Prove that any function can be written as the sum of an even function and an odd function.

Question

I understand some of the basic concepts that surrounds even and odd functions but this question just stumped me and I'm not sure on how to tackle it. Any Starting points/methods would be helpful

Prove that any function can be written as the sum of an even function and an odd function.

Answer 1

Suppose it were true. What could the functions possibly be?

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 $g(x) = \frac {f(x) + f(-x)}2$.

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 $h(x)= f(x) - g(x) = f(x) - \frac {f(x) + f(-x)}2 = \frac {f(x) - f(-x)}2$ 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 $g(x) = \frac {f(x) + f(-x)}2$ and $h(x) = \frac {f(x) - f(-x)}2$.

Answer 2

Notice that you can write any function as:

$f(x)= (1 / 2) f(x) + (1 / 2) f(x) + (1 / 2) f(-x) - (1 / 2) f(-x).$

let $g(x) = (1 / 2) (f(x) + f(-x)).$ notice that $g(-x)=(1 / 2) (f(-x) + f(x))=g(x)$

so $g$ is even.

let $h(x) = (1 / 2) (f(x) - f(-x)).$

notice that $h(-x) = (1 / 2) (f(-x) - f(x))=(1 / 2) (f(x) - f(-x)) = - h(x)$

so $h$ is odd.

Add the 2 functions: $(1 / 2) (f(x) + f(-x))+(1 / 2) (f(x) - f(-x))=(1 / 2) f(x) + (1 / 2) f(x) + (1 / 2) f(-x) - (1 / 2) f(-x)=f(x).$