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I think this question isn't that hard, but I am a bit confused.

Define the linear operator $T_k:H\mapsto H$ by \begin{align} T_ku=\sum^\infty_{n=1}\frac{1}{n^3}\langle u,e_n\rangle e_n+k\langle u,z\rangle z, \end{align} where $z=\sqrt{6/\pi^2}\sum^\infty_{n=1}e_n/n$. For negative $k$, show $T_k$ has at most one negative eigenvalue.

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Please use $\langle, \rangle$ (\langle, \rangle) –  martini Nov 9 '12 at 10:09

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  • A computation shows that $T_k$ is self-adjoint if $k$ is a real number (as a sum of such operators), so if $\lambda$ and $\mu$ are two different eigenvalues of $T_k$, the associated eigenvectors for these eigenvalues are orthogonal.
  • If $u$ is an eigenvector for $\lambda<0$, then for each $j\geqslant 1$, $$\langle u,e_j\rangle+kj^2|z|^{-1}\langle u,z\rangle=\lambda j^3\langle u,e_j\rangle,$$ hence, $$\langle u,e_j\rangle=-\frac{kj^2}{1-\lambda j^3},$$ as we can assume $\langle u,z\rangle =1$ and $\lambda\neq j^{-3}$.

  • Now we conclude. Let $\lambda$ and $\mu$ two distinct eigenvalues of $T_k$ and $u,v$ associated eigenvectors. Then $$0=\langle u,v\rangle =\sum_{j\geq 1}\frac{kj^2}{1-\lambda j^3}\frac{kj^2}{1-\mu j^3},$$ a contradiction.

  • Even if it's not required in the question but in the comments, $A_k$ is a compact operator for each $k$. Let $$T_N:=\sum_{j=1}^N\frac 1{j^3}\langle u,e_j\rangle e_j+k\sum_{j=1}^N\frac 1{j}\langle u,e_j\rangle \frac 1{\lVert z\rVert}z.$$ It's a finite ranked operator, and $\lVert T-T_N\rVert\to 0$.

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So why cant there be more than one negative eigenvalue? –  user27126 Nov 10 '12 at 22:11
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I've added the conclusion. –  Davide Giraudo Nov 10 '12 at 22:14
    
Also, how does one show $T_k$ is compact? –  anegligibleperson Nov 16 '12 at 2:50
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@anegligibleperson I've added it. –  Davide Giraudo Nov 16 '12 at 9:53

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