Inner Product is Sesquilinear

Theorem
Let $\mathbb F$ be a subfield of $\C$.

Let $V$ be a inner product space over $V$ with inner product $\innerprod \cdot \cdot$.

Define the $f : V \times V \to \mathbb F$ by:


 * $\map f {x, y} = \innerprod x y$

for each $x, y \in V$.

Then $f$ is sesquilinear.

Proof
Let $\alpha \in \mathbb F$.

Let $x_1, x_2, y \in V$.

By the definition of the inner product, $f$ is linear in its first argument.

So, we have:


 * $\innerprod {\alpha x_1 + x_2} y = \alpha \innerprod {x_1} y + \innerprod {x_2} y$

From the definition of the inner product, we also have that $f$ is conjugate symmetric, so:

So we have:


 * $\innerprod {\alpha x_1 + x_2} y = \alpha \innerprod {x_1} y + \innerprod {x_2} y$

and:


 * $\innerprod y {\alpha x_1 + x_2} = \overline \alpha \innerprod y {x_1} + \innerprod y {x_2}$

for $\alpha \in \mathbb F$ and $x_1, x_2, y \in V$.

So $f$ is sesquilinear.