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How do one prove the uniqueness of a functional equation. (elementary) not functional analysis class...

For example, if we have $f(x+y)=x+f(y)$ and $f(0)=1$. Letting y=0, we obtain $f(x)=x+1.$ But how do you prove that the function I derived is just the function satisfies the above relationship.

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    $\begingroup$ You've got a necessary condition on your function (ie f(x) = x+1 ) , now you have to check that this very function satisfies all the conditions you've started with. If yes, it is the only one, if not there is no solutions. That's a type of reasoning called (I translate from french) analysis and synthesis. $\endgroup$ – mvggz Oct 18 '14 at 14:59
  • $\begingroup$ Why would you need to check back to see if the derived function does satisfy? I mean, what does it mean by IF a function $f$ satisfies your functional equation, THEN one must have $f(x)=x+1 $ If you proved it, why do you still have to show $f(x)=x+1 $ satisfy the given condition since you already the showed the only function satisfy the given condition is $f(x)=x+1 $ . $\endgroup$ – Kun Oct 18 '14 at 15:12
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    $\begingroup$ No, because there is no equivalency in your reasoning. You proved that if f(x+y) = x + f(y) then by saying y=0 you have f(x) = x + f(0), but this not equivalent to your functional equation at all. It is a consequence, not a sufficient and necessary condition $\endgroup$ – mvggz Oct 18 '14 at 15:15
  • $\begingroup$ So you have to check back, to see if deductions will completely fit your conditions you've started with. $\endgroup$ – mvggz Oct 18 '14 at 15:17
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    $\begingroup$ You got it, the step where you say y=0 reduces the hypothesis, since you don't take into account all real numbers for y, only 0. In that case you can make an equivalency so it's not a very good example, but like my example or certain other problems like these ones, you don't have it $\endgroup$ – mvggz Oct 18 '14 at 15:41
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You just computed it: this is the best possible case, when you can compute the solution explicitly. You have shown that IF a function $f$ satisfies your functional equation, THEN one must have $f(x)=x+1$ for all $x\in \mathbb{R}$. In particular, the solution to that equation is unique.

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