Is the $\operatorname{argmin}$ of a uniform strongly convex function continuous? Let $f:\mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}$ be a continuous function, and assume that $f(x,y)$ is  $\mu$-strongly convex in $x$, for some $\mu>0$ and for any $y \in \mathbb{R}^m$. Define the mapping
$$g(y) =  \operatorname{argmin}_x f(x,y)$$
Can we conclude that  $g:\mathbb{R}^m \rightarrow \mathbb{R}^n$ is continuous?
I know that continuity holds if $f$ is strictly convex in the first argument and the $\operatorname{argmin}$ is computed on a compact domain (see Is the function argmin continuous?). I am wondering if the same holds when $f$ is strongly convex but the domain is unbounded.
By the discussion in the cited post (Is the function argmin continuous?), it seems to me that it would suffice to prove/disprove that, for any bounded sequence $(y_k)_{k\in \mathbb{N}}$, the sequence $(g(y_k))_{k\in\mathbb{N}}$ is also bounded.
I believe the latter is also equivalent to a condition I found in the book [Rockafellar, Wets - Variational analysis], namely that $f$ is level-bounded in $x$ locally uniformly in $y$ (Definition 1.16): for any $\bar {y}\in \mathbb R^m$ and $\alpha \in \mathbb{R}$, there exists a neighbourhood $V \subset \mathbb{R^m}$ such that the set $\{(x,y): y \in V, f(x,y) \leq \alpha \}$ is bounded.
Note that I am assuming that strong convexity is uniform, i.e., the parameter $\mu$ is independent of $y$.
 A: This doesn't exactly answer your question but it may be good enough for your problem.
In "A First Course in Optimization Theory" by Rangarajan K. Sundaram. Theorem 9.17 states:
Suppose that $f$ is a continuous function on $S\times \Theta$ and $D$ is a compact-valued continuous correspondence on $\Theta$. Let
\begin{align}
f^{*}(\theta) &= \max \{f(x,\theta)\ |\ x\in D(\theta)\} \\
D^{*}(\theta) &= \arg \max \{f(x,\theta)\ |\ x\in D(\theta)\} = \{x\in D(\theta)\ |\ f(x,\theta) = f^{*}(\theta)\}
\end{align}
Then if $f(\cdot, \theta)$ is strictly concave in $x$ for each $\theta$, and $D$ is a convex-valued (i.e., $D(\theta)$ is a convex set for each $\theta$) then $D^{*}$ is a continuous function.
You could try and make $D(\theta) = \{ x \ |\ f(x,\theta)\leq q(\theta) \}$ the level sets of $f(x,\theta)$ parameterized by $\theta$. Since in your problem $f(x,y)$ is strongly convex in $x$, your know that the level sets are compact. You just find what sould be $q(\theta)$ and checking that $D$ is a continuous correspondence.
A: I found a proof. Here it is
Let $f:\mathbb R^n\times \mathbb R^m \mathbb \rightarrow \mathbb R : (x,y)\mapsto f(x,y)$ be a  continuous function, and assume that $f(\cdot,y)$ is $\mu$-strongly convex for any $y \in \mathbb R^m$, $\mu >0$. Let $X\subseteq \mathbb R^n$ be a convex closed set. Then the (single valued, full domain) mapping
$y \mapsto g(y) = \operatorname{argmin}_{x\in X} f(x,y)$
is continuous.
Proof:
We show that, for any given sequence $(y_k)_{k \in \mathbb N}$ with $y^k \rightarrow y^\star$ (converging, hence bounded), $x^k := g(y^k) \rightarrow  g(y^\star)=: x^\star$, which is the definition of continuity.
First, we show that $(x^k)_{k \in \mathbb N}$ is bounded. Let $Y$ be a compact set containing $(y^k)_{k\in \mathbb N}$. Let $x_0\in X$ and
\begin{align*}
l_0  := \max_{y\in Y} \  f(x_0, y), \qquad 
l_1  := \min_{x\in \partial B(x_0, 1), y\in Y} \ f(x,y) 
\end{align*}
where $\partial B( x_0,1) = \{x \in \mathbb R^n \mid \|x-x_0\| = 1 \}$ is the boundary of the unit ball centered at $x_0$; the $\min$ and $\max$ are achieved because the domains are compact. Let $d \in \mathbb R^n$ be any unitary vector, i.e., $\|d\| =1$; $x_1 := x_0 +d \in \partial B (x_0,1)$; $x_2 = x_0 + Md$, for some scalar such that
\begin{align}
   M>1, \qquad M > 2 \textstyle \frac{l_0 - l_1}{\mu}+1.
\end{align} Then,
$ x_1 = \frac{M-1}{M}x_0 + \frac{1}{M} x_2. $
By definition of strong convexity, we have, for all $y \in Y$
\begin{align*}
l_1 & \leq f(x_1,y) 
\\
& \leq \textstyle   \frac{M-1}{M} f(x_0,y)+ \frac{1}{M}f(x_2,y) - \frac{1}{2}\mu \frac{M-1}{M} \frac{1}{M}\|x_0 - x_2\|^2
\\
& = \textstyle   \frac{M-1}{M} f(x_0,y)+ \frac{1}{M}f(x_2,y) - \frac{1}{2}\mu (M-1).
\end{align*}
Assume for contradiction that there exists $y \in Y$ such that $f(x_2,y) \leq f(x_0,y)$. Then, since $f(x_0,y) \leq l_0 $, the previous inequality implies
\begin{align*}
l_1 - l_0 \leq - \frac{1}{2}\mu(M-1),
\end{align*} which contradicts the assumption on $M$. Since $d$ is arbitrary, we conclude that, for any $y \in Y$, for all $x$ such that $\| x_0 -x \| >M $, $f(x_0,y) < f(x,y)$. In turn,  for all $y\in Y$, $\| g(y) \| < \| x_0\|+M$, i.e., $g$ is uniformly bounded over $Y$; thus $(x_k)_{k \in \mathbb N}$ is bounded. Hence $(x_k)_{k \in \mathbb N}$ admits an accumulation point, say  $x'$. Let $\bar{K} = (\bar{k}_1,\bar{k}_2,\dots)\subseteq \mathbb N$ be a diverging subsequence such that $x^{\bar k_n} \rightarrow x'$. Since $f(x^{\bar k_n},y^{\bar k _n} ) \leq f(x,y^{\bar k_n}) $ for all $x\in X$, by continuity of $f$, we have $f(x',y^\star) \leq f( x, y^\star)$ for all $x\in X$. Since the minimizer must be unique by strong convexity, we have $x' = x^\star$. In particular, this shows that $x^\star$ is the unique accumulation point of $x^k$: therefore, $x^k \rightarrow x^\star$.
