# Rapidly decreasing functions and convergence

• May 13th 2011, 04:33 AM
measureman
Rapidly decreasing functions and convergence
I am trying to show that, for f a rapidly decreasing function, the mapping from f to f' is continuous. Since the space of RDF's is an F-space, Rudin says that I can use the closed graph theorem for F-spaces, and hence I need to show that the graph of the mapping is closed.

So I wish to show that $\{(f, f') : f \in RDF's\}$ is closed. I think that I can do this if it follows that if $f_k \to f$, then $f_k' \to f'$?

Can someone tell me if this is true, and if so, how would I prove it?
• May 13th 2011, 06:21 AM
girdav
Yes, that what you have to show. What is the topology over the space RDF?
• May 13th 2011, 06:30 AM
Opalg
Quote:

Originally Posted by measureman
I am trying to show that, for f a rapidly decreasing function, the mapping from f to f' is continuous. Since the space of RDF's is an F-space, Rudin says that I can use the closed graph theorem for F-spaces, and hence I need to show that the graph of the mapping is closed.

So I wish to show that $\{(f, f') : f \in RDF's\}$ is closed. I think that I can do this if it follows that if $\color{red}f_k \to f$, then $\color{red}f_k' \to f'$?

Can someone tell me if this is true, and if so, how would I prove it?

The condition in red looks suspiciously like a statement of what you are trying to prove.

I'm guessing that the F-space topology is the topology of uniform convergence on compact sets for the function and all its derivatives. In that case, the continuity of differentiation follows directly from the definition. If $f_k$ and each of its derivatives converges uniformly to the corresponding derivative of f, then it's immediately obvious that the same is true for $f_k'$ and f'.