Let $f:[0,1]\times [0,\infty)\rightarrow R$ be a two-dimensional, real-valued function. Also assume the following
- $f$ is twice continuously differentiable
- For every $t\in[0,1]$, $f(t,x)$ attains its unique maximum at some $x^*(t)$
- $f(t,x)\geq f(t,x')\Rightarrow f(t',x)> f(t',x')$ for $t'>t$ and $x>x'$.
I can show that $x^*(t)$ is non-decreasing. Is $x^*(t)$ also continuous?
Yes. Since $x^*(t)\leq x^*(1)$ for all $t$, we can assume that the domain of $f$ is $[0,1]\times [0,x^*(1)]$. Suppose now that $t\mapsto x^*(t)$ is not continuous. Then, there exists a sequence $\langle t_n\rangle$ converging to $t$ such that $\langle x^*(t_n)\rangle$ does not converge to $x^*(t)$. Since $[0,x^*(1)]$ is compact, we can assume by passing to a subsequence that $\langle x^*(t_n)\rangle$ converges to some $x$. Since $f$ is continuous, $\langle f(t_n,x^*(t_n))\rangle$ converges to $f(t,x)$. By assumption $f(t,x)<f(t,x^*(t))$. But using again the continuity of $f$, this implies $f(t_n,x^*(t))>f(t_n,x^*(t_n))$ for $n$ large enough, which is impossible.
Alternatively, one can use the maximum theorem and the fact that a single-valued upper hemicontinuous correspondence is essentially a continuous function.