Hilbert-Schmidt operator defined by non-orthogonal basis

Question

I have the following operator on $L^2(0,1)$

$$ Tf = \sum_{n \geq 0} 2^{-n}\langle f,v_n\rangle v_n$$

Where $v_n(t) = t^n$.

I was able to prove that it is Hilbert-Schmidt, but now I need to calculate its Hilbert-Schmidt norm, and its integral kernel. The problem is that it is defined through an indipendent system of vectors (the polynomials) which are dense but not orthogonal (if it had been $v_n(t) = \exp(2\pi i nt)$ it would have been much easier. How do I identify the eigenvectors?

I was thinking about setting $t = \exp(2\pi i x)$ and reduce to the orthogonal case somehow but I reach incorrect results, I am not sure of where I am doing wrong.

Can anybody help me?

Answer 1

The operator is given by $$ Tf(x)=\int_0^1 f(t) \sum_{n\ge 0} \frac{(xt)^n}{2^n}\, dt =\int_0^1 f(t)\frac{2}{2-xt}\, dt.$$ Let $K(t, x):=2(2-xt)^{-1}$. This is the kernel you are looking for. The squared Hilbert-Schmidt norm of $T$ is given by the integral $$ \iint_{[0, 1]^2} \frac{4dtdx}{(2-xt)^2}= 2\log(2).$$ (thanks for correcting the mistake in my computations)