If $f\colon\mathbb R\to\mathbb R$ is continuous and satisfies $f(x+y) = f(x) + f(y)$ and $f(1)=1$, then $f(x)=x$ for all $x$.

Question

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!

prove it for $x$ of the form $\frac1n$ first, then for all rationals. Prove $f(n)=n$ for natural numbers $n$. — Mirko, Nov 09 '16 at 01:14
If you did prove $f(x+y)=x+y$ for $\forall x,y$ then it follows that $f(x)=x$ by just letting $y=0$. Maybe you should post your proof, though. — dxiv, Nov 09 '16 at 01:16
See I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ which are continuous and satisfy $f(x+y)=f(x)+f(y)$ (and the posts linked there) or I want to show that $f(x)=x.f(1)$ where $f:R\to R$ is additive (and the posts linked there). — Martin Sleziak, Nov 09 '16 at 06:03

score 12 · Accepted Answer · edited Nov 09 '16 at 06:15

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

$$f(x)=f(\underbrace{1+1+1+\dots+1}_x)=\underbrace{f(1)+f(1)+\dots+f(1)}_x=1+1+1+\dots+1=x$$

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

For positive rational numbers,

$$1=f(1)=f(\underbrace{\frac1x+\frac1x+\frac1x+\dots+\frac1x}_x)=f(\frac1x)+f(\frac1x)+\dots+f(\frac1x)=xf(\frac1x)$$

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

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

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

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

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

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

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. — marty cohen, Nov 09 '16 at 03:33
@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. — Simply Beautiful Art, Nov 09 '16 at 14:51

score 1 · Answer 2 · answered Nov 09 '16 at 02:13

1

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.

answered Nov 09 '16 at 02:13

Kaynex

2,448

1

This is for integers. The OP wants a proof for reals. – marty cohen Nov 09 '16 at 03:31
I suppose I could have used "a" rather than 1. – Kaynex Nov 09 '16 at 03:34
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. – Simply Beautiful Art Nov 09 '16 at 14:29
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. – Kaynex Nov 09 '16 at 17:14
@Kaynex and at that point, you get my answer pretty much. – Simply Beautiful Art Nov 10 '16 at 19:03
Agreed. They're similar. – Kaynex Nov 10 '16 at 21:02

If $f\colon\mathbb R\to\mathbb R$ is continuous and satisfies $f(x+y) = f(x) + f(y)$ and $f(1)=1$, then $f(x)=x$ for all $x$.

2 Answers2