Consider the linear operator $\mathbb{L} : L^2([0,1])\to L^2([0,1])$ defined by $$ (\mathbb{L}f)(x) = \int_0^1 xy(f(x)-f(y)) \mathrm{d}y $$ for all $f\in L^2([0,1])$ and $x \in [0,1]$. Is $\mathbb{L}$ diagonalizable and why?

**Definition:** $\mathbb{L}$ diagonalizable means that there exists eigenvalues $\{\lambda_k\}_{k\in\mathbb{N}}$ and an orthonormal basis $\{f_k\}_{k\in\mathbb{N}}$ such that

$$ \mathbb{L}f = \sum_{k\in\mathbb{N}} \lambda_k\langle f, f_k \rangle f_k $$ for all $f\in L^2([0,1])$.