1
$\begingroup$

If $f(t,u)$ is continuous wrt. $t$ (and $u$), then is $\sup_{u \in H^1(\Omega)} f(t,u)$ continuous wrt. $t$?

I am unable to prove this. Help appreciated.

  • 1
    @paulgarrett thanks2012-12-23

1 Answers 1

1

Let $S= (0,\infty) \times (0,\infty)$. If $f\colon S\to\mathbb R$ is defined by $ f(t,u) = \begin{cases} 0, & \text{if } t\le 1, \\ (t-1)u, & \text{if } t>1 \text{ and } (t-1)u\le 1, \\ 1, & \text{if } t>1 \text{ and } (t-1)u>1, \end{cases} $ then $ \sup_{u\in (0,\infty)} f(t,u) = \begin{cases} 0, & \text{if } t\le 1, \\ 1, & \text{if } t>1 \end{cases} $ is not continuous.

  • 1
    I think it is continuous, actually. The third case of the definition doesn't come in to play infinitesimally to the right of $t=1$; it's always the second case of the definition that ends up being relevant.2012-12-23