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

Question

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.)

Answer 1

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 $y$.