0
$\begingroup$

Let $H$ be a complex hilbert space and $T\in B(H)$.

  1. Prove that If $(Tx,x) \geq 0 ,\forall x\in H$ then $T^*=T$ and $\sigma(T)\subset \mathbb{R}_+$.

  2. Show that $T^*=T$ and $\sigma(T)\subset R_+$ iff $<Tx,x>\geq 0$ for every $x\in H$, using the spectral theorem for self adjoint and normal operators..

I think that the first part in 1 is immediate using another claim. Since $(Tx,x)\geq 0$ for all $x\in H$, so $(Tx,x)\in \mathbb{R}$ for all $x\in H$ and so $T$ is self-adjoint. What about the other part?

This post Spectrum of a positive operator in $B(H)$. suggests a proof for the inverse claim in 1..

$\endgroup$
1

1 Answer 1

1
+50
$\begingroup$

For 2, as you say, $(T x, x) \in \mathbb R$ for all $x \in H$ implies that $T = T^*$ is self-adjoint. By the spectral theorem we have $$ T = \int_{\lambda \in \sigma(T)} \lambda\, dP_\lambda \tag1 $$ with the projection-valued measure $P_\lambda$. Then, $$ (T x, x) = \int_{\lambda \in \sigma(T)} \lambda\,|x(\lambda)|^2 d\mu(\lambda) \geq 0 \qquad \forall x \in H \tag2 $$ where $x(\lambda)$ are the spectral coefficients of the vectors $x$ (w.r.t. the measure $P_\lambda$). Since $|x(\lambda)|^2$ is an arbitrary non-negative function and $d\mu(\lambda)$ is a positive measure the above condition leads to the conclusion $\lambda \geq 0\; \forall \lambda \in \sigma(T)$, i.e. $\sigma(T) \subset \mathbb R_+$.

The inverse is also true because for $\sigma(T) \subset \mathbb R_+$ we have for any non-negative function $|x(\lambda)|^2$ and for positive measure $d\mu(\lambda)$ that $$ \int_{\lambda \in \sigma(T)} \lambda\,|x(\lambda)|^2 d\mu(\lambda) \geq 0. $$ But $x(\lambda)$ can be in particular a spectral representation of a vector $x \in H$ so that (2) holds. Therefore, $T\geq 0$.

For 1, you have given the answer yourself (together with the link for the inverse). What is missing is to show that $\sigma(T) \subset \mathbb R_+$. We first obtain that $T=T^∗$ and hence $\sigma(T) \subset \mathbb R$. Then for each $\lambda \in \sigma(T)$ we have either $x \in H$ such that $T x = \lambda x$ and hence $(Tx,x) = \lambda \|x\|^2 \geq 0$ which implies $\lambda \geq 0$ or a sequence $x_n$ such that $\|x_n\| = 1$ and $\|(T-\lambda)x \| \rightarrow 0$ as $n\rightarrow \infty$. Then $$ 0 \leq (T x_n, x_n) = ((T-\lambda)x_n, x_n) + \lambda (x_n, x_n) \rightarrow \lambda $$ because $|((T-\lambda)x_n, x_n)| \leq \|((T-\lambda)x_n\| \|x_n\| \rightarrow 0$. In both cases the conclusion is the same that $\lambda \geq 0$ and hence $\sigma(T) \subset \mathbb R_+$.

$\endgroup$
7
  • $\begingroup$ Hi @Nikodem, thank you! That's the proof 2 right? (Which is supposed to be done using the spectral theorem..). In 1 I did not get why $\sigma(T)\subset (R_+)$. I did not much understand what is the role of $x(\lambda)$, isn't $(Tx,x)=(\int_{\lambda\in \sigma(T)} \lambda dP\lambda x,x)=(\int_{\lambda\in \sigma(T)} \lambda (dP\lambda x,x)$? (Maybe I am confusing with my symbols). $\endgroup$
    – user864806
    Jun 9, 2021 at 7:40
  • $\begingroup$ @Bestmat, this should be 2, yes. For 1, you have given answer yourself (together with the link for the inverse), I guess. $x(\lambda)$ comes from the scalar product $(y,x) = \int_\lambda y(\lambda) \overline{x(\lambda)} d\lambda$. So $(dP_\lambda x, x) = |x(\lambda)|^2 d\lambda$. $\endgroup$
    – Nikodem
    Jun 9, 2021 at 21:50
  • $\begingroup$ Yeah, the idea that I don't get why $\sigma(T)\subset R_+$ in 1, given that $(Tx,x)\geq 0$, can you explain it shortlt, lease $\endgroup$
    – user864806
    Jun 10, 2021 at 6:21
  • $\begingroup$ @Bestmat In 1. we obtain first that $T=T^*$ and hence $\sigma(T) \subset \mathbb R$. Then for each $\lambda \in \sigma(T)$ we have either $x \in H$ such that $T x = \lambda x$ and hence $(T x, x) = \lambda \| x \|^2 \geq 0$ which implies $\lambda \geq 0$ or a sequence $x_n$ such that the above holds in the limit $n \rightarrow \infty$. In both cases the conclusion is the same that $\sigma(T) \subset \mathbb R_+$. Should I add that part to my answer, too? $\endgroup$
    – Nikodem
    Jun 10, 2021 at 10:10
  • $\begingroup$ Now I manage to unddrsand it better! I think yes, so it can be clearer to future readers:) @Nikodem $\endgroup$
    – user864806
    Jun 10, 2021 at 13:59

You must log in to answer this question.