Theorem: Norm of Hermitian Operator

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

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: 
{{begin-eqn}}
{{eqn | l = \size {\innerprod {A x} x_\HH}
      | o = \le
      | r = \norm {A x}_\HH \norm x_\HH
      | c = Cauchy-Bunyakovsky-Schwarz Inequality for Inner Product Spaces
}}
{{eqn | o = \le
      | r = \norm A \norm x_\HH^2
      | c = Fundamental Property of Norm on Bounded Linear Transformation
}}
{{eqn | r = \norm A
      | c = since $\norm x_\HH = 1$
}}
{{end-eqn}}
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: 
{{begin-eqn}}
{{eqn | l = 2 \innerprod {A u} v_\HH + 2 \innerprod {A v} u_\HH
      | r = 2 \paren {\innerprod {A u} v_\HH + \innerprod {A v} u_\HH}
}}
{{eqn | r = 2 \paren {\innerprod {A u} v_\HH + \innerprod v {A u}_\HH }
      | c = {{Defof|Hermitian Operator}}
}}
{{eqn | r = 2 \paren {\innerprod {A u} v_\HH + \overline {\innerprod {A u} v_\HH} }
      | c = using conjugate symmetry of the inner product
}}
{{eqn | r = 4 \map \Re {\innerprod {A u} v_\HH}
      | c = Sum of Complex Number with Conjugate
}}
{{end-eqn}}
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 Numerical Range 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: 
{{begin-eqn}}
{{eqn | l = 4 \map \Re {\innerprod {A u} v_\HH}
      | r = \innerprod {\map A {u + v} } {u + v}_\HH - \innerprod {\map A {u - v} } {u - v}_\HH
}}
{{eqn | o = \le
      | r = M \norm {u + v}_\HH^2 - M \norm {u - v}_\HH^2
      | c = applying the above inequality with $h = u + v$ and $h = u - v$ 
}}
{{eqn | o = \le
      | r = M \paren {\norm {u + v}_\HH^2 + \norm {u - v}_\HH^2}
      | c = {{Defof|Norm on Vector Space}}
}}
{{eqn | r = 2 M \paren {\norm u_\HH^2 + \norm v_\HH^2}
      | c = Parallelogram Law (Inner Product Space)
}}
{{end-eqn}}
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: 
{{begin-eqn}}
{{eqn | l = \norm v_\HH
      | r = \norm {\frac {\norm u_\HH} {\norm {A u}_\HH} A u}_\HH
}}
{{eqn | r = \frac {\norm u_\HH} {\norm {A u}_\HH} \norm {A u}_\HH
      | c = using positive homogeneity of the norm
}}
{{eqn | r = \norm u_\HH
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \innerprod {A u} v_\HH
      | r = \innerprod {A u} {\frac {\norm u_\HH} {\norm {A u}_\HH} A u}_\HH
}}
{{eqn | r = \frac {\norm u_\HH} {\norm {A u}_\HH} \innerprod {A u} {A u}_\HH
      | c = Inner Product is Sesquilinear
}}
{{eqn | r = \frac {\norm u_\HH} {\norm {A u}_\HH} \norm {A u}_\HH^2
      | c = {{Defof|Inner Product Norm}}
}}
{{eqn | r = \norm u_\HH \norm {A u}_\HH
}}
{{end-eqn}}
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: 
:$\norm A = M = \sup \set {\size {\innerprod {A h} h_\HH}: h \in \HH, \norm h_\HH = 1}$
{{qed}}