Norm of Hermitian Operator

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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

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

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


Then the norm of $A$ satisfies:

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


Corollary

Suppose that:

$\forall h \in \HH: \innerprod {A h} h_\HH = 0$

Then $A$ is the zero operator $\mathbf 0$.


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:

\(\ds \size {\innerprod {A x} x_\HH}\) \(\le\) \(\ds \norm {A x}_\HH \norm x_\HH\) Cauchy-Bunyakovsky-Schwarz Inequality for Inner Product Spaces
\(\ds \) \(\le\) \(\ds \norm A \norm x_\HH^2\) Fundamental Property of Norm on Bounded Linear Transformation
\(\ds \) \(=\) \(\ds \norm A\) since $\norm x_\HH = 1$

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:

\(\ds 2 \innerprod {A u} v_\HH + 2 \innerprod {A v} u_\HH\) \(=\) \(\ds 2 \paren {\innerprod {A u} v_\HH + \innerprod {A v} u_\HH}\)
\(\ds \) \(=\) \(\ds 2 \paren {\innerprod {A u} v_\HH + \innerprod v {A u}_\HH }\) Definition of Hermitian Operator
\(\ds \) \(=\) \(\ds 2 \paren {\innerprod {A u} v_\HH + \overline {\innerprod {A u} v_\HH} }\) using conjugate symmetry of the inner product
\(\ds \) \(=\) \(\ds 4 \map \Re {\innerprod {A u} v_\HH}\) Sum of Complex Number with Conjugate

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 Operator is Hermitian iff Inner Product is Real, we have:

$\innerprod {A h} h_\HH$ is a real number.

So, from the definition of the absolute value, we have:

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

We can therefore see that:

$\innerprod {A \dfrac h {\norm h_\HH} } {\dfrac h {\norm h_\HH} }_\HH \le M$

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:

\(\ds 4 \map \Re {\innerprod {A u} v_\HH}\) \(=\) \(\ds \innerprod {\map A {u + v} } {u + v}_\HH - \innerprod {\map A {u - v} } {u - v}_\HH\)
\(\ds \) \(\le\) \(\ds M \norm {u + v}_\HH^2 - M \norm {u - v}_\HH^2\) applying the above inequality with $h = u + v$ and $h = u - v$
\(\ds \) \(\le\) \(\ds M \paren {\norm {u + v}_\HH^2 + \norm {u - v}_\HH^2}\) Definition of Norm on Vector Space
\(\ds \) \(=\) \(\ds 2 M \paren {\norm u_\HH^2 + \norm v_\HH^2}\) Parallelogram Law (Inner Product Space)


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:

\(\ds \norm v_\HH\) \(=\) \(\ds \norm {\frac {\norm u_\HH} {\norm {A u}_\HH} A u}_\HH\)
\(\ds \) \(=\) \(\ds \frac {\norm u_\HH} {\norm {A u}_\HH} \norm {A u}_\HH\) using positive homogeneity of the norm
\(\ds \) \(=\) \(\ds \norm u_\HH\)

and:

\(\ds \innerprod {A u} v_\HH\) \(=\) \(\ds \innerprod {A u} {\frac {\norm u_\HH} {\norm {A u}_\HH} A u}_\HH\)
\(\ds \) \(=\) \(\ds \frac {\norm u_\HH} {\norm {A u}_\HH} \innerprod {A u} {A u}_\HH\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds \frac {\norm u_\HH} {\norm {A u}_\HH} \norm {A u}_\HH^2\) Definition of Inner Product Norm
\(\ds \) \(=\) \(\ds \norm u_\HH \norm {A u}_\HH\)

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:

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

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Norm_of_Hermitian_Operator&oldid=627987"