Norm of differentiation operator $Tf(t)=f^{'}$..

Consider $T:C^1[0, 1]\rightarrow C[0, 1]$ given by $Tf=f'$ where $$\|f\|_{C^1}=\|f\|_\infty+\|f'\|_\infty$$ and $\|f\|_\infty=\sup_{x\in [0, 1]}|f(x)|$. How to prove $\|T\|=1$? The inequality $\|T\|\leq 1$ is easy, but I can't find a function for showing this upper bound is achieved.

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how can we show above equality ||f||₁=||f′||_{∞}+||f||_{∞} – 1214 May 3 '14 at 13:25

Consider the functions $$f_n(x)=x^n$$ for $n\geq 1$. Then $$\|f_n\|_{C^1}=\|x^n\|_\infty+\|nx^{n-1}\|_\infty =n+1$$ and $$\|Tf_n\|_\infty=\|nx^{n-1}\|_\infty=n.$$ So $$\|T\|\geq \frac{n}{n+1}$$ for all $n$.

Hence $\|T\|\geq 1$ when letting $n$ tend to $\infty$.

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@DavidMitra Oh! Thanks! What a trick. – 1015 Feb 16 '13 at 20:58
Nice Proof, thanks =D – PtF Feb 17 '13 at 11:13

No function can realize the upper bound.

However, we can approach it by functions like $f_n(x):=\frac{\sin(2\pi n x)}n$.

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I was almost sure that approximation would be needed...Thanks for the tip.. – PtF Feb 16 '13 at 20:23