I have come up with the following problem, which I am sure has a simple solution, but I have not been able to find any until now, nor did I found any reference in literature:
Let $f:\mathbb{R}\times K\rightarrow \mathbb{R}$ be a continuous function, with $K$ a compact space. For every $t\in\mathbb{R}$, define $f_t:K\rightarrow\mathbb{R}$ as $f_t(x)=f(t,x)$. Is it true that $\mathrm{min}_{x\in K}f_t(x)$ is continuous as a function $\mathbb{R}\rightarrow\mathbb{R}$ of the $t$ variable?
Is the hypothesis on $K$ sufficient, or are more restrictive ones necessary, e.g. requiring that $K$ be a compact metric space?
I was thinking about using the uniform continuity of the $f_t$, or using some finite cover of $K$ by sufficiently small open balls (in which case it is necessary to assume that K is a metric space), or maybe even a proof by contradiction, but there simply does not seem to be any reasonable way to control the "closeness" of the minimum points of $f_t$ for different values of $t$ (especially because they might vary in cardinality).
Any help would be much appreciated, thank you!
Best Answer
Since $K$ is compact, for all $t\in\mathbb R$ there exists $x_t\in K$ such that $f(t,x_t)=\min_{x\in K}f(t,x)$. Fix $t\in\mathbb R$ and $\varepsilon>0$. For all $x\in K$, by continuity of $f\colon\mathbb R\times K\to\mathbb R$ at $(t,x)$ (and the definition of the product topology) there is an open neighborhood $U_x$ of $t$ in $\mathbb R$ and an open neighborhood $V_x$ of $x$ in $K$ such that $$\forall (s,y)\in U_x\times V_x,\quad|f(s,y)-f(t,x)|\le\varepsilon.$$ As $\bigcup_{x\in K}U_x\times V_x$ is an open cover of the compact $\{t\}\times K$, there exist $x_1,\ldots,x_n\in K$ such that $\{t\}\times K\subseteq\bigcup_{i=1}^nU_{x_i}\times V_{x_i}$.
Now $U:=\bigcap_{i=1}^nU_{x_i}$ is an open neighborhood of $t$ (by finite intersection). Let $s\in U$. For each $x\in K$, by choosing $i\in\{1,\ldots,n\}$ such that $x\in V_{x_i}$, we have $|f(s,x)-f(t,x)|\le\varepsilon$ (because $(s,x)\in U_{x_i}\times V_{x_i}$). Hence $$\forall(s,x)\in U\times K,\quad|f(s,x)-f(t,x)|\le\varepsilon.\tag{$\star$}$$ In particular, for $s\in U$, we have $(s,x_t)\in U\times K$ so $|f(s,x_t)-f(t,x_t)|\le\varepsilon$. Then $$f(t,x_t)\ge f(s,x_t)-\varepsilon\ge f(s,x_s)-\varepsilon$$ and $$f(s,x_s)\le f(s,x_t)\le f(t,x_t)+\varepsilon,$$ thus showing that $|f(t,x_t)-f(s,x_s)|\le\varepsilon$ for all $s\in U$ neighborhood of $t$.
(Useless note. $(\star)$ also shows that $\sup_{x\in K}|f(t,x)-f(s,x)|\xrightarrow[s\to t]{}0$.)