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I have found many threads about the inverse statement but none for this: (Part of exercise 6.24 in Brezis)

Let $T$ be linear, self-adjoint operator on a Hilbert space and assume that the spectrum of $T$ fulfills $\sigma(T) \subset [0,\infty)$. Prove that the operator is positive, i.e. $(Tu,u)\geq 0$.

So from know on I will set $\lambda < 0$ be a real number outside the spectrum, i.e. $T-\lambda I$ will be one-to-one and onto. I have tried the following:

  • Starting with $(Tu,u)$, we set $v$ such that $Tv-\lambda v = u$ (exists by assumption), and we have to prove that $(T^2v-\lambda Tv, Tv-\lambda v) \geq 0$. Now we can isolate a few interesting terms but I can't show positivity

  • To me it looks like the proof should go like this: We have some property which is known to be positive and somewhere inside this term we have a factor of $-\lambda(Tu, u)$ and if we assume $(Tu,u)$ (and thus $-\lambda(Tu, u)$) to be negative, we will get a contradiction for $0>\lambda \to 0$. I started with $0\geq (Tu-\lambda u, Tu- \lambda) = \|Tu\|^2 - 2\lambda (Tu, u) + \lambda^2 \|u\|^2$ but this did not get me anywhere.

Philipp Wacker
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1 Answers1

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In the context of Brezis' book, this is already proven in the text. In Proposition 6.9 he proves that, with $$ m=\inf\{\langle Tv,v\rangle: \|v\|=1\},\ \ M=\sup\{\langle Tv,v\rangle:\ \|v\|=1\}, $$ one has $\sigma(T)\subset [m,M]$, and $m,M\in \sigma(T)$. In your case $\sigma(T)\subset[0,\infty)$, so $m\geq0$.

Martin Argerami
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  • I am not assuming that: I take $\lambda$ to be in the resolvent set, which is defined as the set of real numbers such that $T-\lambda I$ is bijective (i.e. one-to-one and onto). But yes, I completely overlooked that I can use that $m \in\sigma(T)$, thanks! – Philipp Wacker May 22 '18 at 15:18
  • You are right. I have removed that sentence. – Martin Argerami May 22 '18 at 15:28