TITLE: Norm of integral operator in $L_2$
QUESTION [11 upvotes]: What is the norm of integral operators $A$ in $L_2(0,1)$?
$Ax(t)=\int_0^tx(s)ds$

REPLY [4 votes]: Here is a rather direct way to obtain the norm, without previous knowledge of what it should be. We take advantage of the C$^*$-identity $$\tag1\|V\|=\|V^*V\|^{1/2}.$$ Because $V^*V$ is compact and positive, its norm is equal to its greatest eigenvalue.
We have
$$\tag2
V^*Vf(x)=\int_x^1\int_0^tf(s)\,ds\,dt. 
$$
If we differentiate the equality $V^*Vf(x)=\lambda f(x)$ twice, we get the differential equation $-f(x)=\lambda f''(x)$. From $(2)$ we get the initial conditions $f(1)=0$, $f'(0)=0$. Since we know that $\lambda>0$ this gives, up to a multiple,
$$
f(x)=\cos\frac x{\sqrt\lambda},\qquad\text{subject to $f(1)=0$.}
$$
Thus
$$
\frac1{\sqrt\lambda}=\frac{(2k+1)\pi}2,\qquad k\in\mathbb N,
$$
so
$$
\sqrt{\lambda}=\frac2{(2k+1)\pi}.
$$
Using $k=0$ to get the largest $\lambda$, $$\|V\|=\frac2\pi.$$