# Is the $\arg\min$ of a strictly convex function continuous?

Let $$X\subset \mathbb{R}^n$$ and $$Y\subset \mathbb{R}^m$$ be compact and convex sets, and let $$f:X\times Y\rightarrow \mathbb{R}$$ be a continuous function. Suppose that for each $$y$$, $$f(x,y)$$ is strictly convex.

Define the function $$g : Y \to X$$ as follows:

$$g(y) = \arg\min_x f(x,y)$$

Is $$g$$ continuous? If not, are there additional restrictions that we can impose on $$f$$ such that it is? Thanks for any help!

(Two similar questions have been asked but in "Is the function argmin continuous?" there is no assumption of convexity and in the answer to "the continuity of argmin on convex function" it is assumed that $$f$$ is continuously differentiable.)

• For each $y$, $f(x,y)$ is strictly convex so there should be a unique minimizer. Commented Jul 7, 2020 at 1:02
• What if the minimum does not exist ($f(x,y) = e^x$)? Should you also assume $Y$ is convex (a discrete set is compact)? Commented Jul 7, 2020 at 1:06
• Oops that's right. I added convexity as an assumption. Commented Jul 7, 2020 at 1:06

Since $$x \mapsto f(x,y)$$ is strictly convex, there is a unique minimiser, call it $$g(y)$$.

Suppose $$y_k \to y^*$$, and let $$x_k = g(y_k)$$ and $$x^* = g(y^*)$$.

Choose any subsequence $$I \subset \mathbb{N}$$, then $$x_k$$ has an accumulation point $$x'$$. Since $$f(x_k,y_k) \le f(x,y_k)$$ for all $$x \in X$$ and $$k \in I$$ we see that $$f(x',y^*) \le f(x,y^*)$$ for all $$x \in X$$ and since the minimiser is unique, we have $$x'=x^*$$.

In particular, this shows that $$x_k \to x^*$$ and so $$g$$ is continuous.

The proof just relies on $$X$$ being compact and $$x \to f(x,y)$$ having a unique minimiser for each $$y$$.

• Why does $x_k$ have an accumulation point? Commented Jul 7, 2020 at 1:39
• Because $X$ is compact. Otherwise the minimisers could disappear off to infinity :-). Commented Jul 7, 2020 at 2:21
• Sorry if this is a stupid question but how do we know that for any subsequence $x_k$ converges? Commented Jul 7, 2020 at 2:38
• Any subsequence of a compact set has a convergent subsequence. Commented Jul 7, 2020 at 3:09
• @copper.hat Small comment regarding your last point: $f$ also needs to be continuous right? Commented Jul 11, 2020 at 13:50