Integration as inner product

Question

Given two continuous functions, f and g, with domain $[0, 1]$ and target set $[0, 1]$, I need to ascertain if this is a valid inner product: $\langle f,g \rangle=\int_0^1 f(x) g(1 - x) \: dx$.

I suspected $\langle f,g \rangle$ would be different from $\langle g,f \rangle$, but apparently it isn't. How can than be?

Answer 1

Put $y=1-x$ to see that $\int_0^{1} f(x)g(1-x)dx=\int_0^{1} f(1-y)g(y)dy=\int_0^{1} g(x)f(1-x)dx$.

Other properties: $ \langle f, f \rangle \geq 0$ is clear. If $ \langle f, f \rangle =0$ then $\int_0^{1}f^{2}(x)dx=0$. By continuity of $f$ this implies $f(x)=0$ for all $x$. Can you check linearity of the inner product in each variable?

Answer 2

Because if $h$ is a map from $[0,1]$ into $\mathbb R$, then$$\int_0^1h(x)\,\mathrm dx=\int_0^1h(1-x)\,\mathrm dx.$$