Let $f:[0, 1]×[0, 1] \rightarrow \Bbb{R}$ be continuous and assume that for all $x \in [0, 1]$ there is a unique $y_x$ such that $f(x,y_x)$ = $\max\{f(x,y):y∈[0, 1]\}$. If $g(x) =y_x$, then $g:[0, 1] \rightarrow [0, 1]$ is continuous.
Is this true? I believe compactness would play a role.
Any hints would be appreciated.
On
Disclaimer: I started writing this before I noticed the answer by copper.hat. Both answer's are saying pretty much the same thing.
Assumption: $f:X\times Y\rightarrow\mathbb{R}$ is continuous and $X, Y$ are compact.
Let $x_{n}$ be a sequence which converges to some $\hat{x}$ and $y_{n}\equiv g(x_{n})$. Let $y_{n_{k}}$ be a subsequence which converges to some $\hat{y}$. Then, $$ \max_{y}f(\hat{x},y)=\max_{y}\lim_{k}f(x_{n_{k}},y)=\lim_{k}\max_{y}f(x_{n_{k}},y)=\lim_{k}f(x_{n_{k}},y_{n_{k}})=f(\hat{x},\hat{y}). $$ In the above, the first and last equalities use the continuity of $f$. The second equality uses the fact that the function defined by $h(x)\equiv\max_{y}f(x,y)$ is continuous.
We have established that $\hat{y}$ is a maximizer of $y\mapsto f(\hat{x},y)$. But since maximizers are unique, it follows that $g(\hat{x})=\hat{y}$. Using the fact that if every convergent subsequence of a bounded sequence converges to the same point then the original sequence converges to that point also, we have $g(x_{n})=y_{n}\rightarrow\hat{y}=g(\hat{x})$, and the claim is proved.
Let $\psi(x) = \sup_{t \in [0,1]} f(x,t)$, it is not hard to show that $\psi$ is continuous.
Suppose $x_n \to x$, then $\psi(x_n) \to \psi(x)$. Let $g_n$ be such that $\psi(x_n) = f(x_n,g_n)$.
Any subsequence of $g_n$ has an accumulation point $g$ and we must have $\psi(x) = f(x,g)$ by continuity.
By uniqueness, we see that, in fact, $g_n \to g$.
This shows that $g(x_n) \to g(x)$.