convergence of sequences in $l^p$spaces with different p's

Question

I have two sequences in an $l^p$ space, i.e. two sequences of sequences:

1. $x_k$ := sequence with first k entries 1, then only 0's.

2. $y_k$ := sequence with first k entries $k^{-\frac{1}{3}}$, then only 0's.

I want to determine, for which values of $p \in [1, \infty) $ the sequences converge to some limit with the $l^p$ norm $||s_k||_{l^p} = (\sum_{i=1}^{\infty} |s_i|^p)^{\frac{1}{p}}$ for $p< \infty$ and $||s_k||_{l^p} = \sup|x_i| :i\in\mathbb{N}$ for $p= \infty$.

I think $x_k$ never converges and $y_k$ converges to the $0$ sequence for $p>= 4$. My reasoning: for $x_k$ only the sequence $x$ of all $1$'s seems even reasonable but no matter how large I choose $k$, $||x_k-x||_{l^p}$ will always be infinitely large. For $y_k$ for $p>=4$, I will get the increasing $k$ with a negative exponent in the expression $||y_k-y||_{l^p}$ and thus $\lim_{k \to \infty}||y_k-y||_{l^p}$ = 0 and the sequence $y$ of all 0 's should do the trick.

Am I on the right track here?

Answer 1

For all $\ell^p$ spaces, if a sequence $x_n \to x$ (as $n \to \infty$) then for every coordinate $l \in \mathbb{N}$ we have $|(x_n)_l - x_l|^p \le ||x_n - x||^p_p$ so

$$|(x_n)_l - x_l| \le (||x_n -x||_p) \to 0 \,\,(n \to \infty)$$

So if $(x_n)$ converges at all, then the coordinatewise limit is the only candidate limit, as you seem to imply.

So if $x_n$ converges to $x$, it must converge to a sequence $x$ of all $1$'s. But this sequence is not even a member of $\ell^p$ for $1 \le p < \infty$. But $x \in \ell^\infty$, as $||x||_\infty = 1$, but then $||x_n -x||_\infty = 1$ for all $n$ and so $x_n \not\to x$.

For $y$ we have that the sequence $(y_n)_k$ is $0$ for $n < k$, and $n^{-\frac{1}{3}}$ for $n \ge k$. So the coordinatewise limit is the all $0$ sequence. So what is $||y_n - 0||_p$, say for finite $1 \le p$ first:

$$||y_n||_p^p = \sum_{k=0}^\infty |(y_n)_k|^p = \sum_{k=0}^n n^{-\frac{p}{3}} = n\cdot n^{-\frac{p}{3}} = n^{\frac{3-p}{3}} $$ for any finite $p$. So

$$||y_n - 0||_p = n^{\frac{3-p}{3p}}$$ which converges to $0$ iff $3-p <0$ (when the fraction becomes negative), so iff $p>3$.

For $p=\infty$, $||y_n||_\infty = n^{-\frac{1}{3}} \to 0\,\,n \to \infty$, so there we alo have convergence.

So for the $x_n$, there is only one candidate limit, which is either not in the space at hand, or else keeps doesn't have distance to the limit converging to $0$. For the $y_n$ the only candidate is $0$ and there we have convergence iff $p > 3$.