If $l^p=\{\langle x_k \rangle \in R^n | \sum_{k=1}^{\infty} |x_k|^p < \infty \}$ and $1\leq p<q$, then $l^p\subset l^q$.
Prove.
I know that I have to show that if $x=\langle x_k \rangle\in l^p\rightarrow x\in l^q$ and I can use a fact that $l^p\subset c_0$, where $c_0=\{x=\langle x_k \rangle \in R^n | \displaystyle \lim_{k \to +\infty} x_k=0\}$.
One hint is that for a number $x$ less than 1, $x^q$ is less than $x^p$. And the numbers in the sequence are approaching 0.
Suppose $1\le p<q\le\infty.$ We will prove that $\ell^{p}\subseteq \ell^{q}$ by proving that $\|a\|_{q}\le\|a\|_{p}$ for any $a\in \ell^{p}.$
Define $b_n := \frac{a_n}{\|a\|_{p}}$ and observe that $\vert b_n \vert \leq 1$ for all $n \in \mathbb N$. Notice the following chain of equivalent statements: \begin{align*} \vert b_n \vert^{q} &\leq \vert b_n \vert^{p} \\ \sum_{n = 1}^\infty \vert b_n \vert^{q} & \leq \sum_{n = 1}^\infty \vert b_n \vert^{p} \\ \sum_{n = 1}^\infty \bigg \vert \frac{a_n}{\|a\|_{p}} \bigg \vert^{q} & \leq \sum_{n = 1}^\infty \bigg \vert \frac{a_n}{\|a\|_{p}} \bigg \vert^{p} \\ \Bigg(\sum_{n = 1}^\infty \bigg \vert \frac{a_n}{\|a\|_{p}} \bigg \vert^{q}\Bigg)^{1/p} & \leq \Bigg( \sum_{n = 1}^\infty \bigg \vert \frac{a_n}{\|a\|_{p}} \bigg \vert^{p} \Bigg)^{1/p}\\ \frac{1}{\big(\|a\|_{p}\big)^{q/p}} \Bigg(\sum_{n = 1}^\infty \vert a_n \vert^{q} \Bigg)^{1/p} & \leq 1\\ \Bigg(\sum_{n = 1}^\infty \vert a_n \vert^{q} \Bigg)^{1/p} & \leq \big(\|a\|_{p}\big)^{q/p}\\ \sum_{n = 1}^\infty \vert a_n \vert^{q} & \leq \big(\|a\|_{p}\big)^{q}\\ \Bigg(\sum_{n = 1}^\infty \vert a_n \vert^{q} \Bigg)^{1/q} & \leq \|a\|_{p}\\ \|a\|_{q}& \leq \|a\|_{p}. \end{align*}
You didn't ask for the case when $q = \infty$, but it holds there too. Notice that $\vert a_n \vert \leq \bigg(\sum_{n = 1}^\infty \vert a_n \vert^{p} \bigg)^{1/p}$ for all $n \in \mathbb N$, and thus $$\|a_n\|_{q} = \|a_n\|_\infty = \sup_{n \in \mathbb N} \vert a_n \vert \leq \bigg(\sum_{n = 1}^\infty \vert a_n \vert^{p} \bigg)^{1/p} = \|a_n\|_{p}.$$
From $l^p\subset c_o$, for $\epsilon =1$, there is $k_o\in N$ such that for all $k\geq k_0$, $|x_k|<1$.
It follows that $|x_k|^q\leq |x_k|^p$ because $1\leq p<q$. $\rightarrow sum^{\infty}{k=k_0} |x_k|^q\leq sum^{\infty}{k=k_0} |x_k|^p$
But I don't know what happens with $\sum^{k_0-1}_{k=1} |x_k|^q$ and how to show that it converges.
– user23709 May 29 '13 at 16:05