When solving an exercise regarding $\ell^p$ spaces, I came up with the following question.
The exercise said,
Let $1<p\leq \infty$ and let $p'=\frac{p}{p-1}$. Let $b\in \ell^{p'}$ and define $T_{b}:\ell^p \to \Bbb{R}$ as $T_{b}(a)=\sum b_na_n$. Show that $T_b$ is linear, continuous and find its norm.
So when solving the exercise (using Holder's) inequality, I came up with the following;
If $a \in \ell^p$ and $q > p$ then $a \in \ell^q$?
I know the converse is not true. Since taking $p=1$, $q=2$ we have that $(\frac{1}{n}) \in \ell^2$ but $(\frac{1}{n}) \notin \ell^1$. However, I wasn't able to find a counter example as this one for my question. Is this true?
It holds true that $q>p\implies \ell^p\subseteq \ell^q$
Indeed, let $(a_n)_{n\in\mathbb N}$ such that $\sum_n |a_n|^p<+\infty$
Since $q>p$, as soon as $|a_n|<1$, which holds from a certain $N$ onwards, $|a_n|^q<|a_n|^p$.
So $\sum_{n\in\mathbb N}|a_n|^q=\sum_{n=1}^N|a_n|^q+\sum_{n>N}|a_n|^q\le\sum_{n=1}^N|a_n|^q+\sum_{n>N}|a_n|^p<+\infty$
What happens more in general?
In the case $L^p(\Omega,\mu)$, where $\mu$ is a finite measure, it holds the opposite inclusion: $q>p\implies L^p(\Omega,\mu)\supseteq L^q(\Omega,\mu)$
In the general case of a non-finite measure (for instance $\mathbb R^d$ with Lebesgue measure), neither holds.