0
$\begingroup$

Suppose that we have a function $g$ defined as $$ g(x,\alpha) = \lim_{\alpha\to +\infty} \{f(k - \alpha x) - f(-\alpha x)\} :=\lim_{\alpha\to +\infty} F_\alpha(x) $$ with $f$ being a convex function and $k$ a constant. The goal is to find $\frac{\partial g(x,\alpha)}{\partial x}$ (maybe an approximate one w.r.t $f'$ or $\nabla f$). Can we argue something about uniform convergence using convexity so that \begin{equation} \frac{\partial}{\partial x} \lim_{\alpha\to \infty} F_\alpha(x) = \lim_{\alpha\to\infty} \frac{\partial F_\alpha(x)}{\partial x} \tag{1} \end{equation} Any alternative approaches not using equation (1) are also welcome. The original problem that I have is with multivariables (i.e., $f:\mathbb{R}^n \to \mathbb{R}$) but I want to start with the simple case.

$\endgroup$
8
  • 2
    $\begingroup$ Do you mean $g(x)$ rather than $g(x, \alpha)$? $\endgroup$
    – River Li
    Jul 13 at 14:13
  • $\begingroup$ @RiverLi So we have $g$ as a function of two variables $x$ and $\alpha$. $\endgroup$
    – aaka
    Jul 14 at 8:59
  • 1
    $\begingroup$ After taking limit, $\alpha$ disappears? $\endgroup$
    – River Li
    Jul 14 at 10:13
  • $\begingroup$ So, $g(x) = \lim_{\alpha\to +\infty} F_\alpha(x)$. You want $g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} \lim_{\alpha\to +\infty} F_\alpha(x) = \lim_{\alpha\to +\infty} \frac{\partial }{\partial x} F_\alpha(x) $. $\endgroup$
    – River Li
    Jul 14 at 11:05
  • $\begingroup$ @RiverLi Can't tell much about the limiting behaviour of $\alpha$ (not even sure whether or not it will disappear) but we do want to deal with equations that you wrote down. I defined $F_\alpha(x)$ because the function itself will depend on each value of $\alpha$. The problem comes from the convex analysis literature that I'm working on, so sorry if it caused any confusion. Any help would be appreciated. $\endgroup$
    – aaka
    Jul 14 at 11:39
0
+100
$\begingroup$

First, if the equation (1) is well-defined. $g(x,\alpha)$ should be $g(x)$, which is independent of $\alpha$.

Second, to exchange derivative and limit: $$ \frac{\partial}{\partial_x}\lim_{\alpha\rightarrow\infty}F_\alpha(x)=\lim_{\alpha\rightarrow\infty}\frac{\partial}{\partial_x}F_{\alpha}(x), $$ To make this exchange work, a common conclusion is that if $F_\alpha(x)$ and $\frac{\partial}{\partial_x}F_{\alpha}(x)$ converge locally uniformly convergence in $x$ when $\alpha\rightarrow\infty$, then we can do the exchange. In general, I don't see how to make use of convexity here.

For your question from the literature (in your comments), assume $k\neq \vec{0}$. when S is a sphere ($|x|<1$), we can see $$ F_\alpha(x)=|k-\alpha x|-|\alpha x|\,, $$ which implies $$\tag{1} F_{\infty}(x)=\lim_{\alpha\rightarrow\infty}F_{\alpha}(x)=\lim_{\alpha\rightarrow\infty}\frac{|k-\alpha x|^2-|\alpha x|^2}{|k-\alpha x|+|\alpha x|}=\lim_{\alpha\rightarrow\infty}\frac{|k|^2-2\alpha k\cdot x}{|k-\alpha x|+|\alpha x|}=\lim_{\alpha\rightarrow\infty}\frac{\frac{|k|^2}{\alpha}-2k\cdot x}{\frac{|k-\alpha x|+|\alpha x|}{\alpha}}=-k\cdot \frac{x}{|x|}\,. $$ Therefore, it's obvious that $F_{\infty}(x)$ is differentiable at $x\neq \vec{0}$ and $$ \frac{\partial}{\partial_x}F_{\infty}(x)=-\frac{k}{|x|}-\frac{(k\cdot x)x}{|x|^3},\quad x\neq \vec{0} $$ Then we go back to $F_{\alpha}(x)$. Fisrt, $F_{\alpha}(x)$ is differential at $x\neq \vec{0}$ and $x\neq \frac{k}{\alpha}$. This implies when $x\neq \vec{0}$, when $\alpha$ is large enough, $F_{\alpha}(x)$ is always differentiable. Furthermore, $$ \frac{\partial}{\partial_x} F_{\alpha}(x)=\frac{-\alpha(k-\alpha x)}{|k-\alpha x|}-\frac{|\alpha| x}{|x|} $$ It's easy to verify that $$\tag{2} \lim_{\alpha\rightarrow\infty}\frac{\partial}{\partial_x} F_{\alpha}(x)=\frac{-\alpha(k-\alpha x)}{|k-\alpha x|}-\frac{|\alpha| x}{|x|}=-\frac{k}{|x|}-\frac{(k\cdot x)x}{|x|^3},\quad x\neq \vec{0}\,. $$ Therefore, the derivative and limit can exchange in this case.

In this case, it's easy to check that the convergence in (1) and (2) is locally uniformly when $x\neq 0$.

For general convex set $S$, we can prove $$ \lim_{\alpha\rightarrow\infty}F_\alpha(x)=\max_{w\in S_x}w^\top k\,, $$ where $S_x=\mathrm{argmax}_{w\in S}-w^\top x$. But this convergence might not be locally uniformly at some particular point even $\lim_{\alpha\rightarrow\infty}F_\alpha(x_0)$ exists. For example, consider $x\in\mathbb{R}^2$, $S=[-1,1]\times \mathbb{R}$, $k=(1/2,1/2)$, then we can see that $$ \lim_{\alpha\rightarrow\infty}F_\alpha(x)=\left\{ \begin{aligned} &-\frac{1}{2},\quad x_{[1]}>0 \\ &\frac{1}{2},\quad x_{[1]}\leq 0 \end{aligned} \right. $$ where $x_{[1]}$ is the first component of $x$. Because $F_\alpha(x)$ is continuous, the convergence is not locally uniformly around $x_0=(0,x_{[2]})$ for any $x_{[2]}\in\mathbb{R}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.