Prove l^p strict subspace of c0

1. Oct 5, 2012

looserlama

1. The problem statement, all variables and given/known data

For F $\in$ {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 $\in$ [1,∞). Does there exist f $\in$ c0(T;F) such that f $\notin$ lp(T;F) for every p $\in$ [1,∞)

2. Relevant equations

Well we know from class that lp(T;F) = {f $\in$ FT | Ʃ |f(t)|p < ∞} for p $\in$ [1,∞) and c0(T;F) = {f $\in$ FT | $\forall$ε $\in$ R>0, $\exists$ a finite set S $\subseteq$ T such that {t $\in$ T | |f(t)| > ε} $\subseteq$ S} ($\Leftrightarrow$ t $\rightarrow$ ±∞ $\Rightarrow$ f(t) $\rightarrow$ 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 $\in$ lp(T;F) $\Rightarrow$ Ʃ|f(t)|p < ∞
Therefore lim as |t|$\rightarrow$∞ of |f(t)|p = 0 $\Rightarrow$ lim as |t|$\rightarrow$∞ of |f(t)| = 0 $\Rightarrow$ f $\in$ c0(T;F)

Therefore lp(T;F) $\subseteq$ c0(T;F).

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

This is what I though of:

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

Clearly f $\in$ 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 $\in$ [1,∞).

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

Any help would be awesome!

2. Oct 6, 2012

micromass

Staff Emeritus

3. Oct 6, 2012

looserlama

Ok.

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

So this is what I was thinking:

Let f(t) = $\frac{1}{ln(|t| + 1)}$ 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|$\frac{1}{ln(|1| + 1)}$|p = Ʃ1/|ln(|t| + 1)|p≤ Ʃ$\frac{1}{|t|}$ and since that diverges $\Rightarrow$ Ʃ|f(t)|p diverges.

The problem is showing that $\exists$ N $\in$ T such that $\forall$ 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?