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I've been trying with these all the afternoon and I still don't get even a hope. Anything would be really helpfull!

Let $\mathcal{H}$ a hilbert space.

  1. Let $K\in\mathcal{L}(\mathcal{H})$ be a self-adjoint bounded operator. Prove that if $K$ is compact, then $||K||$ or $-||K||$ is a eigenvalue of $K$.

  2. Let $K$ be a self-adjoint bounded operator on $\mathcal{H}$. Prove that if $K^2=K^3$, then $K=K^2$.

Thank you all!

Some extra info (thanks to Mike, the first commenter!): it should be enough with some basic spectrum properties, the spectral theorem and a little bit of functional calculus.

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    $\begingroup$ What kinds of tools do you know of? The spectral theorem? Functional calculus? Also, (1) is false, for instance multiplication by the function $x$ is a self-adjoint operator of norm 1 on $L^2[0,1]$, but has no eigenvalues. Did you mean to say $\|K\|$ or $-\|K\|$ is in the spectrum? $\endgroup$ – Mike F Nov 13 '16 at 4:55
  • $\begingroup$ We are starting now with functional calculus so I think that it should be enough with some basic spectrum properties and the spectral theorem. Every way, by sure the question was badly written, I will edit it, it should be the spectrum. $\endgroup$ – Esteban Gutiérrez Nov 13 '16 at 5:03
  • $\begingroup$ With the comments of @MikeF in mind, for $(1)$ you can show that the spectral radius of $K$ equals $\|K\|$ $\endgroup$ – Aweygan Nov 13 '16 at 5:03
  • $\begingroup$ Dears, I found the question 1 in a book (without the answer), and there the operator is compact, so I think that my teacher just missed a hypothesis. $\endgroup$ – Esteban Gutiérrez Nov 13 '16 at 6:07
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The facts you state do not depend on $K$ being compact, but only on $K$ being selfadjoint. For example, suppose $K^2x=0$. Then $0=\langle K^2x,x\rangle=\langle Kx,Kx\rangle=\|Kx\|^2$ implies $Kx=0$. So, $$ \mathcal{N}(K^2)=\mathcal{N}(K). $$ Therefore, if $K^2=K^3$, then $K^2(I-K)=0$ implies $K(I-K)=0$ or $K^2=K$.

There are several ways to prove that either $\|K\|$ or $-\|K\|$ is in the spectrum of $K$. One way is the prove that the norm of $K$ is the same as the spectral radius, which can be done by showing that $\|K\|=\|K^2\|^{1/2}$ for any selfadjoint operator $K$. This is because the spectral radius is $r_{\sigma}(K)=\lim_{n}\|K^n\|^{1/n}$, and $$ \|K\|=\|K^2\|^{1/2}=\|K^4\|^{1/4}=\cdots=\lim_{n}\|K^{2^n}\|^{1/2^n}=r_{\sigma}(K). $$ To see that $\|K\|=\|K^2\|^{1/2}$ for any selfadjoint $K$, note that $$ \|K^2\| \le \|K\|\|K\|=\|K\|^2, $$ and $$ \|K\|^2=\sup_{\|x\|=1}\|Kx\|^2 = \sup_{\|x\|=1}(K^2x,x) \le \sup_{\|x\|=1}\|K^2x\|\|x\|=\|K^2\|. $$ Every non-zero point of the spectrum $\sigma(K)$ of a selfadjoint compact operator is an eigenvalue. So, either $\|K\|=0$ or at least one of $\|K\|,-\|K\|$ is such a non-zero point of the spectrum because $\|K\|=r_{\sigma}(K)$.

An alternative method for showing that either $\|K\|$ or $-\|K\|$ is in the spectrum relies on the following operator norm equality which holds for selfadjoint operators $K$: $$ \sup_{\|x\|} |\langle Kx,x\rangle| = \|K\|. $$ Either $\sup_{\|x\|=1}\langle Kx,x\rangle =\|K\|$ or $\inf_{\|x\|=1}\langle Kx,x\rangle =-\|K\|$. The first case may be assumed to hold by replacing $K$ with $-K$ if necessary. Let $\{ x_n \}$ be a sequence of unit vectors chosen so that $\langle Kx_n,x_n\rangle$ converges to $\|K\|$. Then $$ \|\|K\|x_n-Kx_n\|^2=\|K\|^2-2\|K\|\langle Kx_n,x_n\rangle+\langle Kx_n,x_n\rangle^2\rightarrow 0. $$ Because $K$ is compact, there is a subsequence $\{ x_{n_k} \}$ such that $\{ Kx_{n_k}\}$ converges to some $y\in \mathcal{H}$. Then $\{ x_{n_k} \}$ also converges: $$ x_{n_k} = \frac{1}{\|K\|}(\|K\|x_{n_k}-Kx_{n_k})+\frac{1}{\|K\|}Kx_{n_k}\rightarrow 0+\frac{1}{\|K\|}y. $$ Hence $\frac{1}{\|K\|}y$ is a unit vector and, by the continuity of $K$, one obtains $K\left(\frac{1}{\|K\|}y\right)=y$, which proves that $\|K\|$ is an eigenvalue.

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  • $\begingroup$ Thank you very much! I really appreciate your answer, but there is something missing :/ I've seen the question better and in question 1 I have to prove that $||K||$ or $-||K||$ is indeed a eigenvalue when $K\in\mathcal{L}(\mathcal{H})$ is compact. $\endgroup$ – Esteban Gutiérrez Nov 13 '16 at 6:36
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    $\begingroup$ @EstebanGutiérrez : Are you aware that every non-zero point of the spectrum of $K$ is an eigenvalue? This is true for selfadjoint compact operators. $\endgroup$ – DisintegratingByParts Nov 13 '16 at 14:26
  • $\begingroup$ @EstebanGutiérrez : I added a second method for proving what you want. $\endgroup$ – DisintegratingByParts Nov 13 '16 at 19:03

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