Although I have not quite studied functional equations, I came upon Cauchy functional equation and tried to prove it. Here is what I have done:

We are given the condition, $f(x+y)=f(x)+f(y)$.

So, for some constant $a$ and another constant $f(a)=b$, we have $f(x+a)=f(x)+b$.

Differentiating both sides wrt $x$, we have $f'(x+a)=f'(x)$.

But this result is valid for any constant $a$, and hence $f'(x)=c$, for some constant $c$. This gives us $f(x)=cx+d$. Putting this result into original condition, we have $c(x+y)+d = cx +d+ cy +d$. Hence $d=0$ and $f(x)=cx$, for some constant $c$.

Is my proof right or are there holes which needs to be filled? I asked here because it is different from the proof I found. My main concern is that I have assumed that the function is differentiable. Is there any elementary way to patch up for non differentiable functions? What about some other points?

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    $\begingroup$ The assumption of differentiability is very strong and indeed unnecessary. Instead, try to derive a solution by assuming only continuity (for the record: without continuity, there are other, wacky solutions) $\endgroup$ – Julien Godawatta Jan 3 '14 at 12:44

There are many conditions you could assume in order to get the linear function. For instance continuity.

Hint: Naturals $\to$ Rationals $\to$ Reals (Continuity)

You could also replace continuity with the following: $f(xy) = f(x)f(y)$

These are the ones I know of that guarantee a linear solution. However as it is, without another assumption, it is wrong to say that only linear functions satisfy above. I'll post a reference when I get one.

Edit: Your proof is correct if you assume differentiability everywhere.

  • $\begingroup$ I have seen that proof using the hint you provided. I wanted to know about the rights and wrongs of my proof. But I will consider that differentiability is very strong and use only continuity in such proofs in future. And are there some elementary means to establish or prove that there exists the wacky solutions. $\endgroup$ – Sawarnik Jan 3 '14 at 13:16
  • $\begingroup$ I have updated my answer. $\endgroup$ – Gautam Shenoy Jan 3 '14 at 13:26
  • $\begingroup$ Also, using the hint you gave, I know only how to prove for rationals, how can i extend it to the reals. The proof over rationals I have studied from the Wikipedia article. $\endgroup$ – Sawarnik Jan 3 '14 at 13:33
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    $\begingroup$ @Sawarnik Prove the more general statement: if $f,g:\;\mathbb{R}\to\mathbb{R}$ are continuous and $f=g$ over $\mathbb{Q}$ then $f=g$ over $\mathbb{R}$, by considering that any real is the limit of a sequence of rationals. Alternatively, we can make an even weaker assumption and still force $f(x)=ax$. Namely, if there is an interval over which $f$ is bounded below, then this is the only solution. $\endgroup$ – Julien Godawatta Jan 3 '14 at 14:18

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