# Real analysis show f is continuous

1. Nov 11, 2007

1. The problem statement, all variables and given/known data
suppose f is a function with the property f(x+y)=f(x)+f(y) for x,y in th reals. suppose f is continuous at 0. show f is continuous everywhere.

2. Relevant equations

3. The attempt at a solution

I do not understand how to show that f is continuous everywhere.

2. Nov 11, 2007

### morphism

Try to see what properties f has. For example, try to prove that f(x-y) = f(x)-f(y). (Hint: Prove first that f(0)=0 and f(-x)=-f(x).)

Then use an epsilon-delta argument to prove that f is continuous everywhere.

3. Nov 11, 2007

### HallsofIvy

Staff Emeritus
Use the definition of continuous: f is continuous if and only if $\lim_{x\rightarrow a}f(x)= f(a)$.

Now, let h= x- a and write that in terms of $\lim_{h\rightarrow 0}$.

Exactly the same proof can be used to show that if such an f is continuous at any specific x, then it is continuous at x=0. Putting the two together, if a function satisfying f(x+y)= f(x)+ f(y) is continuous for any x, then it is contnuous for all x.

And, by the way, f must be of the form f(x)= Cx for some number C. The really remarkable thing is that there exist function satisfying f(x+y)= f(x)+ f(y) that are not continuous!

4. Nov 11, 2007

ok I have deterimed that i can not do this problem. I do not not even understand how two deduce that f(0)=0

5. Nov 11, 2007

### morphism

You know that f(x+y)=f(x)+f(y). What happens if you plug in x=y=0?

6. Nov 11, 2007

you have f(0)=f(0)+f(0)

7. Nov 11, 2007

### morphism

So...

8. Nov 11, 2007

how is this 0

9. Nov 11, 2007

f(0)-f(0)=f(0)+f(0)-f(0)
0=f(0)

10. Nov 11, 2007

### morphism

That's good. Now can you show that f(-x)=-f(x)? (Try to plug in "clever" values for x and y in f(x+y)=f(x)+f(y).)

11. Nov 12, 2007

### CompuChip

If you prefer, I also think an $\epsilon-\delta$ proof is not very hard in this case.

12. Nov 12, 2007

### Kummer

In fact we can solve this functional equation. Suppose that f is continous at 0. Then f(0+0)=f(0)+f(0) so f(0)=0. Now $$\lim_{y\to 0}f(x+y) = \lim_{y\to 0}f(x)+f(y) = f(x)+f(0)=f(y)$$ so $$f$$ is continous everywhere. Let $$f(1)=c$$. But using the argument f(1+...+1) = nf(1) you can show that f(n)=nc. Now consider a rational number p/q with p,q>0 so if x=p/q then qx=p and so f(qx) = p so f(x+...+x)=p thus pf(x)=ap; that means f(p/q) = a*(p/q). We can extend this argument to negative also easily. And so if x is any real number there is a sequence of rationals converging to it so f(x) = a*x because this is true for rationals. Thus, the only possible functions are the lines through the origin.

13. Nov 13, 2007

### HallsofIvy

Staff Emeritus
There do, however, exist non-continuous solutions to that functional equation. In distinction to the continuous solutions, all of the form f(x)= Cx, non-continuous solutions are remarkably messy- there are "dense in the plane".

14. Nov 14, 2007

### Kummer

Trivial? Use axiom of choice and contruct the Hamel Basis.

15. Nov 15, 2007

### HallsofIvy

Staff Emeritus
I didn't say it wasn't easy! I'm not sure I would consider contructing a Hamel basis trivial, though. This is one of those situations where it is easy to tell HOW to do something, just very difficult to do it. "First construct a basis for the real numbers as a vector space over the rational numbers"- It's very easy to see that we can use "1" as a basis element and then $\sqrt{2}$ as another, ... but not clear to me how I could give the entire basis. Now define f(1)= 1, f($\sqrt{2}$)= 2, f(v)= 0 for v any other member of the basis, and define f(x) for any real number x "by linearity". Then we have a function satisfying f(x+ y)= f(x)+ f(y) (and f(rx)= rf(x) for any rational r) but which is not continuous.

(I think we have gotten well off the original question now!)

Last edited: Nov 15, 2007