I'm unsure how to formally prove or disprove the following claim. It came up in trying to prove convergence of a gradient descent-style algorithm. I've tried using different definitions of continuity including the topological and epsilon-delta ones, but haven't quite got it.
Assume $f : \mathbb{R}^d \to \mathbb{R}^d$ is continuous, $U \subset \mathbb{R}^d$ is compact and define $$ U_a=\{x+atf(x)∣t∈[0,1],x∈U\}$$ for any $a \in \mathbb{R}^{\geq0}$. Notice that $U_a$ is just a compact expansion of $U$ with $U_a \subset U_{a'}$ for all $a < a'$. Is the function $L(a)=\sup_{x∈U_a}g(x)$ continuous for any continuous function $g : \mathbb{R}^d \to \mathbb{R}$?
The answer is yes.
Let $g: \mathbb R^d \to \mathbb R$ be a continuous function and $a \in [0,+\infty[$, $\epsilon >0$.
Since $U$ is compact (remark : $U$ is usually used to denote open sets rather than compacts), there is $M>0$ such that $\|f(x) \| \leq M$ for any $x\in U$. Since $U_{a+1}$ is compact, $g$ is uniformly continuous on it. Let $\eta$ be such that for all $x,y\in U_{a+1}$, $\|x-y\|\leq \eta$ implies $\|g(x) -g(y)\|\leq \epsilon$.
Then, if $x \in U_{a + \eta /M}$, there is $y\in U_{a}$ such that $\|x -y\| \leq \eta$. Therefore : $$g(x) \leq g(y) + \epsilon \leq L(a) +\epsilon$$ and by taking the supremum of $x \in U_{a+\eta/M}$ you get : $$L(a) \leq L(a+\eta/M)\leq L(a) +\epsilon $$
The same argument with $a$ replaced by $a-\eta/M$ shows that : $$L(a) - \epsilon \leq L(a-\eta/M) \le L(a)$$
Since $L$ is increasing, this shows that $L$ is continuous.