Operator is Hermitian iff Inner Product is Real

Theorem
Let $\HH$ be a Hilbert space over $\C$ with inner product $\innerprod \cdot \cdot_\HH$.

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

Then $A$ is Hermitian :


 * $\forall h \in \HH: \innerprod {A h} h_\HH \mathop \in \R$

Necessary Condition
Suppose that $A$ is Hermitian.

Then:


 * $A^* = A$

where $A^*$ denotes the adjoint of $A$.

Let $x \in \HH$.

Then, by the definition of the adjoint, we have:


 * $\innerprod {A x} x_\HH = \innerprod x {A x}_\HH$

From the conjugate symmetry of the inner product, we have:


 * $\innerprod x {A x}_\HH = \overline {\innerprod {A x} x_\HH}$

So:


 * $\innerprod {A x} x_\HH = \overline {\innerprod {A x} x_\HH}$

So, from Complex Number equals Conjugate iff Wholly Real:


 * $\innerprod {A x} x_\HH$ is a real number.

Sufficient Condition
Suppose that:


 * $\innerprod {A x} x_\HH$ is a real number for each $x \in \HH$.

Let $\alpha \in \C$.

Let $x, y \in \HH$.

We have:

We have that:


 * $\innerprod {\map A {x + \alpha y} } {x + \alpha y}_\HH$

is a real number.

Note that both:


 * $\innerprod {A x} x_\HH$

and:


 * $\size \alpha^2 \innerprod {A y} y_\HH$

are also real numbers.

So, we must have that:


 * $\overline \alpha \innerprod {A x} y_\HH + \alpha \innerprod {A y} x_\HH$

is a real number.

We therefore have:

Setting $\alpha = 1$, we have:


 * $\innerprod {A x} y_\HH + \innerprod {A y} x_\HH = \innerprod {A^* y} x_\HH + \innerprod {A^* x} y_\HH$

Setting $\alpha = i$, we have:


 * $-i \innerprod {A x} y_\HH + i \innerprod {A y} x_\HH = i \innerprod {A^* y} x_\HH - i \innerprod {A^* y} x_\HH$

Dividing by $i$ we have:


 * $-\innerprod {A x} y_\HH + \innerprod {A y} x_\HH = \innerprod {A^* y} x_\HH - \innerprod {A^* y} x_\HH$

So:


 * $2 \innerprod {A y} x_\HH = 2 \innerprod {A^* y} x_\HH$

giving:


 * $\innerprod {A y} x_\HH = \innerprod {A^* y} x_\HH$

From Inner Product is Sesquilinear:


 * $\innerprod {A y - A^* y} x_\HH = 0$

Setting $x = A y - A^* y$ we have:


 * $\innerprod {A y - A^* y} {A y - A^* y}_\HH = 0$

From the positiveness of the inner product we have:


 * $A y - A^* y = 0$

so:


 * $A y = A^* y$

Since $y \in \HH$ was arbitrary, we have:


 * $A = A^*$

That is:


 * $A$ is Hermitian.