Vtyjkyuk

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?

First, deal with the simpler cases $a = 0$ or $b = 0$. Second, if $a, b neq 0$, then you can factor out $frac1b$ to instead apply the operator to $fracab + x = c + x$.

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.