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Analysis Question on addivity

  1. Oct 25, 2006 #1
    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. jcsd
  3. Oct 25, 2006 #2


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    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?
  4. Oct 25, 2006 #3
    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.
  5. Oct 25, 2006 #4
    Sorry, I have been working on three midterms and am a bit exhausted. It maybe in front of my face, but I can't see it.
  6. Oct 25, 2006 #5


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    I'm not sure I follow your proof, but it seems to be overcomplicating things. Note that to show f is continuous at x, you just need to show that for any e>0, there is a d>0 with |f(x+y)-f(x)|<e for all y with |y|<d. It should be clear that the additivity of f implies this is equivalent to continuity at x=0. For the sub-additive one, do the same thing, and the inequality should work in your favor.
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