n+\sum_{i=f(b/2)}^\infty \epsilon_i<a_n+\frac{b}{2}<b \end{align*}$$

for any $k\in \mathbb {N}$ , and so the result follows.

It is instructive to compare the rate given in Lemma 4.1 above to the initial rate (R): The former is an improvement of the latter, but shares the same basic structure. Generally speaking, Lemma 4.1 is more broadly applicable to algorithms that are quasi-Fejér monotone and enjoy a certain type of regularity property, here represented by the function $\xi $ . The general quantitative study of quasi-Fejér monotone sequences from the perspective of proof mining began explicitly in [Reference Kohlenbach, Leuştean and Nicolae25], was further developed in [Reference Pischke42] (notably under regularity assumptions), and recently lifted to a stochastic setting in [Reference Neri, Pischke and Powell35], where, for example, Theorem 3.4 of [Reference Neri, Pischke and Powell35] is a direct generalisation of Lemma 4.1 applicable to stochastic algorithms.

Our next result now keeps the analysis of (Γ^s ₄ ) intact, but instead adjusts (Γ₁ ), replacing the weaker property $a_{n+1}\leq a_n+\epsilon _n$ with the stronger recurrence inequality satisfied by $\epsilon _n$ -greedy algorithms in uniformly smooth Banach spaces, conversely allowing us to weaken the assumption ${\sum _{i=0}^\infty \epsilon _i<\infty }$ to $\epsilon _n\to 0$ as $n\to \infty $ throughout the proof. To obtain a rate of convergence that is uniform in $(a_n)$ we now require an explicit bound on the whole sequence $(a_n)$ , though boundedness of $(a_n)$ is not required as an assumption in order to establish convergence. The result is inspired by a similar recurrence inequality used in [Reference Powell and Wiesnet46] (cf. Lemma 3.1 of that paper) in the somewhat different context of Krasnoselskii–Mann type algorithms for computing fixpoints of weakly contractive mappings.

Lemma 4.2. Let $(a_n)$ and $(\epsilon _n)$ be sequences of nonnegative reals, and ${\xi :[0,\infty )\to [0,1)}$ be a function with $\xi (0)=0$ and $\xi (b)>0$ for $b>0$ , such that for all $n\in \mathbb {N}$ and $b\geq 0,$

$$\begin{align*}a_{n+1}\leq (1-\xi(b))a_n+\epsilon_n \end{align*}$$

whenever $a_n\geq b$ . Suppose that $\epsilon _n\to 0$ as $n\to \infty $ with rate f. Then $a_n\to 0$ as $n\to \infty $ with rate

$$\begin{align*}\Phi(b):=f(\mu_{\xi,b})+\Big\lceil \frac{L}{\mu_{\xi,b}}\Big\rceil+1 \ \ \ \text{for} \ \ \ \mu_{\xi,b}:=\frac{b}{4}\cdot \xi\left(\frac{b}{2}\right), \end{align*}$$

where $L>0$ is any bound on $(a_n)$ .

Proof Fix $b>0$ and apply Lemma 3.1 for $N:=f(\mu _{\xi ,b})$ and $\delta :=\mu _{\xi ,b}$ , by which there exists $k\in [f(\mu _{\xi ,b});\Theta (b)]$ such that $a_k-a_{k+1},\epsilon <\mu _{\xi ,b}$ for

$$\begin{align*}\Theta(b):=f(\mu_{\xi,b})+\Big\lceil \frac{K_b}{\mu_{\xi,b}}\Big\rceil, \end{align*}$$

where $K_b$ is any bound for $a_{f(\mu _{\xi ,b})}$ . Suppose for contradiction that $a_l\geq b/2$ for all $l\in [f(\mu _{\xi ,b});\Theta (b)+1]$ . Then in particular we have

$$\begin{align*}a_{k+1}\leq (1-\xi(b/2))(a_{k+1}+\mu_{\xi,b})+\mu_{\xi,b} \end{align*}$$

and therefore

$$\begin{align*}\mu_{\xi,b}\leq \frac{a_{k+1}}{2}\cdot\xi\left(\frac{b}{2}\right)< \frac{\xi(b/2)\cdot a_{k+1}}{2-\xi(b/2)}\leq \mu_{\xi,b}. \end{align*}$$

Thus there exists $l\in [f(\mu _{\xi ,b});\Theta (b)+1]$ such that $a_l<b/2$ , and we now show by induction that $a_{l+m}<b$ for all $m\in \mathbb {N}$ . For the induction step, we first note that since $l+m\geq l\geq f(\mu _{\xi ,b})$ we have $\epsilon _{l+m}<\mu _{\xi ,b}$ . We then consider two cases: Either $a_{l+m}<b/2$ , and then

$$\begin{align*}a_{l+m+1}\leq a_{l+m}+\epsilon_{l+m}<\frac{b}{2}+\mu_{\xi,b}<b, \end{align*}$$

or otherwise we have $b/2\leq a_{l+m}<b$ , and then

$$\begin{align*}a_{l+m+1}\leq (1-\xi(b/2))a_{l+m}+\mu_{\xi,b}\leq a_{l+m}-2\mu_{\xi,b}+\mu_{\xi,b}<b. \end{align*}$$

Therefore $a_n\to 0$ as $n\to \infty $ with convergence rate $\Theta (b)+1$ , and setting $K_b:=L$ for some uniform bound L on $(a_n)$ gives the result.

4.2 Handling uniform smoothness

Up until now, we have postponed the fact that we need to deal with uniform smoothness in a computational way: So far, our computational analysis has dealt with nothing beyond sequences of real numbers. Interestingly, and as is very often the case in applied proof theory, these results on the convergence of sequence of reals contain the core computational content of the original proof (see [Reference Franci and Grammatico9] for a comprehensive survey on the general importance of such results in analysis, and [Reference Neri and Powell36] for a recent discussion of the role they play in the context of applied proof theory, which also contains references to some the many places in which quantitative versions of such results have played a role). Indeed, the only computational role that uniform smoothness plays in the proof of Theorem 2.6 is in establishing that $\alpha (b)>0$ for $\alpha $ defined as in Lemma 2.5, which explicitly involves the modulus of smoothness $\rho _X$ , or more precisely, a rate of convergence for $\rho _X(t)/t\to 0$ as $t\to 0$ .

For the vast majority of case studies in applied proof theory that take place in uniformly smooth Banach spaces, such a rate of convergence is essentially all that is required (though see [Reference Findling and Kohlenbach8] for an example where additional properties of the modulus of smoothness are needed). In such cases, it is often convenient to reformulate the proof using the following alternative definition of uniform smoothness: A Banach space is uniformly smooth if and only if for any $\varepsilon>0$ there exists $\delta>0$ such that

(US) $$ \begin{align} \|{x}\|=1\wedge \|{y}\|\leq \delta\implies \|{x+y}\|+\|{x-y}\|\leq 2+\varepsilon\|{y}\| \end{align} $$

for any $x,y\in X$ . This characterisation of uniform smoothness is simpler from a logical perspective, and admits a direct computational interpretation in the form of a so-called modulus of uniform smoothness $\omega :(0,\infty )\to (0,\infty )$ , which is defined to be any function that for any $\varepsilon>0$ returns some witness for $\delta $ in (US). Moduli of uniform smoothness, which are distinct from the (uniquely defined) modulus of smoothness, were first used in [Reference Kohlenbach and Leuştean24] and appear in many other places in proof mining as a quantitative analogue of uniform smoothness.

For the purpose of our case study, our main task is therefore to reformulate the main lemmas on uniform smoothness in terms of the simpler logical modulus. We start with Lemma 2.3.

Lemma 4.3. Let $(X,\omega )$ be a uniformly smooth Banach space with modulus of uniform smoothness $\omega $ . Take $x\neq 0$ and let $F_x:=j(x)/\|{x}\|$ . Then

$$ \begin{align*} \frac{t\|{y}\|}{\|{x}\|}\leq \omega(\varepsilon)\implies \|{x+ty}\|\leq \|{x}\|\left(1+\frac{\varepsilon t\|{y}\|}{\|{x}\|}\right)+tF_x(y) \end{align*} $$

for any $y\in X$ and $t\in (0,\infty )$ .

We now give a computational version of Lemma 2.5, which is formulated in terms of the modulus $\omega $ instead of $\rho _X$ , and implicitly replaces the function $\alpha (b)$ defined there with a function $\xi (b)$ computable in $\omega $ .

Lemma 4.4. Let $(X,\omega )$ be a uniformly smooth Banach space. Suppose that $S\subseteq X$ and $x^\ast \in \overline {\mathrm {co}({S})}$ , that $(x_n)$ is $(\epsilon _n)$ -greedy with respect to $x^\ast $ and S, and $K>0$ is such that $\sup \{ {\|{y-x^\ast }\|} \ | \ {y\in S}\}\leq K$ . Using our notation ${a_n:=\|{x_n-x^\ast }\|}$ , for any $n\in \mathbb {N}$ and $b>0$ , if $b\leq a_n$ then

$$ \begin{align*} a_{n+1}\leq (1-\xi(b))a_n+\epsilon_n \end{align*} $$

for

$$ \begin{align*} \xi(b):=\frac{1}{4}\min\left\{\frac{b}{K}\cdot\omega\left(\frac{b}{2K}\right),1\right\}. \end{align*} $$

Proof Fix $n\in \mathbb {N}$ , $b,\varepsilon>0,$ and $\lambda \in (0,1)$ . Suppose that $b\leq a_n$ . If

(*) $$ \begin{align} \frac{\lambda K}{(1-\lambda)b}\leq \omega(\varepsilon b) \end{align} $$

then analogously to the proof of Lemma 2.5, by Lemma 4.3 we have

$$ \begin{align*} &\|{(x_n-x^\ast)+\lambda (y-x^\ast)/(1-\lambda)}\|\\ &\quad \leq \|{x_n-x^\ast}\|\left(1+\frac{\varepsilon b \lambda\|{y-x^\ast}\|}{(1-\lambda)\|{x_n-x^\ast}\|}\right)+\lambda F_n\left(\frac{y-x^\ast}{1-\lambda}\right) \end{align*} $$

for any $y\in S$ and thus

$$ \begin{align*} \|{(1-\lambda)x_n+\lambda y-x^\ast}\|&\leq (1-\lambda)a_n\left(1+\frac{\varepsilon\lambda K}{1-\lambda}\right)+\lambda F_n(y-x^\ast)\\ &\leq (1-\lambda(1-\varepsilon K))a_n+\lambda F_n(y-x^\ast), \end{align*} $$

where as before we write $F_n:=F_{x_n-x^\ast }$ . Now, if we define

$$\begin{align*}\lambda_\varepsilon:=\frac{1}{2}\min\left\{\frac{b\cdot\omega(\varepsilon b)}{K},1\right\} \end{align*}$$

then (*) holds for $\lambda :=\lambda _\varepsilon $ , and so using again the fact that

$$\begin{align*}\inf\{F_n(y-x^\ast)\mid y\in S\}\leq 0 \end{align*}$$

we have

$$ \begin{align*} &\inf\{\|{(1-\lambda)x_n+\lambda y-x^\ast}\|\mid y\in S, \lambda\in [0,1)\}\\ &\quad\leq \inf\{(1-\lambda_\varepsilon(1-\varepsilon K))a_n+\lambda_\varepsilon F_n(y-x^\ast)\mid y\in S\}\\ &\quad\leq(1-\lambda_\varepsilon(1-\varepsilon K))a_n \end{align*} $$

and so finally, setting $\varepsilon :=1/2K$ we have

$$\begin{align*}a_{n+1}\leq (1-\lambda_\varepsilon/2)a_n+\epsilon_n=(1-\xi(b))a_n+\epsilon_n \end{align*}$$

for $\xi (b)$ defined as in the statement of the lemma.

We can now simply combine Lemma 4.3 with Lemmas 4.1 and 4.2 to get our main result—a computational interpretation (and qualitative strengthening) of Theorem 2.6.

5.1 Rates of convergence for fixed step sizes

A natural question to ask once we have established convergence of greedy approximation schemes for optimal step sizes is whether we can establish an analogous result when the step sizes are fixed in advance. A number of results of this kind, for $\lambda _n:=1/(n+1)$ , are provided in [Reference Darken, Donahue, Gurvits and Sontag6] in the special case that X has modulus of smoothness of the form $\rho (u)\leq \gamma u^t$ for $t>1$ , though no such result is given in the general case, the authors simply noting in reference to the proof analysed in this article that “The stepwise selection of $\lambda =\lambda _n$ in the above proof apparently depends upon the modulus of smoothness $\rho (u)$ ”. We now provide a general convergence theorem for fixed step sizes, in terms of the alternative modulus of smoothness $\omega $ , exploiting our quantitative analysis. We label a sequence $(x_n) (\lambda _n)\text {-}(\varepsilon _n)$ -greedy with respect to $x^\ast $ and S if for all $n\in \mathbb {N}$ :

$$\begin{align*}\|{x_{n+1}-x^\ast}\|\leq \inf\left\{\|{(1-\lambda_n)x_n+\lambda_n y-x^\ast}\|\mid y\in S\right\}+\epsilon_n, \end{align*}$$

where now $(\lambda _n)\subset [0,1)$ is some fixed sequence. We also assume that

$$\begin{align*}\inf\left\{\|{(1-\lambda_n)x_n+\lambda_n y-x^\ast}\|\mid y\in S\right\}\leq \|{x_n-x^\ast}\| \end{align*}$$

for all $n\in \mathbb {N,}$ i.e., step sizes in $[0,1)$ are meaningful in that they do not worsen our estimate. By a simple adaptation of the results of the previous section we obtain the following.

Theorem 5.1. Let $(X,\omega )$ be a uniformly smooth Banach space. Suppose that $S\subseteq X$ and $x^\ast \in \mathrm {co}({S})$ , that $(x_n)$ is $(\lambda _n)\text {-}(\varepsilon _n)$ -greedy with respect to $x^\ast $ and S for some fixed step sizes $(\lambda _n)\subset [0,1)$ , $K>0$ is such that ${\sup \{\|{y-x^\ast }\|\mid y\in S\}\leq K}$ , and $\sum _{i=0}^\infty \epsilon _i\leq L$ with rate f. Finally, suppose that $\xi :(0,\infty )\to (0,1)$ is such that for all $b>0,$

$$\begin{align*}2\xi(b)\leq \lambda_n\leq \frac{1}{2}\min\left\{\frac{b}{K}\cdot \omega\left(\frac{b}{2K}\right),1\right\} \end{align*}$$

for all $n\in [f(b/2);\Phi _\xi (b)]$ , where $\Phi _\xi $ is defined in terms of $\xi $ as in Lemma 4.1. Then $x_n\to x^\ast $ as $n\to \infty $ with rate $\Phi _\xi $ .

Proof Fix $n\in [f(b/2);\Phi (b)]$ and $b>0$ . Suppose that $b\leq a_n$ . Then using the upper bound on $\lambda _n$ we have

$$\begin{align*}\frac{\lambda_n K}{(1-\lambda_n)b}\leq \omega(b/2K) \end{align*}$$

and so just as in the proof of Lemma 4.4 (for $\varepsilon :=1/2K$ ) it follows that

$$\begin{align*}\|{(1-\lambda_n)x_n+\lambda_n y-x^\ast}\|\leq (1-\lambda_n/2)a_n+\lambda_n F_n(y-x^\ast) \end{align*}$$

for $a_n:=\|{x_n-x^\ast }\|$ and $F_n:=F_{x_n-x^\ast }$ as usual, and thus also

$$\begin{align*}a_{n+1}\leq (1-\lambda_n/2)a_n+\epsilon_n\leq (1-\xi(b))a_n+\epsilon_n, \end{align*}$$

where for the second inequality we now use the lower bound on $\lambda _n$ . Since we also have $a_{n+1}\leq a_n+\epsilon _n$ (for any $n\in \mathbb {N}$ ), we have established that the recurrence inequality of Lemma 4.1 is satisfied for $\xi $ (extended by $\xi (0):=0$ ) but now only for $n\in [f(b/2);\Phi (b)]$ . But a simple inspection of the proof of Lemma 4.1 reveals that we only require the full recurrence inequality to hold in that range to establish the existence of some $n\in [f(b/2);\Phi (b)]$ such that $a_n<b/2$ (though we also need $a_{n+1}\leq a_n+\epsilon _n$ for any $n\in \mathbb {N}$ ), and then $a_{n+k}<b$ for any $k\geq \Phi (b)$ follows immediately.

For example, by setting $\omega (\varepsilon ):=\varepsilon ^r$ , $\xi (b):=b^s$ , and $\lambda _n:=n^{-t}$ we can use Theorem 5.1 to obtain conditions on $r,s$ , and t such that convergence of $(x_n)$ is guaranteed with a simple rate of the form $O(n^{-k})$ , a result similar in spirit to, e.g., [Reference Darken, Donahue, Gurvits and Sontag6, Theorem 3.5], but now for more general step-sizes. However, we stress that Theorem 5.1 is not proposed as an improvement to the more refined convergence rates of [Reference Darken, Donahue, Gurvits and Sontag6], but rather a simple illustration of how our proof-theoretic analysis can be readily extended.

5.2 Greedy algorithms in hyperbolic spaces

A far deeper extension of our analysis would involve working over a more general class of spaces. One candidate here is represented by hyperbolic spaces, which informally speaking are metric spaces $(X,d)$ that come equipped with a function $\oplus $ and associated axioms that allow one to reason abstractly about convex combinations $(1-\lambda )x\oplus \lambda y$ of elements $x,y\in X$ (see [Reference Kohlenbach16] for a formal definition widely used in the context of proof mining, and a discussion on the relationship of this definition to various others in the literature). As such, the notion of an $(\epsilon _n)$ -greedy algorithm can be readily extended from normed spaces to this more general setting, as a sequence $(x_n)$ of elements satisfying

$$\begin{align*}d(x_{n+1},x^\ast)\leq \inf\left\{d((1-\lambda)x\oplus \lambda y,x^\ast)\mid y\in S, \lambda\in [0,1)\right\}+\epsilon_n. \end{align*}$$

However, in order to establish convergence we would need a suitable notion of uniform smoothness for hyperbolic spaces, which would in turn presumably require an appropriate notion of duality. Only recently have such generalisations of concepts from normed spaces been explored, notably via the smooth hyperbolic spaces of Pinto [Reference Pinto39] and the dual systems of Pischke [Reference Pischke41]. We therefore propose that the convergence of greedy algorithms in hyperbolic spaces represents an interesting case study that could build on this recent work, though we leave this open for now.