Norm of integral operator in $L_2$ What is the norm of integral operators $A$ in $L_2(0,1)$?
$Ax(t)=\int_0^tx(s)ds$
 A: Here is a rather direct way to obtain the norm of the Volterra operator, 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 $\lambda>0$ (because $V^*V$ is an injective positive operator) this gives, up to a multiple,
$$
f(x)=\cos\frac x{\sqrt\lambda},\qquad\text{subject to $f(1)=0$.}
$$
Thus
$$
\frac1{\sqrt\lambda}=\frac{(2k+1)\pi}2,\qquad k\in\mathbb N\cup\{0\},
$$
so
$$
\sqrt{\lambda}=\frac2{(2k+1)\pi}.
$$
Using $k=0$ to get the largest $\lambda$, $$\|V\|=\frac2\pi.$$
A: It's enough to use Schwarz inequality in the following manner:
$$ \| A x \|^2 = \int_0^1 \left| \int_0^t x(s) \, ds \right|^2 dt = \int_0^1 \left| \int_0^t \sqrt{\cos \frac{\pi}{2}s} \cdot \frac{x(s)}{\sqrt{\cos \frac{\pi}{2}s}} \,ds \right|^2 dt \le \int_0^1 \left( \int_0^t \cos \frac{\pi}{2}s \, ds \int_0^t \frac{|x(s)|^2}{\cos \frac{\pi}{2}s}\right) dt = \frac{2}{\pi} \int_0^1 \int_0^t \sin \frac{\pi}{2}t \, \frac{|x(s)|^2}{\cos \frac{\pi}{2}s} \, ds\,dt = \frac{2}{\pi}\int_0^1 \left( \int_s^1 \sin \frac{\pi}{2} t \, dt \right) \frac{|x(s)|^2}{\cos \frac{\pi}{2}s} \,ds = \left( \frac{2}{\pi} \right)^2 \| x \|^2
$$
Equality holds for $x(s) = \cos \frac{\pi}{2}s$.
A: It is Problem 188 in the book by P. Halmos, "A Hilbert space problem book". In the solution, the author writes that "A direct approach seems to lead nowhere." The norm is indeed $2/\pi$, and is computed through the adjoint $A^*$ and a suitable kernel. It is a rather long proof, so please try to read it on Halmos' book.
A: The norm of the Volterra operator is $2/\pi$.  I will try to recall the proof; the bound suggests that the optimum occurs for some trigonometric polynomial, say $\cos(\pi x/2)$.
