Calculate the operator norm defined on L2 For $f \in L^2([0, 1]) $ and $x \in [0, 1]$ we have the following
$$
A(f)(x) = \int_0^1 (x - y) f(y)dy
$$
In this particular exercise I have to calculate the operator norm $\| A \|$.
We know that from a theorem that if $A$ is compact and self-adjoint then either $\|A\|$ or $-\| A \|$ is an eigenvalue to the operator $A$. 
My attempt: 
We know that $A$ is indeed compact, but it is not self-adjoint. However, by some inspection, we have that $iA$ is self-adjoint (and compact). So we just have to calculate $\|iA\| = \|A\| = |\lambda|$ where $\lambda$ is an eigenvalue satisfying $A(f)(x) = \lambda f(x)$, for some eigenfunction $f$. Now setting $f(x) = a + bx$ we get by calculation that 
$$\lambda(a + bx) = \int_0^1 (x - y) (a + by) dy = -\frac a 2 - \frac b 3 + (a + \frac b 2)x.$$
But from this I'm not sure how to solve for $\lambda$. Maybe there is an another way to solve this?
 A: First, deal with the simpler cases $a = 0$ or $b = 0$. Second, if $a, b \neq 0$, then you can factor out $\frac{1}{b}$ to instead apply the operator to $\frac{a}{b} + x = c + x$.
The more general case, if you would want to do it, is to identify the constant terms of both sides and the coefficient of $x$ on both sides, getting two equations involving $x, a$ and $b$. Then you want to maximize $|\lambda|$ by choosing $a$ and $b$ appropriately.
A: Your operator has finite range (its range is generated by $\{f_0(x)=1,f_1(x)=x\}$. For $f(x)=a+bx$ as eigenfunction, the values 
$a=0$ or $b=0$ gives $a=b=0$. Thus $ab\neq0$. From your derivation we have
\begin{aligned}
\lambda a&=-\big(\frac{a}{2}+\frac{b}{3}\big)\\
\lambda b&=\big(a+\frac{b}{2}\Big)
\end{aligned}
Let $r=b/a$. Then,
\begin{aligned}
\lambda=-\Big(\frac12+\frac{r}{3}\Big)=\Big(\frac{1}{r}+\frac{1}{2}\Big)
\end{aligned}
This gives $r_1=\frac{1}{2}\big(-3+i\sqrt{3}\big)$ and $r_2=\frac{1}{2}\big(-3-i\sqrt{3}\big)$. Then you get
$\lambda=\pm i\frac{\sqrt{3}}{6}$. The norm of your operator is then $\sqrt{3}/6$.
