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I am trying to solve the following problem:

Find all the continuous functions $f:\mathbb{R} \to \mathbb{R}$ that satisfy if $a$, $b$ and $c$ are nonzero real numbers such that $a+b+c=0$ then $f(a)+f(b)+f(c)=0$.

I have only proved that $f(0)=0$: Let $f$ be a function that holds the condition in the statement, and let $x$ be a nonzero real number. Because of the hypothesis on $f$, we have that $$f(2x) + f(-x) + f(-x) = 0.$$ Thus, $f(2x) = -2f(-x)$. Since $f$ is continuous, we get that $f(0) = -2f(0)$ as $x\to 0$. It follows that $f(0)=0$.

I guess that the only function that fulfills the statement is $f(x)=mx$ for any real $m$, but I don't know how to show it.

Lord Vader
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  • Consider that $f(x) + f(-x) + f(0) = 0$. Then consider $c = -a-b$. – eyeballfrog Apr 09 '23 at 20:16
  • @eyeballfrog It follows that $f$ is an odd function. Since $f(x+y) = -f(-x) - f(-y)$, we have that $f(x+y) = f(x)+f(y)$. Therefore, it can be shown that $f(x) = \alpha x$ for some $\alpha\in\mathbb{R}$. – Lord Vader Apr 09 '23 at 20:20
  • @eyeballfrog However, I don't think that we can make sure that $f(x) + f(-x) + f(0) = 0$, because the hypothesis says that $a$, $b$ and $c$ are nonzero. – Lord Vader Apr 09 '23 at 20:22
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    Note that $f(x)+f(-x-\varepsilon)+f(\varepsilon)=0$ for $x\not=0$ and all positive $\varepsilon$ sufficiently small. Then use continuity. – Jens Schwaiger Apr 09 '23 at 20:29
  • Oh I missed the nonzero part. – eyeballfrog Apr 09 '23 at 21:59

3 Answers3

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You have already proved that $f(0)=0$ and (by @JensSchwaiger's comment) $f(-x)=-f(x).$ Therefore, your equation boils down to $$f(a+b)=f(a)+f(b)$$ (whenever $a,b,a+b\ne0,$ but also when at least one of them is $0,$ by the two previous properties). This is Cauchy's functional equation, whose continuous solutions are indeed the linear functions $f(x)=mx,$ as you guessed.

Anne Bauval
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It's easy to prove that $$\begin{align} &f(0) = 0\\ &f(-x) = -f(x) \end{align}$$ And for all $n\in\mathbb{N}$, we have $$f(nx)+f(-(n-1)x)+f(-x)=0$$$$ \Longrightarrow f(nx) = f(x)+f((n-1)x)=...=nf(x) \tag{1}$$ From $(1)$, we deduce easily, by replace $x = \frac{y}{n}$ that $$f\left(\frac{1}{n}y\right)=\frac{1}{n}f\left(y\right) \tag{2}$$ From $(1),(2)$, all $q \in \mathbb{Q}^+$, there exists a way to write $q = \frac{m}{n}$ with $m,n \in \mathbb{N}$ $$f(q) = f\left(\frac{m}{n}\right)=\frac{m}{n}f(1) = q f(1)\tag{3}$$ Finally, for all $x\in \mathbb{R}$, as $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists a sequence $(q_i)_{i=1,...,+\infty}$ such that $$\lim_{i\infty}q_i=r$$ By the continuity of the function $f$, and by using $(3)$, we deduce $$f(r) = f(\lim_{i\infty}q_i)=\lim_{i\infty}f(q_i)=\lim_{i\infty}\left(q_i f(1)\right) = rf(1) $$

Conclusion: the solution is $f(x) = a\cdot x$.

NN2
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In the question we are given a functional equation

Find all the continuous functions $f:\mathbb{R} \to \mathbb{R}$ that satisfy if $a$, $b$ and $c$ are nonzero real numbers such that $a+b+c=0$ then $f(a)+f(b)+f(c)=0$.

For convenience, define the sequence $\,s_n := f(n x)\,$ where $\,x\ne0\,$ is real. The functional equation implies $\,s_2 = -2s_{-1}\,$ from $\,(a,b,c)=(-x,-x,2x).\,$ Similarly $\,s_{-2} = 2s_1,\;\; s_4 = -2s_{-2} = 4s_1.\,$ Next, $\,s_3 = 2s_1-s_{-1}\,$ from $\,(a,b,c)=(-x,-2x,3x)\,$ and $\,s_4 = s_1-3s_{-1}\,$ from $\,(a,b,c)=(-x,-3x,4x).\,$ The two values of $\,s_4 = 4s_1 = s_1-3s_{-1}\,$ imply that $\,s_{-1} = -s_1.\,$ Next, $\,s_n = s_1 - s_{-n+1}\,$ from $\,(a,b,c) = (-x,(-n+1)x,n\,x).\,$ Similarly, $\,s_{-n} = -s_1 - s_{n-1}.\,$ By induction this implies that $\,s_n = n\,s_1\,$ for all integer $n\ne0.\,$ In other words, $\,f(n\,x) = n\,f(x).$

Since this is true for all $\,x\ne0\,$ and $\,f\,$ is continuous, by Cauchy's functional equation, we get that $\,f(x) = xf(1)\,$ for all real $\,x.$

Somos
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