Which is finer(larger) between the sequence spaces $l_{p}$ & $l_{p+1}$

Question

Prove that, $l_{3}\subset l_{7}$ & $L_{9}[0,1]\subset L_{6}[0,1]$, where $l_{p}$ & $L_{p}[0,1]$ are of usual notation.

Are the converses hold for both cases?

Can these two results generalized for any $p(1\le p<\infty)$ ?

I only know that, $l_{p}(1<p<\infty)$ is not a subset of $l_{1}$.

For ex. consider the sequence, $\{x_{n}\}=\{\dfrac{1}{n}\}$. Then, $\{x_{n}\}\in l_{p}$ for any $p(1< p<\infty)$ but $\{x_{n}\}\notin l_{1}$.

How I prove these directly?

Answer 1

I think you've already basically worked out the logic. The case for $l_1$ can be easily generalised. Suppose $\{ x_n \}_{n \in \mathbb{N}} \in l_p$, then $(\sum^{\infty}_{n=1}|x_n|^p)^{1/n}<\infty $ and so $\sum^{\infty}_{n=1}|x_n|^p<\infty $. Therefore there exists an $N$ such that for arbitrary $n>N$, $|x_n|^p<1$. i.e $|x_n|<1$. And so for any $n>N$, $|x_n|^{p+1} < |x_n|^p$. Therefore $\sum^\infty_{n=N+1}|x_n|^{p+1} < \sum^\infty_{n=N+1}|x_n|^{p} < \infty $ and so $\{ x_n \}_{n \in \mathbb{N}} \in l_{p+1}$.

EDIT: I initially claimed here that the $L_p$ case was analogous which is completely wrong, viewers should read the comments for the answer to the rest of the question.