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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that: $$f(x+y) = f(x) + f(y)$$ $$f(1) = 1$$ Show that $$f(x) = x$$ I have been having trouble approaching this problem. I have shown, through a system of equations, that $f(x+y) = x + y$, but that's about as far as I can get. Appreciate any help anyone has to offer!


marked as duplicate by Asaf Karagila real-analysis Nov 9 '16 at 6:03

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See that

$$f(0)=f(0)+f(0)=2f(0)\implies f(0)=0$$

$$\begin{align}f(a+b+c+\dots+z) & =f(a)+f(b+c+\dots+z) \\&=f(a)+f(b)+ f(c+\dots+z)\\&=\dots\\&=f(a)+f(b)+f(c)+\dots+f(z)\end{align}$$

For natural numbers $x$, we have


$$f(x)=x\ \forall\ x\in\mathbb N$$

For positive rational numbers,


$$1=xf(\frac1x)\implies f(\frac1x)=\frac1x\ \forall\ x\in\mathbb N$$

$$f(\frac yx)=f(\frac1x+\frac1x+\dots+\frac1x)=f(\frac1x)+f(\frac1x)+\dots+f(\frac1x)=yf(\frac1x)=\frac yx\\f(\frac yx)=\frac yx\ \forall\ (x,y)\in\mathbb N\times\mathbb N$$

$$f(x)=x\ \forall\ x\in\mathbb Q^+$$

For positive real numbers, every real number is the cauchy sequence of rational numbers, which, since $f(x)$ is continuous, gives us

$$f(x)=x\ \forall\ x\in\mathbb R^+$$

Finally, for all negative numbers, we have


$$0=f(x)+f(-x)\implies f(-x)=-f(x)=-x$$

  • 2
    $\begingroup$ The standard proof: First for integers, then for rationals, finally, by continuity, for reals. btw, weaker assumptions than continuity are sufficient, for example bounded in some interval. $\endgroup$ – marty cohen Nov 9 '16 at 3:33
  • $\begingroup$ @martycohen You honestly don't even need to prove for the integers TBH, as the rationals take care of that. I just put it in because it follows a good line of thinking. $\endgroup$ – Simply Beautiful Art Nov 9 '16 at 14:51
  • $\begingroup$ I think that you need integers to prove for rationals. $\endgroup$ – marty cohen Nov 9 '16 at 22:23

If we let $y = 1$: $$f(x + 1) = f(x) + 1$$ This is a recurrence relation and can be solved with simple techniques. However, we already know what the solution is supposed to be. Proving that $f(x) = x$ is a solution:

LS: $$f(x + 1) = x + 1$$

RS: $$f(x) + 1 = x + 1$$

Because LS = RS, $f(x) = x$ is a solution to our problem. Note that this does not prove that it's the only solution.

  • 1
    $\begingroup$ This is for integers. The OP wants a proof for reals. $\endgroup$ – marty cohen Nov 9 '16 at 3:31
  • $\begingroup$ I suppose I could have used "a" rather than 1. $\endgroup$ – Kaynex Nov 9 '16 at 3:34
  • $\begingroup$ The reason this works is because $f(1)=1$. If you want to do this for $a$, you need to prove $f(a)=a$, which is the problem at hand. $\endgroup$ – Simply Beautiful Art Nov 9 '16 at 14:29
  • $\begingroup$ Oh, that is true. I suppose the above only works for integers. However, we also know that $f(ax) = af(x)$, which can extend it to rationals. $\endgroup$ – Kaynex Nov 9 '16 at 17:14
  • $\begingroup$ @Kaynex and at that point, you get my answer pretty much. $\endgroup$ – Simply Beautiful Art Nov 10 '16 at 19:03

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