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Let A be a positive semi-defnite operator on an infnite-dimensional Hilbert space H. Suppose there exist nonnegative numbers ${{\lambda }_{0}},{{\lambda }_{1}},...,{{\lambda }_{n}}$ such that the operator $K={{\lambda }_{0}}I+{{\lambda }_{1}}A+...+{{\lambda }_{n}}{{A}^{n}}$ is polynomial operator with these nonnegative coefficients. K is a nonzero compact operator. Prove that ${{\lambda }_{0}}=0$ and that ${{A}^{n}}$ is compact.

I've tried to prove it using spectrum of compact operator but without success. I would appreciate some help.

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  • $\begingroup$ What is the meaning of λ0? $\endgroup$ – José Carlos Santos Jun 4 '17 at 8:52
  • $\begingroup$ Sorry, its λ with index 0. $\endgroup$ – XYZ Jun 4 '17 at 8:53
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    $\begingroup$ Then the question becomes: what is $\lambda_0$? $\endgroup$ – José Carlos Santos Jun 4 '17 at 8:54
  • $\begingroup$ This is operator K, $K={{\lambda }_{0}}I+{{\lambda }_{1}}A+...+{{\lambda }_{n}}{{A}^{n}}$ $\endgroup$ – XYZ Jun 4 '17 at 9:46
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Not sure if there is a more elementary argument, but here is one:

Since $A$ is positive semidefinite, $\sigma(A) \subset \mathbb{R}_+$. Let $p(t) = \lambda_0 + \lambda_1t + \ldots + \lambda_nt^n$, and $0\leq i\leq n$ be any integer such that $\lambda_i\neq 0$. Then, we claim that $A^i$ is compact. From this, one can conclude that $\lambda_0 = 0$ since $I$ is not compact, and $A^n$ is compact since $\lambda_n \neq 0$.

To prove the claim, let $f(t) = \lambda_it^i$, then it suffices to show that $f(A)$ is compact. Note that $$ p(t) \geq f(t) \geq 0 \quad\forall t\in \sigma(A) $$ So by the spectral theorem $$ K = p(A) \geq f(A) \geq 0 $$ Let $P_n$ be the sequence of finite rank projections such that $P_nS\to S$ for every compact operator $S$. Then $$ (I-P_n)K(I-P_n) \geq (I-P_n)f(A)(I-P_n) \quad\forall n\in \mathbb{N} $$ Hence, $\|(I-P_n)K(I-P_n)\| \geq \|(I-P_n)f(A)(I-P_n)\|$ for all $n\in \mathbb{N}$. Using the identity $\|S^{\ast}S\| = \|S\|^2$, it follows that $$ \|f(A)^{1/2}(I-P_n)\| \leq \|K^{1/2}(I-P_n)\| $$ Since $K^{1/2}$ is compact, this second term goes to $0$, and so $f(A)^{1/2}$ is compact, whence $f(A)$ is compact.

Hope this helps.

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  • $\begingroup$ Thank you very much. It does help a lot. $\endgroup$ – XYZ Jun 5 '17 at 13:39
  • $\begingroup$ If it has helped you a lot, you should validate the answer of @Prahlad Vaidyanathan $\endgroup$ – Jean Marie Jun 11 '17 at 17:48

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