# Orthogonal complement of subspace of continuous functions and odd in the interval $[-1,1]$.

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?

• a) Prove that even functions are orthogonal to odd functions. b) Prove that every continuous function is the sum of an even and an odd continuous function. Feb 8, 2015 at 13:50
• I tried make this, but I have not had success to prove that $(f|g)=0$ when $f$ is odd and $g$ is pair. I found problems to solve the integral. Feb 8, 2015 at 14:00
• $g(-t)f(-t)=-g(t)f(t)$.So,when you split the integral into $\int_{-1}^{0} g(t)f(t)dt+\int_{0}^{1} g(t)f(t)dt$ , the two will cancel each other. We can say that odd functions lie in the orthogonal complement of even functions. Feb 9, 2015 at 8:16

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 $$V=U \overset{\perp}{\oplus} W$$