Spectral radius as an upper bound to norm

Question

The operator norm of a matrix is an upper bound for the spectral radius, and equality holds in particular when the matrix is Hermitian. One proof of this uses the spectral theorem. Is there a direct (simple) proof, specially one that generalizes to infinite dimensional spaces?

Answer 1

Recall the spectral radius formula $r(T) = \lim_{n\to\infty} \|T^n\|^{\frac 1 n}$ which is valid in any $C^*$algebra. Then, also $$r(T) = \lim_{n\to\infty} \|T^{2^n}\|^{\frac{1}{2^n}}= \lim_{n\to\infty} \|T\|^{\frac{2^n}{2^n}}=\|T\|,$$

where $\|T^{2^n}\|^{\frac{1}{2^n}}= \|T\|^{\frac{2^n}{2^n}}$ holds by induction for self-adjoint $T$.

Now, we want to prove the spectral radius formula without using the spectral theorem.

Let $T$ be a bounded operator. Then $r(T) = \lim_{n\to\infty} \|T^n\|^{\frac 1 n}$.

Proof. If $\lambda\in\sigma(T)$, then $\lambda^n\in \sigma(T^n)$ for any $n\in\mathbb N$. So, $|\lambda^n|\leq \|T^n\|$ and thus, $r(T)\leq \inf_{n} \|T^n\|^{\frac 1n}$.

Now consider the resolvent $R: \rho(T)\to L(H), \lambda \mapsto R(\lambda) = \frac{1}{\lambda}(1-\frac {x}{\lambda})$ for $\lambda\neq 0$.

For any $\Lambda\in L(H)^*$, $\Lambda \circ R$ is analytic on $\rho(T)\supset \{\lambda\in \mathbb C: |\lambda| > r(T)\} \supset \{\lambda\in\mathbb C: |\lambda|>\|T\|\}$. (easy to check)

So, $\Lambda \circ R$ has a Laurent expansion on $\{\lambda: |\lambda|>r(T)\}$. For $|\lambda|>\|T\|$, $$\Lambda\circ T = \sum_{n=0}^\infty \frac{\Lambda(T^n)}{\lambda^{n+1}}.$$

But then, this must also be valid on $\{\lambda: |\lambda| > r(T)\}$ (c.f. course in complex analysis). Now, fix $|\lambda|>r(T)$. Then $\frac{\Lambda(T^n)}{\lambda^{n+1}}\to 0$, so, $\frac{|\Lambda(T)^n|}{|\lambda|^{n+1}} \leq C$ for some $C>0$. Hence, $\limsup_n |\Lambda(T^n)|^{\frac 1n}\leq |\lambda|$ and thus, $\limsup_n |\Lambda(T^n)|^{\frac 1n}\leq r(T)$.

Now, fix $D>r(T)$, so $\limsup_n \left(|\frac{|\Lambda(T^n)|}{D^n}\right)^{\frac 1n} <1$. This implies that $\left(\Lambda(\frac{T^n}{D^n})\right)_{n=1}^\infty$ is bounded. So $\{\frac{T^n}{D^n}:n\in\mathbb{N}\}$ is weakly bounded and hence norm bounded. So there exits a constant $E>0$ such that $\|\frac{T^n}{D^n}\|\leq E$. Hence, $\limsup_n\|T^n\|^{\frac 1n} \leq D$. This gives us now $$\limsup_n \|T^n\|^{\frac 1n} \leq r(T)\leq\inf_n\|T^n\|^{\frac 1n} \leq \liminf_n \|T^n\|^{\frac 1n},$$ which proves the claim.

Answer 2

A proof can be obtained by looking at bilinear forms. For any bounded linear operator $B$ on a complex Hilbert space $H$, the norm of $B$ is $$ \|B\|=\sup_{\|x\|=\|y\|=1}|(Bx,y)| $$ In the case of a Hermitian operator $A$, the above reduces to $$ \|A\|=\sup_{\|x\|=1}|(Ax,x)| $$ You can find these facts in almost any elementary treatment of Functional Analysis; these facts do not rely on anything advanced, but the proofs use several nice standard tricks coming from the polarization identity $b(x,y)=\frac{1}{4}\sum_{n=0}^{3}i^nb(x+i^ny,x+i^ny)$ for sesquilinear forms (linear in the first, conjugate linear in the second.)

If $A$ is Hermitian, define real numbers $m_l$, $m_r$ by $$ m_l = \inf_{\|x\|=1}(Ax,x),\;\;\; m_r = \sup_{\|x\|=1}(Ax,x). $$ You can show that $\lambda \in\mathbb{R}\setminus[m_l,m_r]$ is in the resolvent set of $A$. And you can show that $m_l$, $m_r$ are in the spectrum of $A$. What you want follows from these facts because $$ \|A\|=\sup_{\|x\|=1}|(Ax,x)| = \max\{|m_l|,|m_r|\} $$ Hence, either $-\|A\|$ or $\|A\|$ is in the spectrum of $A$, and nothing outside $[\,-\|A\|,\|A\|\,]$ is in the spectrum.

These topics are generally covered at the beginning of a course on Functional Analysis.