Norm of Hermitian Operator

Theorem
Let $\mathbb F \in \set {\R, \C}$.

Let $\HH$ be a Hilbert space over $\mathbb F$.

Let $A : \HH \to \HH$ be a bounded Hermitian operator.

Let $\innerprod \cdot \cdot_\HH$ denote the inner product on $\HH$.

Let $\norm \cdot_\HH$ be the inner product norm on $\HH$.

Then the norm of $A$ satisfies:


 * $\norm A = \sup \set {\size {\innerprod {A h} h_\HH}: h \in \HH, \norm h_\HH = 1}$

Proof
Let:


 * $M = \sup \set {\size {\innerprod {A h} h_\HH}: h \in \HH, \norm h_\HH = 1}$

To show that $M = \norm A$ we first show that:


 * $M \le \norm A$

We will then show that:


 * $\norm A \le M$

Let $x \in \HH$ be such that:


 * $\norm x_\HH = 1$.

Then we have:

So taking the supremum over:


 * $\set {x \in \HH : \norm x_\HH = 1}$

we have:


 * $M \le \norm A$

We will now show that:


 * $\norm A \le M$

Let $x, y \in \HH$.

Since $A$ is linear operator, we have:


 * $\innerprod {\map A {u + v} } {u + v}_\HH = \innerprod {A u + A v} {u + v}_\HH$

and:


 * $\innerprod {\map A {u - v} } {u - v}_\HH = \innerprod {A u - A v} {u - v}_\HH$

Using Inner Product is Sesquilinear, we have:


 * $\innerprod {\map A {u + v} } {u + v}_\HH = \innerprod {A u} u_\HH + \innerprod {A u} v_\HH + \innerprod {A v} u_\HH + \innerprod {A v} v_\HH$

and:


 * $\innerprod {\map A {u - v} } {u - v}_\HH = \innerprod {A u} u_\HH - \innerprod {A u} v_\HH - \innerprod {A v} u_\HH + \innerprod {A v} v_\HH$

We therefore obtain:


 * $\innerprod {\map A {u + v} } {u + v}_\HH - \innerprod {\map A {u - v} } {u - v}_\HH = 2 \innerprod {A u} v_\HH + 2 \innerprod {A v} u_\HH$

We have:

so we have:


 * $4 \map \Re {\innerprod {A u} v_\HH} = \innerprod {\map A {u + v} } {u + v}_\HH - \innerprod {\map A {u - v} } {u - v}_\HH$

Recall that for each $h \in \HH$ with $\norm h_\HH = 1$, we have:


 * $\size {\innerprod {A h} h_\HH} \le M$

by the definition of supremum.

From the definition of the absolute value, we have:


 * $\innerprod {A h} h_\HH \le M$

We can therefore see that:


 * $\ds \innerprod {A \frac h {\norm h_\HH} } {\frac h {\norm h_\HH} }_\HH \le M$

for each $h \in \HH \setminus \set 0$.

So from Inner Product is Sesquilinear, we obtain:


 * $\innerprod {A h} h_\HH \le M \norm h_\HH^2$

Note that if $h = 0$, from Inner Product with Zero Vector we have:


 * $\innerprod {A h} h_\HH = 0$

and:


 * $\norm h_\HH = 0$

so the inequality also holds for $h = 0$, and we obtain:


 * $\innerprod {A h} h_\HH \le M \norm h_\HH^2$

for all $h \in \HH$.

So, we obtain:

Now, take $u \in \HH$ with $A u \ne 0$.

Let:


 * $v = \dfrac {\norm u_\HH} {\norm {A u}_\HH} A u$

Then, we have:

and:

Since from the definition of a norm:


 * $\norm u_\HH \norm {A u}_\HH$ is a real number

we have that:


 * $\innerprod {A u} v$ is a real number.

So we have:


 * $4 \map \Re {\innerprod {A u} v} = 4 \norm u_\HH \norm {A u}_\HH$

giving:


 * $4 \norm u_\HH \norm {A u}_\HH \le 2 M \paren {\norm u_\HH^2 + \norm v_\HH^2}$

Since $\norm u = \norm v$, we have:


 * $4 \norm u_\HH \norm {A u}_\HH \le 4 M \norm u_\HH^2$

That is:


 * $\norm u_\HH \norm {A u}_\HH \le M \norm u_\HH^2$

for all $u \in \HH$ with $A u \ne 0$.

Note that we have:


 * $M \norm u_\HH^2 \ge 0$

so the inequality also holds for $u \in \HH$ with $A u = 0$.

So, for $u \in \HH \setminus \set 0$, we have:


 * $\norm {A u}_\HH \le M \norm u_\HH$

Since $\map A 0 = 0$, this inequality also holds for $u = 0$.

So:


 * $M \in \set {c > 0: \forall h \in \HH: \norm {A h}_\HH \le c \norm h_\HH}$

From definition 4 of the operator norm, we have:


 * $\norm A = \inf \set {c > 0: \forall h \in \HH: \norm {A h}_\HH \le c \norm h_\HH}$

so, from the definition of infimum we have:


 * $\norm A \le M$

Since:


 * $M \le \norm A$

and:


 * $\norm A \le M$

we have:


 * $\norm A = M = \sup \set {\size {\innerprod {A h} h_\HH}: h \in \HH, \norm h_\HH = 1}$