Showing an operator is not bounded.

Question

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?

Answer 1

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.

Answer 2

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.