Norm of a Positive Operator

Question

For a positive operator $A\in B(\mathcal{H})$ on a complex Hilbert space $\mathcal{H}$, I want to prove that $\|A\|=\sup\{\lambda: \lambda \in \sigma(A)\}$, where $\sigma(A)$ is the spectrum of $A$.

So, in effect, I need to show that the spectral radius for a positive operator on a complex Hilbert space is equal to it's norm. I began by showing that the spectral radius $\rho(T)$ of any self-adjoint operator is less than or equal to $\|T\|$. I am stuck in the reverse inequality.

Answer 1

In context of the above comment, it suffices to show that the spectral radius $\rho(T)=\|T\|$ for a self-adjoint operator $T\in B(\mathcal{H})$. That $\rho(T)\leq \|T\|$ is a trivial observation. For the reverse inequality, we need to show that either $\lambda$ or $-\lambda$ is in $\sigma(T)$, where $\lambda= \|T\|$. This is same as showing that $\lambda I-T$ or $-\lambda I-T$ is not invertible. To this end, we prove that $T^2-\lambda^2I$ is not invertible, for if it is not then at least one of its factors $\lambda I-T$ or $-\lambda I-T$ is not invertible.

The proof is by the method of contradiction. Suppose, if possible, $T^2-\lambda^2I$ is invertible.Then there exists $\alpha >0$ such that $$\alpha I\leq(T^2-\lambda^2I)^*(T^2-\lambda^2I)=(T^2-\lambda^2I)^2$$.

Now,\begin{multline*} 0\leq <(\lambda^2I-T^2)^2x,x>= \|(\lambda^2I-T^2)x\|^2\leq\|\lambda^2I-T^2\|\|(\lambda^2I-T^2)x\|^2\\ \leq 2\lambda^2\|(\lambda^2I-T^2)x\|^2\leq 2\lambda^2<(\lambda^2I-T^2)x,x>. \end{multline*} This yields $$2\lambda^2(\lambda^2I-T^2)\geq (\lambda^2I-T^2)^2\geq\alpha I$$ which implies $$(2\lambda^4-\alpha)I\geq 2\lambda^2T^2$$. Observe that $$(2\lambda^4-\alpha)\|x\|^2=<(2\lambda^4-\alpha)Ix,x>\geq<(2\lambda^2T^2)x,x>\geq2\lambda^2\|Tx\|^2$$ for each $x\in \mathcal{H}$

This means that $\|Tx\|\leq \sqrt{\lambda^2-\frac{\alpha}{2\lambda^2}}$ which contradicts the assumption that $\|T\|=\lambda$. This proves that $T^2-\lambda^2I$ is not invertible which in turn completes the proof.

Answer 2

The spectral radius of $T$ is $\lim_n \|T^{n}\|^{1/n}$. However, $\|T^{2}\|=\|T\|^{2}$ for any selfadjoint $T$. This is because $$ \|T\|^{2}=\sup_{\|x\|=1}\|Tx\|^{2}=\sup_{\|x\|=1}(T^{2}x,x) \le \sup_{\|x\|=1}\|T^{2}x\|\|x\|=\|T^{2}\|. $$ On the other hand, $$ \|T^{2}\|\le \|T\|\|T\|=\|T\|^{2}. $$ You get the final result by noticing that $\|T^{2^{n}}\|^{1/2^{n}}=\|T\|$.