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?

  • $\begingroup$ 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. $\endgroup$ Feb 8, 2015 at 13:50
  • $\begingroup$ 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. $\endgroup$
    – Pitágoras
    Feb 8, 2015 at 14:00
  • 1
    $\begingroup$ $g(-t)f(-t)=-g(t)f(t)$.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. $\endgroup$
    – Srinivas K
    Feb 9, 2015 at 8:16

1 Answer 1


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


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