Friday 6 September 2019

My proof of Cauchy functional equation?



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?


Answer



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.


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