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Prove l^p strict subspace of c0

  1. Oct 5, 2012 #1
    1. The problem statement, all variables and given/known data

    For F [itex]\in[/itex] {R,C} and for an infinitie discrete time-domain T, show that lp(T;F) is a strict subspace of c0(T;F) for each p [itex]\in[/itex] [1,∞). Does there exist f [itex]\in[/itex] c0(T;F) such that f [itex]\notin[/itex] lp(T;F) for every p [itex]\in[/itex] [1,∞)

    2. Relevant equations

    Well we know from class that lp(T;F) = {f [itex]\in[/itex] FT | Ʃ |f(t)|p < ∞} for p [itex]\in[/itex] [1,∞) and c0(T;F) = {f [itex]\in[/itex] FT | [itex]\forall[/itex]ε [itex]\in[/itex] R>0, [itex]\exists[/itex] a finite set S [itex]\subseteq[/itex] T such that {t [itex]\in[/itex] T | |f(t)| > ε} [itex]\subseteq[/itex] S} ([itex]\Leftrightarrow[/itex] t [itex]\rightarrow[/itex] ±∞ [itex]\Rightarrow[/itex] f(t) [itex]\rightarrow[/itex] 0) are both vector spaces.

    3. The attempt at a solution

    Well as I said above, we know that both lp(T;F) and c0(T;F) are vector spaces, so to show that one is a subspace of the other it is sufficient to show that one is a subset of the other.

    So,
    Let f [itex]\in[/itex] lp(T;F) [itex]\Rightarrow[/itex] Ʃ|f(t)|p < ∞
    Therefore lim as |t|[itex]\rightarrow[/itex]∞ of |f(t)|p = 0 [itex]\Rightarrow[/itex] lim as |t|[itex]\rightarrow[/itex]∞ of |f(t)| = 0 [itex]\Rightarrow[/itex] f [itex]\in[/itex] c0(T;F)

    Therefore lp(T;F) [itex]\subseteq[/itex] c0(T;F).

    Now this is the hard part, showing that it is a strict subset.

    This is what I though of:

    Define f [itex]\in[/itex] FT by f(t) = [itex]\frac{1}{t1/p}[/itex] if t ≠ 0 and 0 if t = 0.

    Clearly f [itex]\in[/itex] c0(T;F) as it's limit goes to 0.

    And it's easy to show that f is not in any lp(T;F) for any specific p [itex]\in[/itex] [1,∞).

    But to properly do this problem I need to find a function that's not in lp(T;F) for every p [itex]\in[/itex] [1,∞).

    i.e., my problem is, I have to chose a p first, then this works, but whatever p I chose, f will not be in it's space, but it will be in the p+1 space. So it doesn't work for for every p, only a specific p.

    So pretty much I can't think of a function that would be in c0 but not in lp for EVERY p [itex]\in[/itex] [1,∞).

    Any help would be awesome!
     
  2. jcsd
  3. Oct 6, 2012 #2

    micromass

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    Think about using logarithms.
     
  4. Oct 6, 2012 #3
    Ok.

    That makes sense as ln(t) increase much slower than t.

    So this is what I was thinking:

    Let f(t) = [itex]\frac{1}{ln(|t| + 1)}[/itex] if t ≠ 0 and 0 if t = 0.

    So clearly lim as |t|→∞ of f(t) = 0, so it is in c0.

    But then showing it isn't in lp for every p is a bit harder.

    This is how I tried:

    Basically I wanted to use the comparison test to show Ʃ|[itex]\frac{1}{ln(|1| + 1)}[/itex]|p = Ʃ1/|ln(|t| + 1)|p≤ Ʃ[itex]\frac{1}{|t|}[/itex] and since that diverges [itex]\Rightarrow[/itex] Ʃ|f(t)|p diverges.

    The problem is showing that [itex]\exists[/itex] N [itex]\in[/itex] T such that [itex]\forall[/itex] t ≥ N |ln(|t| + 1)|p ≤ |t|.

    I tried differentiating both sides but it ends up giving p|ln(|t| + 1)|p - 1 ≤ |t| +1 which is essentially the same thing as before.

    It makes sense to me as ln(t) increases much slower than t, and we can always find an N for which ln(t)p will be less than t for any t ≥ N. But I don't know how to show that.

    I also thought of using the ratio or root test, but that seems like it wouldn't work very well...

    Any thoughts?
     
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