Norm of Operator : $Af(x)=f'(x)$ Problem
Let :
$$A~~:~C^{1}\left([0,1],||.||\right)\to~C\left([0,1],||.||_{\infty}\right)$$
$$Af(x)=f'(x)$$
Where : $||f||_{C^{1}}=||f||:=||f||_{\infty}+||f'||_{\infty}$
I was prove that $A$ operator Linear and continue
$$|Af(x)|=|f'(x)|\leq~|f'(x)+|f(x)|\leq ||f||$$
$$\implies ||Af||_{\infty}\leq||f||$$
So : $||A||\leq1$
I need now prove that $||A||=1$
I tired to find a function  $f\in C^{1}$ such that $||f||=1$ and $||Af||=1$
I need idea for this

Thanks!
 A: In the codomain $\mathcal{C}[0,1]$, you have the usual supremum norm:
$$\|f\|_{\infty}=\sup_{x\in[0,1]}|f(x)|=\max_{x\in[0,1]}|f(x)|,$$
where we can write "$\max$" instead of "$\sup$" since we're dealing with continuous functions on a compact interval.
In the domain $\mathcal{C}^1[0,1]$, you have the $\mathcal{C}^1$-norm:
$$\|f\|_{\mathcal{C}^1}=\|f'\|_{\infty}+\|f\|_{\infty}=\sup_{x\in[0,1]}|f'(x)|+\sup_{x\in[0,1]}|f(x)|=\max_{x\in[0,1]}|f'(x)|+\max_{x\in[0,1]}|f(x)|.$$
Now, the operator norm in this case is:
$$\|A\|=\sup\frac{\|Af\|_{\infty}}{\|f\|_{\mathcal{C}^1}}=\sup_{\|f\|_{\mathcal{C}^1}\leqslant1}\|Af\|_{\infty},$$
and here "$\sup$" may be effectively necessary, not replaceable by "$\max$".
Indeed, if $\|Af\|_{\infty}=\|f'\|_{\infty}=1$, then $\|f\|_{\mathcal{C}^1}=\|f'\|_{\infty}+\|f\|_{\infty}=1+\|f\|_{\infty}$, and that would be equal to $1$ iff $\|f\|_{\infty}=0$, which makes $f$ the identically zero function, which clearly doesn't work. So it's impossible to

… find a function $f\in\mathcal{C}^1$ such that $\|f\|=1$ and $\|Af\|=1$.

Instead, you need to come up with examples showing that the expression in the supremum (in either definition) can be arbitrarily close to $1$. For example, how about constructing a sequence of functions $f_n(x)$ such that, even though $\frac{\|Af_n\|_{\infty}}{\|f_n\|_{\mathcal{C}^1}}<1$ for each of them, the limit of this ratio is $1$? Hint: power functions.
