\[\left[ 0,1 \right]\]
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:
\[\begin{equation} \begin{aligned} \langle \alpha f(x), g(x) \rangle &= \int^1_0 (\alpha f(x)) g(x) dx \\ &= \int^1_0 f(x) (\alpha g(x)) dx \\ &= \langle f(x), \alpha g(x) \rangle \end{aligned} \end{equation}\]