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 $f\in L^2([0,1])$.

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  • 1
    $\begingroup$ This question is probably better suited to math.stackexchange.com $\endgroup$ – Neal Feb 11 at 15:59
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    $\begingroup$ Can you explain the connection of your question with graph laplacian ? $\endgroup$ – Piyush Grover Feb 11 at 16:07

Your operator is a rank one perturbation of the multiplication operator $(Mf)(x) = (x/2)f(x)$, which has (purely) absolutely continuous spectrum equal to $[0,1/2]$. Since the ac spectrum is invariant under trace class perturbations (so certainly under rank one perturbations), your operator $L$ still has the same ac spectrum, so doesn't even come close to having pure point spectrum (and thus it isn't "diagonalizable," if you want to put it this way, though I personally don't think it's very good terminology).


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