Functions s.t. f(x+y) = f(x) + f(y)?

• Feb 3rd 2010, 08:00 PM
tunaaa
Functions s.t. f(x+y) = f(x) + f(y)?
How can find all real continuous functions s.t. f(x+y) = f(x) + f(y), and prove it? So far I've tried to do it by a process of elimination, leaving only functions of the form f(x) = ax and the trivial case f(x) = 0, but I think I'm missing a trick here as I don't think it's practical to systematically eliminate every possible function.

Thanks
• Feb 3rd 2010, 08:08 PM
Drexel28
Quote:

Originally Posted by tunaaa
How can find all real continuous functions s.t. f(x+y) = f(x) + f(y), and prove it? So far I've tried to do it by a process of elimination, leaving only functions of the form f(x) = ax and the trivial case f(x) = 0, but I think I'm missing a trick here as I don't think it's practical to systematically eliminate every possible function.

Thanks

Claim: The only functions that satisfy this are $\displaystyle f(x)=f(1)x$

Proof: We make the following observations

1. $\displaystyle f(0)=f(0+0)=f(0)+f(0)\implies f(0)=0$

2. $\displaystyle 0=f(0)=f(x+-x)=f(x)+f(-x)\implies -f(x)=f(-x)$

3. It follows by induction that $\displaystyle f(zx)=zf(x),\text{ }z\in\mathbb{Z}$

4.$\displaystyle f(1)=f\left(\frac{q}{q}\right)=qf\left(\frac{1}{q} \right)\implies f\left(\frac{1}{q}\right)=\frac{f(1)}{q},\text{ }q\in\mathbb{Z}$

5. $\displaystyle f\left(\frac{p}{q}\right)=pf\left(\frac{1}{q}\righ t)=f(1)\frac{p}{q},\text{ }\frac{p}{q}\in\mathbb{Q}$

So, clearly we have that $\displaystyle f(x)=f(1)x,\text{ }x\in\mathbb{Q}$

and since $\displaystyle \mathbb{Q}$ is dense in $\displaystyle \mathbb{R}$ we see that given any $\displaystyle x\in\mathbb{R}-\mathbb{Q}$ there exists a sequence of rational numbers $\displaystyle \left\{q_n\right\}_{n\in\mathbb{N}}$ such that $\displaystyle q_n\to x$ and since $\displaystyle f$ is continuous $\displaystyle f(x)=\lim\text{ }f(q_n)=\lim\text{ }f(1)q_n=f(1)\lim\text{ }q_n=f(1)x$.

The conclusion follows.
• Feb 4th 2010, 05:13 AM
HallsofIvy
Excellent, Drexel28!

By the way, tunaaa, there exist non-continuous functions that satisfy f(x+y)= f(x)+ f(y) but they are really nasty! Not only are they discontinuous, they are not bounded in any interval, no matter how small. In fact, their graph is dense in the plane. No matter what x, y, and $\displaystyle \epsilon> 0$ are, there exist a point of the graph within distance $\displaystyle \epsilon$ of (x, y).