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We have the functional equation, $$ f(x + y) = f(x) + g(y). \tag{1}\label{eq1} $$ For all $a, b \in \mathbf{R}$, $f(x) = ax + b\ $and$\ g(x) = ax$ satisfy \eqref{eq1}.

My question is:

  • If we only require that $f$ satisfies \eqref{eq1}, then is $\left(f(x),\ g(x)\right) = \left(ax + b,\ ax\right)$ the only solution?
  • If we only require that $f$ is continuous and satisfies \eqref{eq1}, then is $\left(f(x),\ g(x)\right) = \left(ax + b,\ ax\right)$ the only solution?

I did ask this question here, but I was very confused about the question back then and couldn't formulate it properly. So, the answers there don't answer my questions.

Mostafizur Rahman
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    Note that with $x=0$, we get that $f(y)=f(0)+g(y)$ for any $y\in \mathbb{R}$, so $f$ and $g$ are only different by a constant. More specifically, $f(x+y)=f(x)+f(y)-f(0)$ for any $x,y\in \mathbb{R}$. So we do not need $g$. Moreover, if we define $h(x)=f(x)-f(0)$, then $$h(x+y)=f(x+y)-f(0)=[f(x)+f(y)-f(0)]-f(0)=h(x)+h(y).$$ So we can work with $h$ instead of $f$, and then translate back once finished. So your question amounts to: If $h(x+y)=h(x)+h(y)$ for all reals $x,y$, then does there exist a real number $a$ such that $h(x)=ax$ for all $x\in \mathbb{R}$? – user114263 Oct 26 '23 at 14:34
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    The answer is no generally, but yes if we assume $f$ is continuous. – user114263 Oct 26 '23 at 14:34
  • @user114263 Thank you. It's exactly what I needed. – Mostafizur Rahman Oct 26 '23 at 23:52

1 Answers1

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This turns out to be a fairly deep question, and the answer is no unless you make some assumptions on $f$.

First, as pointed out in the comments, by setting $x=0$ one sees that $f$ and $g$ differ by a constant, so it suffices to consider the functional equation $h(x+y) = h(x) + h(y)$, where $h(x) = f(x)-f(0)$.

Now, if $h$ is continuous, then $h(x) = h(1)x$. Here’s a quick proof: One can show that for all natural numbers $n,m$ we have $h(n/m)= (n/m)h(1)$. So if we take a sequence of rational numbers $r_n$ converging to some arbitrary $x\in\mathbb{R}$, by continuity we have $\lim_{n\to\infty} h(r_n) = h(\lim_{n\to\infty}r_n) = h(x)$. But $h(r_n) = r_n h(1)$, so taking limits on both sides shows that $h(x)=xh(1)$, as desired.

However, using the axiom of choice one can construct discontinuous $h$ satisfying $h(x+y)=h(x)+h(y)$, and these won’t be linear. However, if one assumes that $h$ is Lebesgue measurable, then one can show $h$ is continuous and hence linear.

kieransquared
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