Let $x:[0,1]\rightarrow \mathbb{R}$ with the supremum norm $\displaystyle{\|x\|=\sup_{t\in [0,1]}|x(t)|}$ and $T: C^1[0,1]\rightarrow C[0,1]$ with $T(x)(t)=\int_0^tsx(s)\, ds$.
I want to calculate $\|T\|:=\sup_{\|x\|\leq 1}\|T_1(x)\|$.
I have found the following: $$\|T(x)\|=\sup_{t\in [0,1]}|T(x)(t)|=\sup_{t\in [0,1]}\left |\int_0^tsx(s)\, ds\right |\leq \sup_{t\in [0,1]}\int_0^t|sx(s)|\, ds \\ =\int_0^1|sx(s)|\, ds=\int_0^1s\cdot |x(s)|\, ds \leq \int_0^1s\cdot \sup_{\sigma \in [0,1]}{|x(\sigma)|}\, ds=\int_0^1s\cdot \|x\|\, ds$$
And so $$\|T\|=\sup_{\|x\|\leq 1}\|T(x)\|\leq \sup_{\|x\|\leq 1}\int_0^1s\cdot \|x\|\, ds=\int_0^1s\cdot 1\, ds=\left [\frac{s^2}{2}\right ]_0^1=\frac{1}{2}$$
So we have an inequality $\|T\|\leq \frac{1}{2}$. How can we get an equality?
Take $x=1$. Then $\|x\|=1$ and $\bigl\|T(x)\bigr\|=\dfrac12$.