Functional equation high school problem $f(x) = f(-x) - 2x $

Question

Let \begin{cases} f: \mathbb R \to \mathbb R\\ f(x) = f(-x) - 2x \end{cases}

I can see that one solution is: $f(x) = -x$ , but I have no idea how to prove that this is the only solution.

Thank you for your help

Answer 1

Separate $f$ in its even and odd parts: $$f_0(x):={f(x)+f(-x)\over 2},\qquad f_1(x):={f(x)-f(-x)\over2}\ .$$ Then $$f(x)=f_0(x)+f_1(x),\qquad f(-x)=f_0(x)-f_1(x)\ .$$ In terms of $f_0$ and $f_1$ your functional equation looks as follows: $$f_0(x)+f_1(x)=f_0(x)-f_1(x)-2x\ ,$$ or $f_1(x)=-x$, and $f_0(x)$ is arbitrary. It is easy to check that $$f(x):=-x+ f_0(x)$$ with an arbitrary even $f_0$ is a solution to your problem.

Answer 2

Hint: what happens when you try $f(x) = -x + x^2$? Another way to look at it: what part of the function, $f$, is $\frac{f(x) - f(-x)}{2}$? Final hint: a function, $g$, that satisfies: $$g(x) = g(-x),$$ is said to be even. What is $\frac{g(x) + g(-x)}{2}$?

Answer 3

$f(x) - f(-x) = [f(-x) - 2x]- f(-x) = -2x$

i.e. $f(-x) = f(x) + 2x$

Suppose

for $x > 0$, $f(x) = g(x)$ where $g(x)$ is any function we darned well please.

Then $x < 0$, $f(x) = g(|x|) + 2|x|= g(-x) - 2x$.

If $x = 0$ ... well, $f(0) = f(-0) - 2*0 = f(0)$ so $f(0)$ can be anything at all.

$f(x)$ satisfies the conditions just fine.

So for example we can have

$f(x) = \begin{cases} f(x) = (e^x + x^3 + x^2)\sin x; x > 0;\\ 10^{100}*\pi; x = 0;\\ (e^{-x} + (-x)^3 + (-x)^2)\sin(-x) - 2x; x < 0 \end{cases}$

will also work.

In general any function:

$f(x) = \begin{cases} f(x) = g(x); x > 0;\\ M; x = 0;\\ g(-x) - 2x; x < 0 \end{cases}$

will do.

For symmetry and aesthetics we can do this as:

$f(x) = \begin{cases} f(x) = h(x)-x; x > 0;\\ M; x = 0;\\ h(-x) - x; x < 0 \end{cases}$

where the new $h(x) = g(x) + x$.

Now if the function needs to be continuous are isn't allowed to be stepwise defined, that's another story.

In either case we can combine:

$f(x) = \begin{cases} f(x) = h(x)-x; x > 0;\\ h(0) = h(0) - 0; x = 0;\\ h(|x|) - x; x < 0 \end{cases}$

as $f(x) = h(|x|) - x = j(x) - x$ where $j(x) = h(|x|)$ is any function where $j(-x) = j(x)$. (If $f$ is required to be continuous, so will $j$.)