The follow question was found on the Hoffman's book.
Let $V$ be the real inner product space consisting of the space of real-valued continuous functions on the interval, $-1\leq t \leq 1$, with the inner product
$(f|g)=\displaystyle \int_{-1}^{1} {f(t)g(t)}dt$
Let $W$ be the subspace odd functions, ie, functions satisfying $f(-t)=-f(t)$. Find the orthogonal complement of $W$.
I suppose that orthogonal complement of $W$ is the subspace of functions that satisfy $f(t)=f(-t)$. Someone have ideas to proof that?
Best Answer
First notice that for all $f\in V$ we have
$$f(t)=\underbrace{\frac12(f(t)+f(-t))}_{g(t)}+\underbrace{\frac12(f(t)-f(-t))}_{h(t)}$$ where obviously $g$ and $h$ are even and odd respectively. Moreover, since the only function which is simultaneously even and odd is the zero function then we get
$$V=U\oplus W$$ where $U$ is the subspace of even functions. Finally for $f\in U$ and $g\in W$ and with the change of variable $t=-x$ we easily get
$$(f\mid g)=0$$ hence we obtain $$V=U \overset{\perp}{\oplus} W$$