# Constructing an interpolated function for the class of smooth strongly convex functions

Shuvomoy Das Gupta

December 1, 2021

In this blog, we study constructing an interpolated smooth and strongly convex function from a set of points. The problem is studied by Yoel Drori and Adrien Taylor from[4].

## Notation and notions.

All norms are 2-norm in this blog. A function $f:\mathbf{R}^{d}\to\mathbf{R}$ is $L$-smooth convex if and only

$\left(\forall x,y\in\mathbf{R}^{d}\right)\quad f(y)\geq f(x)+\langle\nabla f(x)\mid y-x\rangle+\frac{1}{2L}\|\nabla f(x)-\nabla f(y)\|^{2}.$

On the other hand, a function is $\mu$​-strongly convex if and only if

$\left(\forall x,y\in\mathbf{R}^{d}\right)\quad f(y)\geq f(x)+\langle\nabla f(x)\mid y-x\rangle+\frac{\mu}{2}\|x-y\|^{2}\qquad (\text{SCVX})$

​ where $f^{\prime}(\cdot)$​ denotes a subgradient of $f$​ at $(\cdot)$​.

On $\mathbf{R}^{d},$ the set of all $L$-smooth convex functions is denoted by $\mathcal{F}_{0,L}(\mathbf{R}^{d})$, the set of all $\mu$-strongly convex functions is denoted by $\mathcal{F}_{\mu,\infty}(\mathbf{R}^{d})$, and the set of all $\mu$-strongly convex and $L$-smooth functions is denoted by $\mathcal{F}_{\mu,L}(\mathbf{R}^{d})$. Finally the set of all lower-semicontinuous, proper, and convex functions is denoted by $\mathcal{F}_{0,\infty}(\mathbf{R}^{d})$.

## Helper results.

### Alternative characterization of smooth convex functions

An alternative characterization of $f\in\mathcal{F}_{0,L}(\mathbf{R}^{d})$, that we will use multiple times, is as follows.

## Theorem 1: Alternative characterization of smooth convex functions[1, Definition 2.6, Theorem 2.27, Theorem 2.28] Theorem

For a function $f:\mathbf{R}^{d}\to\mathbf{R}$​ the following statements are equivalent:

(a) $f\in\mathcal{F}_{0,L}(\mathbf{R}^{d})$​.

(b)$f$​ is convex and it satisfies $\left(\forall x,y\in\mathbf{R}^{d}\right)\quad f(y)\leq f(x)+\langle\nabla f(x)\mid y-x\rangle+\frac{L}{2}\|x-y\|^{2}.$

(c) $f$ is convex and satisfies $\left(\forall x,y\in\mathbf{R}^{d}\right)\quad\|\nabla f(x)-\nabla f(y)\|\leq L\|x-y\|$.

(d) $\frac{L}{2}\|\cdot\|^{2}-f\in\mathcal{F}_{0,\infty}(\mathbf{R}^{d})$​​.

In $(b)$ and $(c)$ of the theorem above, saying that $f$ is convex is necessary, as the inequalities only in $(b),(c)$ are not sufficient to establish $L$-smooth convexity.

We now record the notion of an interpolable function.

## Definition 1: Interpolable function. Definition

Suppose we are given the set of triplets $\{(x_{i},g_{i},f_{i})\}_{i\in I}\subseteq\mathbf{R}^{d}\times\mathbf{R}^{d}\times\mathbf{R}$ where $I$ is a finite index set. Let $\mathcal{F}(\mathbf{R}^d)$ be some class of functions. Then the set $\{(x_{i},g_{i},f_{i} ) \}_{i\in I}$ is $\mathcal{F}(\mathbf{R}^{d})$-interpolable if and only if there exists a function $f\in\mathcal{F}(\mathbf{R}^{d})$ such that for all $i\in I$ we have $f_{i}=f(x_{i})$ and $g_{i} \in \partial f(x_{i})$.

### Equivalence between smooth strongly convex and smooth convex function class

To establish our main result, we will use the following equivalence result.

## Lemma 1: Equivalence between function classes Lemma

We have $f\in\mathcal{F}_{\mu,L}(\mathbf{R}^{d})\Leftrightarrow\frac{L}{2}\|\cdot\|^{2}-f\in\mathcal{F}_{0,L-\mu}(\mathbf{R}^{d})$.

Proof.

First, we prove $f\in\mathcal{F}_{\mu,L}(\mathbf{R}^{d})\Rightarrow\frac{L}{2}\|\cdot\|^{2}-f\in\mathcal{F}_{0,L-\mu}(\mathbf{R}^{d})$​​. Define $h\coloneqq\frac{L}{2}\|\cdot\|^{2}-f$​​, so $f=\frac{L}{2}\|\cdot\|^{2}-h$​​ and $\nabla f(x)=Lx-\nabla h(x)$​​. Clearly, $f\in\mathcal{F}_{0,L}(\mathbf{R}^{d})$​​ also, so due to Theorem 1$(a),(d),$​​we have $h\in\mathcal{F}_{0,\infty}(\mathbf{R}^{d})$​​. Because $f$​​ is $\mu$​​-strongly convex and differentiable, for any $x,y\in\mathbf{R}^{d}$​​ we have from $(\text{SCVX})$:

\begin{aligned} & f(y)\geq f(x)+\left\langle \nabla f(x)\mid y-x\right\rangle +\frac{\mu}{2}\|x-y\|^{2}\\ \Leftrightarrow & \frac{L}{2}\|y\|^{2}-h(y)\geq\frac{L}{2}\|x\|^{2}-h(x)+\left\langle Lx-\nabla h(x)\mid y-x\right\rangle +\frac{\mu}{2}\|x-y\|^{2}\\ \Leftrightarrow & -\frac{L}{2}\|y\|^{2}+h(y)\leq-\frac{L}{2}\|x\|^{2}+h(x)-\left\langle Lx-\nabla h(x)\mid y-x\right\rangle -\frac{\mu}{2}\|x-y\|^{2}\\ \Leftrightarrow & h(y)\leq\frac{L}{2}\|y\|^{2}-\frac{L}{2}\|x\|^{2}+h(x)-L\left\langle x\mid y-x\right\rangle +\left\langle \nabla h(x)\mid y-x\right\rangle -\frac{\mu}{2}\|x-y\|^{2}\\ & \quad\quad\quad=\frac{L}{2}\|y\|^{2}-\frac{L}{2}\|x\|^{2}+h(x)-L\left\langle x\mid y\right\rangle +L\|x\|^{2}+\left\langle \nabla h(x)\mid y-x\right\rangle -\frac{\mu}{2}\|x-y\|^{2}\\ & \quad\quad\quad=\frac{L}{2}\left(\|y\|^{2}+\|x\|^{2}-2\left\langle x\mid y\right\rangle \right)+h(x)+\left\langle \nabla h(x)\mid y-x\right\rangle -\frac{\mu}{2}\|x-y\|^{2}\\ & \quad\quad\quad=h(x)+\left\langle \nabla h(x)\mid y-x\right\rangle +\frac{L}{2}\|x-y\|^{2}-\frac{\mu}{2}\|x-y\|^{2}\\ & \quad\quad\quad=h(x)+\left\langle \nabla h(x)\mid y-x\right\rangle +\frac{L-\mu}{2}\|x-y\|^{2}.\end{aligned}

So we have proven that $h$ is convex and for any $x,y\in\mathbf{R}^{d},$

$h(y)\leq h(x)+\left\langle \nabla h(x)\mid y-x\right\rangle +\frac{L-\mu}{2}\|x-y\|^{2}$

which this is equivalent to saying that $h\in\mathcal{F}_{0,L-\mu}(\mathbf{R}^{d})$ due to Theorem 1$(a),(b)$.

Next, we prove $h\coloneqq\frac{L}{2}\|\cdot\|^{2}-f\in\mathcal{F}_{0,L-\mu}(\mathbf{R}^{d})\Rightarrow f\in\mathcal{F}_{\mu,L}(\mathbf{R}^{d})$. Due to Theorem 1$(a),(b)$, we have: $h$ convex and for any $x,y\in\mathbf{R}^{d},$

\begin{aligned} & h(y)\leq h(x)+\left\langle \nabla h(x)\mid y-x\right\rangle +\frac{L-\mu}{2}\|x-y\|^{2}\\ \Leftrightarrow & \frac{L}{2}\|y\|^{2}-f(y)\leq\frac{L}{2}\|x\|^{2}-f(x)+\left\langle Lx-\nabla f(x)\mid y-x\right\rangle +\frac{L-\mu}{2}\|x-y\|^{2}\\ \Leftrightarrow & -\frac{L}{2}\|y\|^{2}+f(y)\geq-\frac{L}{2}\|x\|^{2}+f(x)-\left\langle Lx-\nabla f(x)\mid y-x\right\rangle -\frac{L-\mu}{2}\|x-y\|^{2}\\ \Leftrightarrow & f(y)\geq f(x)+\left\langle \nabla f(x)\mid y-x\right\rangle +\frac{L}{2}\|y\|^{2}-\frac{L}{2}\|x\|^{2}-L\left\langle x\mid y\right\rangle +L\|x\|^{2}-\frac{L-\mu}{2}\|x-y\|^{2}\\ & \quad\quad\quad=f(x)+\left\langle \nabla f(x)\mid y-x\right\rangle +\frac{L}{2}\underbrace{\left(\|y\|^{2}+\|x\|^{2}-2\left\langle x\mid y\right\rangle \right)}_{=\|x-y\|^{2}}-\frac{L-\mu}{2}\|x-y\|^{2}\\ & \quad\quad\quad=f(x)+\left\langle \nabla f(x)\mid y-x\right\rangle +\frac{\mu}{2}\|x-y\|^{2},\end{aligned}

i.e., we have shown that for all $x,y\in\mathbf{R}^{d}$​​​

$f(y)\geq f(x)+\left\langle \nabla f(x)\mid y-x\right\rangle +\frac{\mu}{2}\|x-y\|^{2},$

​​​ hence by defintion $f$​​​ is $\mu$​​​-strongly convex. Finally, we have

\begin{aligned} \|\nabla f(x)-\nabla f(y)\| & =\|Lx-\nabla h(x)-Ly+\nabla h(y)\|\\ & =\|L(x-y)+(\nabla h(y)-\nabla h(x))\|\\ & \overset{a)}{\leq}L\|x-y\|+\|\nabla h(y)-\nabla h(x)\|\\ & \overset{b)}{\leq}L\|x-y\|+(L-\mu)\|x-y\|\\ & =L\|x-y\|,\end{aligned}

where $a)$ follows from triangle inequallity and $b)$ follows from Theorem 1$(a),(c)$. So, $f$ is $L$-smooth besides being $\mu$-strongly convex, i.e., $f\in\mathcal{F}_{\mu,L}(\mathbf{R}^{d})$. This completes the proof.

### Interpolation equivalence for smooth strongly convex and smooth convex functions

Also, we are going to use the following interpolation result.

## Theorem 2: Interpolation equivalence for smooth strongly convex and smooth convex functions Theorem

Suppose we are given the set of triplets set of triplets $\{(x_{i},g_{i},f_{i})\}_{i\in I}\subseteq\mathbf{R}^{d}\times\mathbf{R}^{d}\times\mathbf{R}$ where $I$​ is a finite index set. Then the following are equivalent.

(a) $\{(x_{i},g_{i},f_{i})\}_{i\in I}$​ is $\mathcal{F}_{\mu,L}(\mathbf{R}^{d})$​​-interpolable.

(b) $\{(\widetilde{x}_{i},\widetilde{g}_{i},\widetilde{f}_{i})\}_{i\in I}\coloneqq\{(x_{i},Lx_{i}-g_{i},\frac{L}{2}\|x_{i}\|^{2}-f_{i})\}_{i\in I}$​ is $\mathcal{F}_{0,L-\mu}(\mathbf{R}^{d})$​​-interpolable.

Proof.

First, we prove $(a)\Rightarrow(b).$​ If $(a)$​ holds, then by definition there exists a function $f\in\mathcal{F}_{\mu,L}(\mathbf{R}^{d})$​ that satisfies for all $i\in I$: $f(x_{i})=f_{i}$​ and $\nabla f(x_{i})=g_{i}$​. If we define, $h=(L/2)\|\cdot\|^{2}-f,$​ then $h\in\mathcal{F}_{0,L-\mu}(\mathbf{R}^{d})$​ due to Lemma 1, and for all $i\in I$​ we have $h(x_{i})=\frac{L}{2}\|x_{i}\|^{2}-f(x_{i})=\frac{L}{2}\|x_{i}\|^{2}-f_{i}$​ and $\nabla h(x_{i})=Lx_{i}-\nabla f(x_{i})=Lx_{i}-g_{i}$​. This proves $(b)$​.

Next, we prove $(b)\Rightarrow(a)$​​. If $(b)$​​ holds, there exists a function $h\in\mathcal{F}_{0,L-\mu }(\mathbf{R}^{d})$​​ that satisfies for all $i\in I$: $h(x_{i})=\frac{L}{2}\|x_{i}\|^{2}-f_{i}$​​ and $\nabla h(x_{i})=Lx_{i}-g_{i}$​​. If we define, $f=(L/2)\|\cdot\|^{2}-h$​​, then $f\in\mathcal{F}_{\mu,L}(\mathbf{R}^{d})$​​ due to Lemma 1, and for all $i\in I$​​ we have $f(x_{i})=\frac{L}{2}\|x_{i}\|^{2}-h(x_{i})=\frac{L}{2}\|x_{i}\|^{2}-(\frac{L}{2}\|x_{i}\|^{2}-f_{i})=f_{i}$​​ and $\nabla f(x_{i})=Lx_{i}-\nabla h(x_{i})=Lx_{i}-(Lx_{i}-g_{i})=g_{i}.$​​​ This proves (a).

### Drori's interpolated function for smooth convex function class

Finally, we present the following interpolation result due to Yoel Drori from[3]. I showed a detailed proof of this result in the previous blog post: here.

## Theorem 3: Interpolation of smooth convex functions. Theorem

If $\{(\widetilde{x}_{i},\widetilde{g}_{i},\widetilde{f}_{i})\}_{i\in I}\subseteq\mathbf{R}^{d}\times\mathbf{R}^{d}\times\mathbf{R}$​ is $\mathcal{F}_{0,L}(\mathbf{R}^{d})$​-interpolable with $L>0$​, then one interpolated function $\widetilde{f}\in\mathcal{F}_{0,L}(\mathbf{R}^{d})$​ that interpolates $\{(\widetilde{x}_{i},\widetilde{g}_{i},\widetilde{f}_{i})\}_{i\in I}$​ is:

$\widetilde{f}(y)=\min_{\alpha\in\Delta}\left[\frac{L}{2}\|y-\sum_{i\in I}\alpha_{i}\left(\widetilde{x}_{i}-\frac{1}{L}\widetilde{g}_{i}\right)\|^{2}+\sum_{i\in I}\alpha_{i}\left(\widetilde{f}_{i}-\frac{1}{2L}\|\widetilde{g}_{i}\|^{2}\right)\right],$

​ where $\Delta=\{\beta\in\mathbf{R}^{\vert I\vert}\mid\beta\geq0,\sum_{i\in I}\beta_{i}=1\}.$

## Main result.

Now we are in a position to state our main result. The results appeared first in the paper by Yoel Drori and Adrien Taylor from[4, Theorem 1]. The proof we present here is based on interpolation argument.

## Theorem 4: Interpolation of smooth strongly convex functions Theorem

If $\{(x_{i},g_{i},f_{i})\}_{i\in I}\subseteq\mathbf{R}^{d}\times\mathbf{R}^{d}\times\mathbf{R}$​ is $\mathcal{F}_{\mu,L}(\mathbf{R}^{d})$​-interpolable with $L>0$​, then one interpolation function $f\in\mathcal{F}_{\mu,L}(\mathbf{R}^{d})$​ that interpolates $\{(x_{i},g_{i},f_{i})\}_{i\in I}$​ is:

\begin{align*} f(y) & :=\max_{\alpha\in\Delta}\Big[\frac{L}{2}\|y\|^{2}-\frac{L-\mu}{2}\|y-\frac{1}{L-\mu}\sum_{i}\alpha_{i}(g_{i}-\mu x_{i})\|^{2}\\ & \quad +\sum_{i\in I}\alpha_{i}(f_{i}+\frac{1}{2(L-\mu)}\|g_{i}-Lx_{i}\|^{2}-\frac{L}{2}\|x_{i}\|^{2})\Big], \end{align*}

where $\Delta=\{\beta\in\mathbf{R}^{\vert I\vert}\mid\beta\geq0,\sum_{i\in I}\beta_{i}=1\}.$

Proof. The proof sketch is as follows comprising two steps.

1. First, we find an interpolated function $\widetilde{f}\in\mathcal{F}_{0,L-\mu}(\mathbf{R}^{d})$ that interpolates $\{(\widetilde{x}_{i},\widetilde{g}_{i},\widetilde{f}_{i})\}_{i\in I}\coloneqq\{(x_{i},Lx_{i}-g_{i},\frac{L}{2}\|x_{i}\|^{2}-f_{i})\}_{i\in I}$ using Theorem 3.

1. Then due to the proof of Theorem 2, the function $f=(L/2)\|\cdot\|^{2}-\widetilde{f}$ will be in $\mathcal{F}_{\mu,L}(\mathbf{R}^{d})$ and will interpolate $\{(x_{i},g_{i},f_{i})\}_{i\in I}$.

(1) From Theorem 3 recall that, if $\{(\widetilde{x}_{i},\widetilde{g}_{i},\widetilde{f}_{i})\}_{i\in I}\subseteq\mathbf{R}^{d}\times\mathbf{R}^{d}\times\mathbf{R}$​ is $\mathcal{F}_{0,L-\mu}(\mathbf{R}^{d})$​-interpolable, then one interpolation function $\widetilde{f}\in\mathcal{F}_{0,L}(\mathbf{R}^{d})$​ that interpolates $\{(\widetilde{x}_{i},\widetilde{g}_{i},\widetilde{f}_{i})\}_{i\in I}$​ is:

$\widetilde{f}(y)=\min_{\alpha\in\Delta}\left[\frac{L-\mu}{2}\|y-\sum_{i\in I}\alpha_{i}\left(\widetilde{x}_{i}-\frac{1}{L-\mu}\widetilde{g}_{i}\right)\|^{2}+\sum_{i\in I}\alpha_{i}\left(\widetilde{f}_{i}-\frac{1}{2(L-\mu)}\|\widetilde{g}_{i}\|^{2}\right)\right],$

where $\Delta=\{\beta\in\mathbf{R}^{\vert I\vert}\mid\beta\geq0,\sum_{i\in I}\beta_{i}=1\}.$​​ In our setup, we have $\{(\widetilde{x}_{i},\widetilde{g}_{i},\widetilde{f}_{i})\}_{i\in I}\coloneqq\{(x_{i},Lx_{i}-g_{i},\frac{L}{2}\|x_{i}\|^{2}-f_{i})\}_{i\in I}$​​, so lets put that in the last equation:

\begin{aligned} \widetilde{f}(y) & =\min_{\alpha\in\Delta}\left[\frac{L-\mu}{2}\|y-\sum_{i\in I}\alpha_{i}\Big(\underbrace{x_{i}-\frac{1}{L-\mu}(Lx_{i}-g_{i})}_{=x_{i}(1-\frac{L}{L-\mu})+\frac{1}{L-\mu}g_{i}=-\frac{\mu}{L-\mu}x_{i}+\frac{1}{L-\mu}g_{i}=\frac{1}{L-\mu}(g_{i}-\mu x_{i})}\Big)\|^{2}+\sum_{i\in I}\alpha_{i}\left(\frac{L}{2}\|x_{i}\|^{2}-f_{i}-\frac{1}{2(L-\mu)}\|Lx_{i}-g_{i}\|^{2}\right)\right]\\ & =\min_{\alpha\in\Delta}\left[\frac{L-\mu}{2}\|y-\frac{1}{L-\mu}\sum_{i\in I}\alpha_{i}(g_{i}-\mu x_{i})\|^{2}-\sum_{i\in I}\alpha_{i}\left(f_{i}+\frac{1}{2(L-\mu)}\|g_{i}-Lx_{i}\|^{2}-\frac{L}{2}\|x_{i}\|^{2}\right)\right].\end{aligned}

(2) Hence, $f=(L/2)\|\cdot\|^{2}-\widetilde{f}$​​​, which will be in $\mathcal{F}_{\mu,L}(\mathbf{R}^{d})$​​​ and will interpolate $\{(x_{i},g_{i},f_{i})\}_{i\in I}$​​​ has the following form:

\begin{aligned} f(y) & =(L/2)\|y\|^{2}-\widetilde{f}(y)\\ & =(L/2)\|y\|^{2}-\min_{\alpha\in\Delta}\left[\frac{L-\mu}{2}\|y-\frac{1}{L-\mu}\sum_{i\in I}\alpha_{i}(g_{i}-\mu x_{i})\|^{2}-\sum_{i\in I}\alpha_{i}\left(f_{i}+\frac{1}{2(L-\mu)}\|g_{i}-Lx_{i}\|^{2}-\frac{L}{2}\|x_{i}\|^{2}\right)\right]\\ & \overset{a)}{=}(L/2)\|y\|^{2}+\max_{\alpha\in\Delta}\left[-\frac{L-\mu}{2}\|y-\frac{1}{L-\mu}\sum_{i\in I}\alpha_{i}(g_{i}-\mu x_{i})\|^{2}+\sum_{i\in I}\alpha_{i}\left(f_{i}+\frac{1}{2(L-\mu)}\|g_{i}-Lx_{i}\|^{2}-\frac{L}{2}\|x_{i}\|^{2}\right)\right]\\ & =\max_{\alpha\in\Delta}\left[(L/2)\|y\|^{2}-\frac{L-\mu}{2}\|y-\frac{1}{L-\mu}\sum_{i\in I}\alpha_{i}(g_{i}-\mu x_{i})\|^{2}+\sum_{i\in I}\alpha_{i}\left(f_{i}+\frac{1}{2(L-\mu)}\|g_{i}-Lx_{i}\|^{2}-\frac{L}{2}\|x_{i}\|^{2}\right)\right],\end{aligned}

where $a)$​​​ uses $\min(\cdot)=-\max(-\cdot)$​​​. This completes the proof. ■

## References.

[1] Taylor, Adrien B. Convex interpolation and performance estimation of first-order methods for convex optimization. Diss. Catholic University of Louvain, Louvain-la-Neuve, Belgium, 2017.

[2] R Tyrell Rockafellar. Convex Analysis. Princeton University Press, 1996.

[3] Drori, Yoel. The exact information-based complexity of smooth convex minimization. Journal of Complexity 39 (2017): 1-16.

[4] Drori, Yoel and Taylor, Adrien, 2021. On the oracle complexity of smooth strongly convex minimization. arXiv preprint arXiv:2101.09740.