3
$\begingroup$

Is $f \in C^1([0,T]\times \Omega) \iff f \in C([0,T]; C^1(\Omega))$

My guess is no, I think the RHS is a bit stronger. But I can't show it. Can someone help me please?

  • 1
    This is more or less a duplicate of http://math.stackexchange.com/questions/226039/continuous-function-spaces-question/263887 where I gave another explicit example (before finding this thread).2012-12-25

1 Answers 1

2

If $f \in C^1([0,T]\times \Omega)$, then $\frac{\partial}{\partial t} f \in C^([0,T] \times \Omega)$. Functions in the set on the right do not have this property.

On the other hand, if $\Omega$ is compact, then $\nabla f C^([0,T] \times \Omega$ is uniformly continuous, hence $\nabla f \in C([0,T],C(\Omega))$ and therefore $f \in C([0,T],C^1(\Omega))$.

So the left set is usually smaller.

  • 0
    Make that "not continuous at $t = 1$".2012-12-17