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Given a bilinear symmetric form $b(u,v)$ on a Hilbert space. I need to know some very basic facts. A reference where these are discussed would be greatly appreciated.

1) There exists a symmetric bounded linear operator $S$ such that $b(u,v)=\langle Su,v\rangle$

2) Also, I would like to know if it is true that the spectrum of a symmetric bounded linear operator is closed.

3) If 1 and 2 are true, it seems to me that there should not be a difference for a symmetric bounded bilinear form between being coercive or positive definite. Am I right?

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  • $\begingroup$ Norbert is right: the spectrum of every bounded operator is closed. However, the spectrum is not the same thing as the set of eigenvalues. Norbert's example shows that the set of eigenvalues need not be closed, which is part of the reason why coercive and positive definite symmetric operators are distinct concepts. $\endgroup$ – Martin May 5 '13 at 22:42
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1) There exist bijection between bounded bilinear operators and bounded opeartors. The proof of this fact requires Banach-Steinhaus theorem. If bilinear form is symmetric, then the respecitive opeartor is (obviously) symmetric too

2) Spectrum of any bounded operator is compact, and as the consequence closed

3) No, consider bilinear form $$ b:\ell_2\times\ell_2\to\mathbb{R}:(x,y)\mapsto\sum\limits_{i=1}^\infty n^{-1}x_iy_i $$ It is symmetric positive but not coercive.

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  • $\begingroup$ Actually, you can prove the bijection in 1) without Banach-Steinhaus, just using the Riesz representation theorem and Cauchy-Schwarz. The point is that you can construct contractions in both directions. In 3) you may want to add that $b$ is positive. $\endgroup$ – Martin May 5 '13 at 0:50
  • $\begingroup$ @Martin, you always catching me! $\endgroup$ – Norbert May 5 '13 at 1:05
  • $\begingroup$ Take it as a compliment :-) I look closely where I think it's worth it and I generally like your answers. $\endgroup$ – Martin May 5 '13 at 13:58
  • $\begingroup$ @MArtin, thank you. I learn a lot from your answers :-) $\endgroup$ – Norbert May 5 '13 at 14:44
  • $\begingroup$ @Norbert Dear Norbert, thank you very much for your response. However, I have a question about your example. It seems to me that the symmetric bounded operator corresponding to your bilinear form has, as a spectrum, the values $\frac{1}{n}$. It seems like it is not closed as it does not include the limiting value $0$. What am I missing? $\endgroup$ – Stephane May 5 '13 at 20:16
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although the last answer was roundabout 2 months ago, I would like to add something: Regarding your example: $$ b:\ell_2\times\ell_2\to\mathbb{R}:(x,y)\mapsto\sum\limits_{i=1}^\infty n^{-1}x_iy_i $$ for a given n. It would be $$ (x,x)=\frac{1}{n}\sum\limits_{i=1}^\infty x_i^2 $$ and so coercive, too, since you would not regard the limit of $$n\rightarrow\infty$$.

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