Q:

Norm of differentiation operator $Tf(t)=f'$

Consider $T:C^1([0, 1])\to C([0, 1])$ 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.

A:

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 $\infty$.

A:

No function can realize the upper bound. 
However, we can approach it by functions like $f_n(x):=\frac{\sin(2\pi n x)}n$. 

