# Find the norm of the operator $A(x(t))=x'(t)$ and prove that it is well-defined

I figured out that it is well-defined and got that: $A(x(t))=x'(t)$ and $A:C^1[a,b] \to C[a,b]$ $$\|A(x(t))\|_{C[a,b]} \leq \|x(t)\|_{C^1[a,b]}$$

But I cannot find a function where this is the norm, so I assume that the norm is less than one. I need to prove this, and find a function to which this applies. All input is well received.

The norm in $C^k[a,b]$ is $$\|f(x)\|_{C^k[a,b]}=\sum_{k}\max_{x\in(a,b)}|f^{(k)}(x)|$$

You can find functions $x$ such that $\|Ax\|_{C}/\|x\|_{C^1}$ is as close to $1$ as you wish.
Let $\epsilon>0$ small. Define $$x_\epsilon(t)=\begin{cases}\dfrac{(t-a)(a+2\,\epsilon-t)}{\epsilon^2} & \text{if }a\le t\le a+\epsilon,\\1 & \text{if }a+\epsilon<t\le b.\end{cases}$$ Then $x_\epsilon\in C^1[a,b]$ and \begin{align} \|x_\epsilon\|_{C^1[a,b]}&=\frac{2}{\epsilon}+1,\\ \|(A\,x_\epsilon)\|_{C[a,b]}&=\frac{2}{\epsilon}. \end{align}