# Showing that $\sigma(T)\subset \mathbb{R}_+$

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 $$\geq 0$$ 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..

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 $$\sigma(T) \subset \mathbb R_+$$.

• 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).
– user864806
Jun 9, 2021 at 7:40
• @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 $(dP_\lambda x, x) = |x(\lambda)|^2 d\lambda$. Jun 9, 2021 at 21:50
• 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
– user864806
Jun 10, 2021 at 6:21
• @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? Jun 10, 2021 at 10:10
• Now I manage to unddrsand it better! I think yes, so it can be clearer to future readers:) @Nikodem
– user864806
Jun 10, 2021 at 13:59