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 $y$.
Best Answer
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, $x^k \rightarrow x^\star$.