A class note that I'm reading states that
$ \mathcal{l}_p(\mathbb{R}) \subset \mathcal{l}_\infty(\mathbb{R}) \; \forall p \ge 1$.
And it asks me to prove that is also valid to:
$ \mathcal{l}_r(\mathbb{R}) \subset \mathcal{l}_p(\mathbb{R}) \; \; 1 \le r \le p \le +\infty$.
I cannot even understand why the first is true. Where is the relationship?
What I have done by now:
Let $x \in \mathcal{l}_1 $ that means $ \sum_i^n | x_i | \le M $ for all $n \in \mathbb{N}$. (Series that converge absolutely)
Let $y \in \mathcal{l}_\infty $ that means $ \{| y_n |\} \le M $. (Bounded sequences where $|y_n|$ is the absolute maximum of sequence.)
I know that all of absolutely convergent series are also bounded. But how to relate the spaces (just to start).
If $(x_n) \in l_p$, then $\sum\limits_{n=1}^\infty |x_n|^p < \infty$. This means that $|x_n|$ is bounded as a sequence. Why? If it wasn't, then there would be a subsequence $|x_{n_k}| \to \infty$, and certainly we'd have $\sum\limits_{n=1}^\infty |x_n|^p \geq \sum\limits_{k=1}^\infty |x_{n_k}|^p = \infty$, a contradiction. Thus, $|x_n|$ is bounded uniformly by a constant $B$, and $\sup\limits_n |x_n| \leq b < \infty$, so that $(x_n) \in l_\infty$ as well.
For your second question, it basically amounts to the fact that if $(x_n) \in l_r$, then the $x_n$ shrink fast enough for the series $\sum\limits_{n=1}^\infty |x_n|^r$ to converge. Since $1 \leq r \leq p$, the sequence $|x_n|^p$ converges to $0$ even faster than $|x_n|^r$, so it must also be summable. Rigorously,
Let $N$ be such that $|x_n| \leq 1$ for all $n \geq N$. Then \begin{align*} \sum\limits_{n=1}^\infty |x_n|^p = \sum\limits_{n=1}^{N-1} |x_n|^p + \sum\limits_{n=N}^\infty |x_n|^p \leq \sum\limits_{n=1}^{N-1} |x_n|^p + \sum\limits_{n=N}^\infty |x_n|^r < \infty \end{align*} Thus, if $(x_n) \in l_r$, then necessarily $(x_n) \in l_p$. Note that the opposite inclusion doesn't hold. For example, the harmonic sequence is $l_2$ but not $l_1$.