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Let $C^1([0,1])$ be the set of all continuously differentiable functions on [0,1], i.e. the derivitave functions are all continuous. Let $C([0,1])$ be the set of all continous functions on [0,1]. Within the sup norm i.e. $\|f\|=\sup_{x\in X} |f(x)|$.

Suppose a linear map $T:C^1([0,1])\to C([0,1])$. Given $Tf=f'$. Suppose I try to prove $T$ is not bounded. I have no idea what is the space? why it is not the same as the set of the derivative of all $f\in C^1([0,1])$

If $f_n:x\to nx \in C^1([0,1])$, then is not $lim_{n\to \infty} f'_n=\infty$?

Halk
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1 Answers1

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The usual example to show differentiation is unbounded is $f_n(x) = \sin(2\pi nx)$ when you equip both spaces with the infinity norm (in this context this is equal to the $C[0,1]$ norm). Here $T(f_n)(x) = 2\pi n\cos(2\pi n x)$. Note that $\|f_n\|_{\infty}=1$ for all $n$, but $\|Tf_n\|_{\infty}=2\pi n\to\infty.$ This implies that $T$ is unbounded.

Edit to add: It seems maybe your confusion is not understanding operator norms. The norm of $T$ as an operator from $(C^1[0,1],\|\cdot\|_\infty)\to (C[0,1],\|\cdot\|_\infty)$ is $$\sup \frac{\|Tf\|_{\infty}}{\|f\|_{\infty}}.$$ If you find a sequence $f_n$ such that the denominator is 1 but the numerator blows up, this means that $T$ is unbounded.

e.g., https://en.wikipedia.org/wiki/Unbounded_operator

Keaton
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