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Let $C^1[0,1]$ be space of all real valued continuous function which are continuously differentiable on $(0,1)$ and whose derivative can be continuously extended to $[0,1]$. For $f$ in that set define the norm of $f = \max{\{\|f(t)\|, \|f'(t)\|\}}$ where $t$ is in $[0,1]$. can you show this space is complete under this norm? and let a linear operator $T:C^1[0,1] \longrightarrow C[0,1]$ defined by $T(f)=f'$, show that $T$ is continuous and norm of $T=1$.

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1 Answer 1

You probably mean $\|f\| =\max{\{\|f\|_{\infty},\|f'\|_{\infty}\}}$ where $\|f\|_{\infty} = \sup_{t \in [0,1]} |f(t)|$.

Hint: Recall the basic fact that if $f_{n}$ is a sequence of $C^{1}$ functions such that $f_{n}$ converges uniformly (or only pointwise) to a continuous function $f$ and $f_{n}'$ converges uniformly to a continuous function $g$ then $f \in C^{1}$ and $f' = g$. It's probably a good idea to try to prove this on your own without consulting your books.

That $\|T\| \leq 1$ is just the definition of the operator norm. Consider also $f(x) = x$.

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I think it deserves emphasis that, in effect by_design of the topology on $C^1$, the differentiation map from $C^1$ to $C^0$ is continuous. That is, rhetorically, "what other topology would/should $C^1$ have?"... since, presumably, we care that those functions have $C^0$ derivatives exactly so we can differentiate them and have a continuous outcome. Naturally, this should be a continuous map $C^1\rightarrow C^0$. (I am resisting ranting about the time "real analysis" courses in the U.S. waste on fretting over measure theory while forgetting how to differentiate anything...) –  paul garrett Jul 26 '11 at 0:32

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