1. Oct 25, 2006

I have a HW problem that asks in two parts to prove i) that if an additive function is continuous at 0 then it is continuous in R, and ii) if a subadditive function is continuous at 0 and f(0)=0 then the subadditive function is on R. I did the first part but dont know how to do the second. Any help would be great.

2. Oct 25, 2006

### StatusX

You need to show some work to get help. How did you do the first part? It seems like a pretty simple extension to get the second part, so where are you getting stuck?

3. Oct 25, 2006

Ok for the first part I showed that for any z (real number) and m (natural number) f((m+1)z)=f(mz) + f(z)=(m+1)f(z). Next f(0) has to equal zero since f(0 + 0)=f(0) + f(0)=f(0). Now f(mz-mz)=f(mz) + f(-mz) which implies f(-mz)=-f(mz) therefore f(mz)=mf(z) for every m (integer). Now for any z' (real number) f(z')=f(nz'/n)=nf(z'/n). Therefore f(mz'/n)=mf(z'/n)=m/n[f(z')] for m (integer) and n (natural number). If we take z'=1 we get f(r)=rf(1) for r (rational number). Finally since f is continuous at 0, there exists a delta such that abs[f(x)]<epsilon. This implies x' (real number) is also less than epsilon in the delta nbd of x. Therefore f(x) is continuous for any x' (real number).

Okay for the second part, I don't how to deal with the inequality.

4. Oct 25, 2006