Let $X$ be the Banach space $C([a,b])$ of complex-valued continuous functions on the closed real interval $[a,b]$ with the sup norm. For each $x\in X$, define $Tx$ be a function on $[a,b]$ and for $t\in[a,b]$
$$Tx(t)=\int_{a}^t x(s)ds$$
Prove that $T$ is a bounded operator from $X$ to itself and $\lVert T \rVert =b-a$.
It is clear that $T$ is a linear operator. However, I don't know how to prove it is bounded and calculate its norm.
To prove that $T$ is bounded means to prove that there exists some $k\ge0$ such that :
$$\forall x\in X,\Vert T(x)\Vert_\infty\le k\,\Vert x\Vert_\infty$$
But, for all $t\in[a,b]$, we have :
$$\left|\int_a^tx(s)\,ds\right|\le\int_a^t\left|x(s)\right|\,ds\le(t-a)\Vert x\Vert_\infty\le(b-a)\Vert x\Vert_\infty$$
Hence, by taking the supremum :
$$\Vert T(x)\Vert_\infty\le(b-a)\Vert x\Vert_\infty$$
So $T$ is a bounded linear operator (which also means that it is a continuous endomorphism of $X$).
Now, computing its norm $\Vert T\Vert$ means finding the lowest possible $k$ ... And the previous calculation shows that $\Vert T\Vert\le b-a$.
To show that $\Vert T\Vert=b-a$ we have two possibilities :
1) finding some $x\in X-\{0\}$ wuch that $\Vert T(x)\Vert_\infty=(b-a)\Vert x\Vert_\infty$
2) finding some sequence $(x_n)$ in $X-\{0\}$ such that $\displaystyle{\lim_{n\to\infty}\frac{\Vert T(x_n)\Vert_\infty}{\Vert x_n\Vert_\infty}}=b-a$
The former is easy : just take $x:t\mapsto 1$.
