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
    Thank you. Doesn't $\nabla f \in C([0,T]\times \Omega)$ automatically imply that $\nabla f \in C(\Omega)$ for a.e $t$ (i.e. that $\nabla f \in C([0,T], C(\Omega))$) regardless of uniformly continuous? Then why do we need uniform continuity to say that $\nabla f \in C([0,T], C(\Omega))$?2012-12-17
  • 0
    I'm only talking about the case when $\Omega$ is compact in the above comment, btw.2012-12-17
  • 0
    You're right. The point is that if $\Omega$ is compact, then uniform continuity is an additional automatic consequence and therefore $t \mapsto \nabla f(t, \cdot)$ is a continuous curve with values in $C(\Omega)$. But if $\Omega$ is not compact, then this need not be a continuous curve. Take e.g. $T = 1, \Omega = \mathbb{R}$, and $f(t,x) = 1/((x-c(t))^2 + 1)$ with $c(t) = 1/(1-t)$ for $0 < t < 1$, $f(x,1) = 0$. Then $f \in C^1([0,T]\times \Omega$, but $\nabla f(t, \cdot)$ is not continuous at $t = 0$ as a curve with values in $C(\mathbb{R})$.2012-12-17
  • 0
    Make that "not continuous at $t = 1$".2012-12-17