Consider the function defined by
$$f(t) := \min_{x\in \bar{B}(0,t)}|g(x)|$$
Where $\bar{B}(0,t) = \{ x\in \mathbb{R}^n \ | \ |x| \leq t \}$ and $g \in C^1$ and bounded. I am wondering if there is an easy way to show that $f$ is continuous. I think I can do it by breaking it into cases as to whether the minimum occurs in $\partial\bar{B}(0,t)$ or in the interior of $B(0,t)$ and then prove right and left continuity. Furthermore, I want to know if there is a faster, more clever way of doing this that will follow immediately from the continuity/differentiability of $g$.
Here it was shown for $n=1$ and this had to be broken into cases in a similar way, but now more attention must be taken when we consider the boundary of the ball.
$g$ is $C^1$ and in particular it is locally Lipschitz.
Let's show that $f$ is locally Lipschitz.
Fix $T> 0$ and let $L$ be a Lipschitz constant for $g$ on $\bar B(0,T)$. Now let $s,t\in [0,T)$. We can assume that $s>t$. Since $\bar B(0,s)$ is compact there exists a $x_\star$ such that $f(s) = |g(x_\star)|$. Now let $y_\star$ be the orthogonal projection of $x_\star$ on $\bar B(0,t)$. Clearly $|y_\star - x_\star|\leq |s-t|$ by construction.
By definition $f$ is non increasing and $f(t)\geq |g(y_\star)|$, since $y_\star \in \bar B(0,t)$. So we have $$|f(s)-f(t)| = f(t) - f(s) \leq |g(y_\star)| - |g(x_\star)| \leq |g(y_\star) - g(x_\star)|\leq L|y_\star - x_\star| \leq L|s-t|\,.$$ So $f$ is locally Lipschitz and hence continuous.