Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to find the all continuous functions $f\colon \mathbb{R}\to\mathbb{R}$ such that $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$

share|cite|improve this question
It's easy to see that $f(0)=0$. The solutions form a linear space, and include $f(x)=x$ and $f(x) = x^2$ but no other powers. I would guess that $f(x) = ax + bx^2$ is the general solution. – Robert Israel Aug 23 '12 at 20:12
A more descriptive title would be useful: as it stands, it is almost unconnected to the question! – Mariano Suárez-Alvarez Aug 23 '12 at 20:25
Setting $x=y=z$ and differentiating the equality many times and evaluating at zero shows Robert's solutions are all the entire ones. – Mariano Suárez-Alvarez Aug 23 '12 at 20:31
@MarianoSuárez-Alvarez Is it obvious then any continuous function satisfying that equality is differentiable? – JSchlather Aug 23 '12 at 20:38
If it were obvious I would have said «Robert's solutions are all solutions» and not what I wrote! – Mariano Suárez-Alvarez Aug 23 '12 at 20:39

This is just to supplement Mariano and Shaun's answer and remove the differentiability assumption. The idea is to use finite differences. Let $\delta_h f(x)=f(x+h)-f(x)$. Then writing the functional equation with $x$ replaced by $x+h$ and subtracting the original equation from it, we derive $$ \delta_hf(x)+\delta_hf(x+y+z)=\delta_hf(x+y)+\delta_hf(z+x). $$ Now doing the same with $y$, and subsequently with $z$, we get $$ \delta_\ell\delta_k\delta_hf(x)=0, $$ for all $h,k,\ell\in\mathbb{R}$ and $x\in\mathbb{R}$. In particular, the following limit exists $$ \delta_k\delta_hf'(x)=\lim_{\ell\to0}\frac{\delta_\ell\delta_k\delta_hf(x)}{\ell}=0. $$ Proceeding similarly, we conclude that $f'''(x)$ exists and equal to $0$ everywhere.

share|cite|improve this answer

All continuous solutions are of the form $f(x) = a x + b x^2$.

For given $z$, let $g(x) = f(x+z) - f(x) - f(z)$. The functional equation becomes $g(x+y) = g(x) + g(y)$. This is the Cauchy functional equation, and it is known that its only continuous solutions are $g(x) = c x$. Of course $c$ can depend on $z$. Now we need to solve $f(x+z) - f(x) - f(z) = c(z) x$. Since the left side is symmetric in $x$ and $z$, $c(z) x = c(x) z$, so $c(z) = k z$ for some constant $k$. Taking $z=-x$, and using $f(0)=0$, we get $- f(x) - f(-x) = - k x^2$, or $f(x) + f(-x) = k x^2$.

Note that if $f(x)$ is a solution of our equation, so is $f(-x)$, and by linearity so are the even and odd parts $ (f(x) + f(-x))/2$ and $(f(x) - f(-x))/2$. Thus it suffices to consider the two cases $f$ even and $f$ odd.

If $f$ is even, $f(x) + f(-x) = 2 f(x) = k x^2$, so $f(x) = (k/2) x^2$.

If $f$ is odd, $f(x) + f(-x) = 0$ so $k=0$. now we have $f(x+z) - f(x) - f(z) = 0$, which is again Cauchy's functional equation, and so $f(x) = a x$ for some constant $a$.

share|cite|improve this answer

Consider the left and right as functions of three variables.

  1. Take $\partial / \partial x$ of both sides (keeping track of the chain rule when necessary):

$$ f'(x) + f'(x+y+z) = f'(x+y) + f'(z+x) $$

  1. Take $\partial / \partial y$ of both sides of the result:

$$ f''(x+y+z) = f''(x + y) $$

  1. Take $\partial / \partial z$ of both sides of that:

$$ f'''(x+y+z) = 0 $$

Now substitute $y=z=0$ to find that $f'''(x)=0$. This shows that the only possible functions would be of the form $f(x) = ax^2 + bx + c$ for constants $a, b, c$. Then apply Robert's observation (see comments above) that $f(0) = 0$ (hence $c=0$), and that the set of functions form a linear space (hence all choices of $a, b \in \mathbb{R}$ are valid).

Hope this helps!

share|cite|improve this answer
I just saw Mariano's comment. :) – Shaun Ault Aug 23 '12 at 20:34
And what if $f$ is not differentiable? – Chris Eagle Aug 23 '12 at 20:37

Let $f: \mathbb R \to \mathbb R$ be a continous solution of the functional equation and let $g(x) := ax^2+bx$ where $a$ and $b$ are chosen in such a way that $f(-1) = g(-1)$ and $f(1) = g(1)$. Then $\tilde f := f-g$ is also a solution and we have $\tilde f(-1) = \tilde f(1) = 0$. We will show $\tilde f \equiv 0$, so that $f = g$.

By replacing $f$ with $\tilde f$ way may assume that $f$ is a solution with $f(-1) = f(1) = 0$, and we will show $f \equiv 0$. Let $Z = \{x \in \mathbb R: f(x) = 0 \}$. A priori we have $\{-1,+1\} \subseteq Z$. Plugging $(1,-1,x)$ into the functional equation yields $$f(x) = \frac{f(x+1) + f(x-1)}{2},$$ which implies $\mathbb Z \subseteq Z$. If we can show $$2x,x \in Z \Rightarrow \frac{x}{2} \in Z$$ we are finished since then every dyadic rational $a/2^n$ is in $Z$ and these are dense in $\mathbb R$ (here the continuity of $f$ is needed). So let $2x \in Z$ and $x \in Z$. Plug $(\frac{x}{2},\frac{x}{2},x)$ into the functional equation to get $$2f\left(\frac{x}{2}\right) + f(x) + f(2x) = f(x) + 2f\left(\frac{3}{2}x\right),$$ hence $f\left(\frac{x}{2}\right) = f\left(\frac{3}{2}x\right)$. Then plugging in $(\frac{x}{2},\frac{x}{2},\frac{x}{2})$ yields $$3f\left(\frac{x}{2}\right) + f\left(\frac{3}{2}x\right) = 3f(x),$$ hence $f(\frac{x}{2}) = 0$ and $\frac{x}{2} \in Z$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.