# Bounding the p-norms on l_p sequence subspaces!

1. Aug 21, 2009

### ELESSAR TELKONT

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

Let $$1\leq r<\infty$$ and $$x\in\ell_{r}=\left\{ x \text{ is a sequence with } \sum_{n=1}^{\infty}\left\vert x_{n}\right\vert^{r} \text{ converges.}\right\}$$, then

$$\left\vert\left\vert x\right\vert\right\vert_{\infty}=\lim_{r\rightarrow\infty}\left\vert\left\vert x\right\vert\right\vert_{r}$$

2. Relevant equations

3. The attempt at a solution

I have already proven that for $$1\leq s\leq r\leq\infty$$ I can bound below by $$\left\vert\left\vert x\right\vert\right\vert_{\infty}\leq \left\vert\left\vert x\right\vert\right\vert_{r}$$. The other part wher I'm stuck is to bound above the r-norm. I haven't found something to do it. Obviously, the case for $$\mathbb{R}^{n}$$ (the finite case if you considere the sequences case as "infinituples") is solved because you have something to do a bounding almost immediatly since

$$\sum_{i=1}^{n}\left\vert x_{n}\right\vert^{r}=\left\vert\left\vert x\right\vert\right\vert_{r}^{r}\leq\sum_{i=1}^{n}\left\vert x_{m}\right\vert^{r}=n\left\vert x_{m}\right\vert^{r}=n\left\vert\left\vert x\right\vert\right\vert_{\infty}^{r}$$

Obviously in sequences case $$n$$ makes not sense, and approximation like i have done above is not possible because a series of a constant sequence don't converge.

Proving what I can't prove the result of the problem follows immediatly because of the "Sandwich theorem" since if I have some quantity bounded above and below like this

$$\left\vert\left\vert x\right\vert\right\vert_{\infty}\leq\left\vert\left\vert x\right\vert\right\vert_{r}\leq c^{\frac{1}{r}}\left\vert\left\vert x\right\vert\right\vert_{\infty}$$ with $$c>0$$ and $$c\in\mathbb{R}$$

and if I take the limit when $$r\rightarrow\infty$$ then

$$\left\vert\left\vert x\right\vert\right\vert_{\infty}\leq\lim_{r\rightarrow\infty}\left\vert\left\vert x\right\vert\right\vert_{r}\leq \left\vert\left\vert x\right\vert\right\vert_{\infty}$$

Then I would like you to say some ideas to do bounding above.

2. Aug 22, 2009

### Dick

Since the sum of |x_i|^r converges, there is an N such that |x_i|^r<1/2 for all i>=N. That means |x_i|^(k*r)<(1/2)^(k-1)*|x_i|^r, right? So if A is the sum |x_i|^r for i from N to infinity, the sum of |x_i|^(k*r)<(1/2)^(k-1)*A. Now let k->infinity. I.e. the sum of the infinite tail of the sequence |x_i|^r vanishes as r->infinity. So ignore it. Now it's just like proving the finite case.