Norm of Compact Hermitian Operator is Equal to Greatest Modulus of Eigenvalue

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Theorem

Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ be a Hilbert space over $\C$.

Let $T : \HH \to \HH$ be a compact Hermitian operator.

Let $\map {\sigma_p} T$ be the point spectrum of $T$.


Then:

$\norm T_{\map \BB \HH} = \max \set {\cmod \lambda : \lambda \in \map {\sigma_p} T}$

where $\norm \cdot_{\map \BB \HH}$ is the norm on the space of linear transformations.


Proof

Suppose that $T = 0$.

Then $\norm T_{\map \BB \HH} = 0$ from Norm Axiom $\text N 1$: Positive Definiteness.

Then $-\lambda I$ is injective if and only if $\lambda \ne 0$.

So $\map {\sigma_p} T = \set 0$ in this case and so:

$\max \set {\cmod \lambda : \lambda \in \map {\sigma_p} T} = 0 = \norm T_{\map \BB \HH}$.


Now let $T \ne 0$ so that $\norm T_{\map \BB \HH} \ne 0$.

From Norm of Hermitian Operator:

$\norm T_{\map \BB \HH} = \sup \set {\cmod {\innerprod {T x} x} : x \in \HH, \, \norm x = 1}$

So by the definition of supremum, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ with $\norm {x_n} = 1$ for each $n \in \N$ such that:

$\ds \norm T_{\map \BB \HH} \ge \cmod {\innerprod {T x_n} {x_n} } \ge \norm T_{\map \BB \HH} - \frac 1 n$

Then by the Squeeze Theorem:

$\cmod {\innerprod {T x_n} {x_n} } \to \norm T_{\map \BB \HH}$

From Operator is Hermitian iff Inner Product is Real, $\innerprod {T x_n} {x_n} \in \R$ for each $n \in \N$.

Since $\cmod {\innerprod {T x_n} {x_n} } \to \norm T_{\map \BB \HH}$, a convergent subsequence of $\sequence {\innerprod {T x_n} {x_n} }_{n \mathop \in \N}$ converges to either $\norm T_{\map \BB \HH}$ or $-\norm T_{\map \BB \HH}$, from Modulus of Limit.

Now, either $\innerprod {T x_n} {x_n} \ge 0$ or $\innerprod {T x_n} {x_n} \le 0$ for infinitely many $n$.

Let $\sequence {x_{n_j} }_{j \mathop \in \N}$ be a subsequence of $\sequence {x_n}_{n \mathop \in \N}$ such that $\innerprod {T x_{n_j} } {x_{n_j} }$ is the same sign for each $j$.

Then:

$\innerprod {T x_{n_j} } {x_{n_j} } \to \alpha$

where $\alpha \in \set {\norm T_{\map \BB \HH}, - \norm T_{\map \BB \HH} }$.

Since $T$ is compact and $\sequence {x_{n_j} }_{j \mathop \in \N}$ is bounded, there exists a subsequence $\sequence {x_{n_{j_k} } }_{k \mathop \in \N}$ such that $\sequence {T x_{n_{j_k} } }_{k \mathop \in \N}$ converges.

Say:

$T x_{n_{j_k} } \to y$ as $n \to \infty$ for some $y \in \HH$.

Now we have:

\(\ds \norm {T x_{n_{j_k} } - \alpha x_{n_{j_k} } }^2\) \(=\) \(\ds \innerprod {T x_{n_{j_k} } - \alpha x_{n_{j_k} } } {T x_{n_{j_k} } - \alpha x_{n_{j_k} } }\) Definition of Inner Product Norm
\(\ds \) \(=\) \(\ds \innerprod {T x_{n_{j_k} } } {T x_{n_{j_k} } } - \alpha \innerprod {x_{n_{j_k} } } {T x_{n_{j_k} } } - \alpha \innerprod {T x_{n_{j_k} } } {x_{n_{j_k} } } + \alpha^2 \innerprod {x_{n_{j_k} } } {x_{n_{j_k} } }\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds \norm {T x_{n_{j_k} } }^2 - 2 \alpha \innerprod {T x_{n_{j_k} } } {x_{n_{j_k} } } + \alpha^2 \norm {x_{n_{j_k} } }^2\) Definition of Inner Product Norm, Definition of Hermitian Operator
\(\ds \) \(=\) \(\ds \alpha^2 - 2 \alpha \innerprod {T x_{n_{j_k} } } {x_{n_{j_k} } } + \alpha^2\) Fundamental Property of Norm on Bounded Linear Transformation
\(\ds \) \(=\) \(\ds 2 \alpha \paren {\alpha - \innerprod {T x_{n_{j_k} } } {x_{n_{j_k} } } }\)

We have:

$\innerprod {T x_{n_{j_k} } } {x_{n_{j_k} } } \to \alpha$ as $n \to \infty$

so:

$\norm {T x_{n_{j_k} } - \alpha x_{n_{j_k} } }^2 \to 0$ as $n \to \infty$

So:

$T x_{n_{j_k} } - \alpha x_{n_{j_k} } \to 0$ as $n \to \infty$.

Since $T x_{n_{j_k} } \to y$, we have:

$\alpha x_{n_{j_k} } \to y$ as $n \to \infty$.

So, since $\alpha \ne 0$, we have:

$\ds x_{n_{j_k} } \to \frac y \alpha$ as $n \to \infty$.

Let:

$\ds x = \frac y \alpha$

Since $\norm {x_n} = 1$ for each $n \in \N$, we have $\norm x = 1$ from Modulus of Limit: Normed Vector Space.

Now, from Continuity of Linear Transformations, $T$ is continuous.

From Sequential Continuity is Equivalent to Continuity in Metric Space, $T$ is sequentially continuous.

So, we have:

\(\ds T x\) \(=\) \(\ds \lim_{j \mathop \to \infty} T x_{n_{j_k} }\)
\(\ds \) \(=\) \(\ds y\)
\(\ds \) \(=\) \(\ds \alpha x\)

So we have $\alpha \in \map {\sigma_p} T$.

So from Spectrum of Bounded Linear Operator contains Point Spectrum, we have $\alpha \in \map \sigma T$.

On the other hand, we have:

$\map \sigma T \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm T_{\map \BB X} }$

as shown in Spectrum of Bounded Linear Operator is Compact.

So we must have:

$\alpha = \max \set {\cmod \lambda : \lambda \in \map \sigma T} = \max \set {\cmod \lambda : \lambda \in \map {\sigma_p} T}$

$\blacksquare$


Sources

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