Reversing: Properties of a continous Function.

[kioria]

I have read on some websites that if f: R -> R is continous for every x in R, then f(x+y) = f(x) + f(y) defines f as a linear function.

Now,

I am given:

Suppose f is continuous at 0, and that for all x, y in R, f(x+y) = f(x) + f(y).
a) Show that f(0) = 0.
b) Prove that f is continuous at every point a in R.

Solution for a)

f(0+0) = f(0) + f(0)
f(0) = 2f(0)
0 = 2f(0) - f(0)
0 = f(0)

I am confused as to how to go about with part b)? (note: this question was under the topc of limits and continuity.) So I was planning to use limits as part of the solution to part b).

Thanks in Advance :wink:

In fact, there are series of questions following this, that is in a similar format, but with f(x+y) = f(x)f(y). And once again, continuity of the function must be proved. If you guys can help me with the first one, I will try to do the second one by myself, but if there are any tips or tricks involved in the second proof, please hint me. Thank you.

[Hurkyl]

The first thing I tend to do when I don't see a clear way to attack a problem is to write down definitions. Here, you're told that f is continuous at zero, and your goal is to prove that f is continuous at a for every real number a, so I would write down those definitions and see if any leads present themselves.

Sure, it doesn't sound like much, but you'd be amazed what you see when things are written on paper and not in your head. :smile:

[arildno]

Here's one hint:
Since f is continuous in 0, what do you know about f's values close to 0?

[jdavel]

kioria,

This might help.

Draw a graph of a function that satisfies the condition f(x+y) = f(x) + f(y). Is this this the only type of function that satisfies the condition? What could change and stil have f be linear?

Now use the definition that hurkyl told you to write down to make your graph discontinuous. Does the linearity condition on f still hold?

[kioria]

I still can't seem to get there... I get the slightest idea, but I am struggling to present them as a hard copy proof.

Arildno: since f is c.t.s. at x = 0, f values close to 0 tend to x = 0. I just can't see where abouts to go with this fact. Can you extend this idea to me?

Jdavel/Hurkyl: I get the idea, but as I said I am having trouble with providing a hard copy proof. Any starters?

Thanks

[kioria]

Actually, I have came across this idea:

To prove, f: R -> R is c.t.s for every a in R,
I have to show:
\lim_{\substack{x\rightarrow a}} f(x) = f(a)
re-writing this idea:
\lim_{\substack{x\rightarrow a}} f(x) = \lim_{\substack{a+\epsilon\rightarrow a}} f(a+\epsilon)
that is:
\lim_{\substack{a+\epsilon\rightarrow a}} f(a+\epsilon) = \lim_{\substack{\epsilon\rightarrow 0}} f(a+\epsilon)

But we know, that:
f(a+\epsilon) = f(a) + f(\epsilon)
so,
\lim_{\substack{\epsilon\rightarrow 0}} f(a+\epsilon) = \lim_{\substack{\epsilon\rightarrow 0}} (f(a) + f(\epsilon)) = f(a)
As f(0) = 0 from Part a).

Can this be a correct proof?

[Hurkyl]

Looks right to me!

[kioria]

Thank you!