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Real Analysis: Additive function proof

  1. Mar 26, 2009 #1
    Ok this is a proof I'm having trouble with. I've given the solution for the first part I got but I'm stuck and I'm not sure if the first part it correct. any help is appreciated

    1. The problem statement, all variables and given/known dataThe purpose of this project is to ascertain under what conditions an additive function has a limit at each point in R. We say that f:R[tex]\rightarrow[/tex]R is additive if for all x,y[tex]\in[/tex]R, f(x+y)=f(x)+f(y)

    (absolute valvue of[tex]\rightarrow[/tex]abv)

    1)Show that for each positive integer n and each real number x, f(nx)=nf(x)

    2)Suppose f is such that there are m>0 and a>0 such that if x[tex]\in[/tex] [-a,a], then abv(f(x))[tex]\leq[/tex]m. Choose [tex]\epsilon[/tex]>0. There is a positive integer N such that [tex]\frac{m}{N}[/tex]<[tex]\epsilon[/tex]. Show that if abv(x-y)<[tex]\frac{a}{N}[/tex], then abv(f(x)-f(y))<[tex]\epsilon[/tex].

    3)Prove that if there are m>0 and a>0 such that if x[tex]\in[/tex][-a,a], then abv(f(x))[tex]\leq[/tex]m, then f has a limit at each x in R and the limit (as t moves towards x) of f(t)=f(x).

    4)Prove that if f has a limit at each x in R, then there are m>0 and a>0 such that if x[tex]\in[/tex][-a,a], then abv(f(x))[tex]\leq[/tex]m.

    2. The attempt at a solution

    1)mathematical induction
    n=1 says f(1x)=1f(x). true by inspection
    n=k says f(kx)=kf(x)=f(x)sub1+f(x)sub2+...+f(x)subk. assume this is true
    n=k+1 says f((k+1)x)=(k+1)f(x)?
    f((k+1)x)=f(x)sub1+f(x)sub2+...+f(x)subk+f(x)sub(k+1)=kf(x)+f(x)=(k+1)f(x).
    therefore if f is additive, then f(nx)=nf(x)
     
  2. jcsd
  3. Mar 26, 2009 #2

    Dick

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    If your induction hypothesis is f(kx)=k*f(x), then f((k+1)x)=f(kx+x)=f(kx)+f(x)=k*f(x)+f(x)=(k+1)f(x). What are all these subk's? Doesn't that prove it? Now what about 2), 3) and 4)?
     
  4. Mar 26, 2009 #3
    in mathematical induction the objective is to prove that n=k being true implies that n=k+1 is true. if you can show this, the statement is considered true. at least thats what we were taught. BTW this is weak mathematical induction.

    Now I've figured out something with part 2. If m/N is less than epsilon and abv(x-y) is less than a/N that would show that abv(f(x)-f(y)) is less than epsilon because N is a positive integer so abv(x-y) is less than a so abv(x-y) belongs to [-a,a] or more precisely [-a/N,a/N]. It makes sense to me but im not sure how to write it as a proof
     
  5. Mar 26, 2009 #4
    Also when I say subk, I mean subscript k. The subscript formating on here doesnt seem to work right for me. Im trying to show that I had added f(x) to itself k times
     
  6. Mar 27, 2009 #5
    Dick's demonstration of the inductive hypothesis and inductive step is the clearest way to do the proof.

    Hmmm, well noting that the distance between x and y is confined to the internal a/N does not change the fact that [tex]|f(x)| \leq m[/tex] for all x in [-a,a]. From this bound, we can basically apply a form of the triangle inequality (or think logically) to deduce that [tex]|f(x)-f(y)| \leq 2m < 2 \epsilon N[/tex].
     
  7. Mar 31, 2009 #6
    Alright, this is what I've got for part 2)
    If I let abv(x-y)<[tex]\frac{a}{N}[/tex], then I can say that abv(Nx-Ny)<a, meaning abv(Nx-Ny)[tex]\in[/tex][-a,a].
    Now if I let abv(F(x)-F(y))<[tex]\frac{m}{N}[/tex], I can say that abv(NF(x)-NF(y))<m because of part one of the proof. Since [tex]\frac{m}{N}[/tex]<[tex]\epsilon[/tex], I can then say that abv(F(x)-(F(y))<[tex]\epsilon[/tex].

    Just a note: I was attacked this weekend and have a bit of a concussion. If I've done some thing stupid I'll blame it on the head injury. Please let me know if this proof is any good.
     
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