1
$\begingroup$

$f(x,t)$ is a function defined on the set $S = \cup_{t \in [0,T]} A(t) \times \{t\}$ where $A(t)$ is an open subset of $\mathbb{R}^n$ that depends on $t$ in some way.

I am told that $f \in C^1(S)$. Since the set $S$ is not closed (I think, even though $\{t\}$ is closed), we cannot say that $f$ is bounded above and below. Am I right? Also we cannot say anything about its divergence either can we (if $f$ is vector valued)? Or maybe I am missing something.

If I were given that $f \in L^\infty(S)$, then I can write $|f(x,t)| \leq \text{const}$ almost everywhere?

1 Answers 1

0

Taking $A(t)=(0,1)$, and $f(x,t) = \frac{1}{x}-\frac{1}{1-x}$ gives a $C^1$ $f$ that is neither bounded below or above on $S$.

If $f \in L^{\infty} (S)$ then $|f(x,t)| \leq ||f||_{\infty}$ ($\mathbb{ess} \sup$ norm) a.e. on $S$.