## An Inner Product on the Set of Functions

We can define an inner product on the set of real valued continuous function defined on the interval
$\left[ 0,1 \right]$
by

$\langle f(x), g(x) \rangle = \int^1_0 f(x) g(x) dx$
We check that this definition satisfies the conditions for an inner product.
The inner product is symmetric:
$\langle f(x), g(x) \rangle = \int^1_0 f(x) g(x) dx = \int^1_0 g(x) f(x) dx = \langle g(x), f(x) \rangle$

The inner product is positive definite:
$\langle f(x), f(x) \rangle = \int^1_0 f(x) f(x) dx \geq \int^1_0 0 dx =0$
and
$\langle f(x), f(x) \rangle =0 \leftrightarrow f(x)=0$
since if
$f(x) \neq 0$
on some subset
$\left(a,b \right) \in \left[0,1 \right]$
then
$\left| f(x) \right| \geq \epsilon \ gt 0$
so
$\langle f(x), f(x) \rangle \geq (b-a) \epsilon^2 \gt 0$

The inner product is linear in both arguments:
\begin{aligned} \langle \alpha f(x), g(x) \rangle &= \int^1_0 (\alpha f(x)) g(x) dx \\ &= \int^1_0 f(x) (\alpha g(x)) dx \\ &= \langle f(x), \alpha g(x) \rangle \end{aligned}