# A Continuity Problem

1. Jul 4, 2008

### e(ho0n3

The problem statement, all variables and given/known data
Suppose f(x + y) = f(x) + f(y) and f is continuous at 0. Show that f is continuous at a.

The attempt at a solution
Since f is continuous at 0, for any e > 0 there is a d > 0 such that |f(x) - f(0)| < e for all x with |x - a| < d. Writing 0 as -a + a, |f(x) - f(0)| = |f(x) - f(-a) - f(a)|. By the Triangle Inequality, |f(x) - f(a)| - |f(-a)| ≤ |f(x) - f(-a) - f(a)| so |f(x) - f(a)| < e + |f(-a)|. This is as far as I've gotten. Am I proceeding the right direction?

2. Jul 4, 2008

### quasar987

First note that f(x + y) = f(x) + f(y) implies f(0)=0, because f(0)=f(0+0)=f(0)+f(0), hence f(0)=0.

So the statement of continuity at 0 reads, "for any e > 0 there is a d > 0 such that |f(x)| < e for all x with |x| < d".

This said, do you see more easily now why continuity at 0 implies continuity everywhere?

3. Jul 5, 2008

### e(ho0n3

Nice tip. I think I got it now: Given an e > 0, I need to find a d > 0 such that |f(x) - f(a)| < e for all x with |x - a| < d. Using the Triangle Inequality, |f(x) - f(a)| ≤ |f(x)| + |f(a)| so it suffices to find a d such that |f(x)| + |f(a)| < e for all x with |x - a| < d. |f(x)| + |f(a)| < e is equivalent to |f(x)| < e - f(a). Since f is continuous at 0, there is definitely a d that will satisfy |f(x)| < e - f(a) for all x with |x - a| < d. Thus, f is continuous at a. This completes the proof. Right?

4. Jul 5, 2008

### Dick

Noooo. If |f(x)-f(a)|<e and |f(x)-f(a)|<|f(x)|+|f(a)| does NOT IMPLY that |f(x)|+|f(a)|<e. At all. Not even a little. Just take quasar987's hint and replace x with x-a.

5. Jul 5, 2008

### e(ho0n3

Perhaps you misunderstood me. I was trying to convey that |f(x)| + |f(a)| < e and |f(x) - f(a)| ≤ |f(x)| + |f(a)| implies that |f(x) - f(a)| < e.

I never thought of replacing x with x - a. That's a lot simpler than what I was trying to do.

6. Jul 5, 2008

### Dick

That implication is at least true, but it's not of much help proving the limit. Take an example of f(x)=x and say, a=3. The values of e you really care about are very close to zero. If x is close to 3, then |f(x)|+|f(a)| is around 6. Having |f(x)-f(a)|<6 is not going to be very useful as e gets small.

7. Jul 6, 2008

### boombaby

|f(x)|+|f(a)| can not be arbitrarily small except when |f(a)|=0 . Hence trying to prove "|f(x)| + |f(a)| < e" will fail unless you can show that f(a)=0 for any given a, but this is not the case

8. Jul 6, 2008

### e(ho0n3

I understand now. Thanks guys.

9. Jul 6, 2008

### e(ho0n3

Now that I've written things down, I'm noticing that replacing x with x - a is not enough for f(x - a) ≠ f(x) - f(a). Is there some way of proving that f(-a) = -f(a)?

10. Jul 6, 2008

### e(ho0n3

Nevermind: 0 = f(0) = f(a - a) = f(a) + f(-a). Adding -f(a) to both sides yields that -f(a) = f(-a).

11. Jul 6, 2008

### Dick

Sure. You've got it.