# Showing an operator is not bounded.

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?

• You can't find a counterexample consisting of just one function, it has to be a sequence. – Michh Jun 17 '20 at 23:48

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.
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.