Proving orthogonality of odd and even functions.?

Question

The question basically asks to prove that the space of square integrable functions on the interval (-pi,pi) can be expressed as an orthogonal direct sum of odd and even functions. (I.e. given any function f belonging to the space, the function may be written as f(x) = k(x) + m(x), where k(x) is even and m(x) is odd, and where k(x) is orthogonal to m(x). )

My approach so far: I attempted to prove the orthogonality of k(x) and m(x) by expressing k(x) = (f(x) + f(-x)) /2 and m(x) = (f(x) - f(-x))/2 and taking their dot product (i.e. taking the integral of k(x)*m(x) over the interval of negative infinity to infinity). This approach did not seem to work as I ended up with the integral taken from -infinity to infinity of (f(x))^2 - (f(-x))^2, which I am not sure how to show is equal to zero.

Answer 1

If we have two functions f(x) and g(x), such that f(x) is symmetric with respect to x = 0, and g(x) is antisymmetric, etc, then obviously the product f(x) g(x) is also antisymmetric, which means the integral from negative infinity to infinity is going to be 0.

Let's say f(x) = (h(x) + (h(-x)) and g(x) = (h(x) - h(-x)), so that f(x) g(x) = (h(x))^2 - (h(-x))^2. Notice that this is also an antisymmetric function? The integral from negative infinity to infinity is the same for both (h(x))^2 and (h(-x))^2, so they cancel each other out.

Answer 2

it form of feels to me that I heard somebody say sooner or later: "If we've 2 applications f(x) and g(x), such that f(x) is symmetric with appreciate to x = 0, and g(x) is antisymmetric, etc, then for sure the product f(x) g(x) is likewise antisymmetric, which potential the necessary from detrimental infinity to infinity is going to be 0. permit's say f(x) = (h(x) + (h(-x)) and g(x) = (h(x) - h(-x)), so as that f(x) g(x) = (h(x))^2 - (h(-x))^2. observe that it is likewise an antisymmetric function? The necessary from detrimental infinity to infinity is the comparable for the two (h(x))^2 and (h(-x))^2, so they cancel one yet another out. "