Question

Prove that any arbitrary function Ψ(x) can always be written as a sum of an even and an odd function.

Answer 1

Here, = f(x)

Let f(x) = g(x)+h(x) ,where g is even and h is odd.

Then f(−x) = g(x)−h(x)

And f(x)+f(−x)= 2g(x) and so, g(x)=(f(x)+f(−x))/2

The is indeed an even function that will work, and is apparently a unique such even function.

If we can verify that h(x)=f(x)−g(x) is odd we will have found that not only is this possible, but we will have found a unique odd/even pair for which this can be true.

Now h(x)=f(x)−g(x)=f(x)−(f(x)+f(−x)/2) is indeed an odd function.

So we are done:

f(x)=g(x)+h(x) where g is even and h is odd is uniquely expressed when g(x)=(f(x)+f(−x))/2 and h(x)=(f(x)−f(−x))/2.