# Graph Laplacian Operator

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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• This question is probably better suited to math.stackexchange.com – Neal Feb 11 at 15:59
• Can you explain the connection of your question with graph laplacian ? – 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).