# 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?