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There are two spaces $C^1 [0,1]$ and $C[0,1]$ with supremum norm, which is defined by $$ \|f\| = \sup_{x\in[0,1]} |f(x)|$$ for any $f$. I have to show that if the operator $A:C^1[0,1] \rightarrow C[0,1]$ is defined by $Af=f'$, then $A$ is not bounded.

I tried to find some counterexample function $f\in C^1[0,1]$ not satisfying $\|Af\| \le C\|f\|$ for some uniformly $C$ . But I failed. How can I show that?

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  • $\begingroup$ You can't find a counterexample consisting of just one function, it has to be a sequence. $\endgroup$ – Michh Jun 17 '20 at 23:48
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Boundedness of a linear map is equivalent to continuity. Let $f_n(x)=\frac {x^{n}} n, f(x)=0$. Then $f_n \to f$ uniformly but $f_n'$ does not tend to $f'$ uniformly.

Also $\|f_n\|=\frac 1 n$ and $\|f_n'\|=1$ so your constant $C$ does not exist.

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Consider $f_n(x) = \sin(2\pi n x)$ for $x \in [0,1]$. Then $\|f_n \| = 1$ but $\|Af_n \| = \|f_n'\| = n$ showing that no such constant $C > 0$ can exist.

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