proves and disproves about inner product spaces $l_p=\{[{a_n}]_{n=1}^ {\infty}|\sum_{n=1}^{\infty}|a_n|^p < \infty \}$
with the norm $||a_n||_p = (\sum_{n=1}^{\infty}|a_n|^p)^\frac{1}{p} $
prove or disprove:

*

*$L_2\subset L_1$


*

*I know its true for functions but is it also for sequnces?



*If $\lim\limits_{n \to \infty}a_n=0$ then $a_n\in l_1$


*

*I think this is not true but can't think of a counter example



*If $a_n , b_n \in l_2$ then $\sum_{n=1}^{\infty}(a_nb_n)^2 \le \sum_{n=1}^{\infty}|a_n|^2 \sum_{n=1}^{\infty}|b_n|^2$


*

*Is it just saying $||a_nb_n||_2^2 \le||a_n||_2^2||b||_2^2$ using cauchy schwartz?

every clue is a big help, thanks in advance.
 A: *

*The first statement is false. As you remarked, it is true for functions but only on certain conditions: if $f$ is in $L^2(\Omega)$ with $\Omega$ a bounded set, then $f$ is in $L^1(\Omega)$. This is because of Cauchy-Scharz inequality:

$$
\int_\Omega |f| \leq \left( \int_\Omega |f|^2 \right)^{\frac{1}{2}} \left( \int_\Omega 1^2 \right)^{\frac{1}{2}} < \infty
$$
because $\Omega$ is bounded so $\int_\Omega 1 = \lambda(\Omega)$ is finite.
However, it is usually false otherwise. Here the equivalent would be to write
$$
\sum_n |a_n| \leq \left( \sum_n |a_n|^2 \right)^{\frac{1}{2}} \left( \sum_n 1^2 \right)^{\frac{1}{2}}
$$
but obviously this does not tell us anything since $\sum_n 1 = \infty$ when $n$ runs through the positive integers.
A simple counter-example example is $a_n = \frac{1}{n}$ which is in $L^2$ but not in $L^1$.


*This is also false: take the same counter-example $a_n = \frac{1}{n}$.


*Suppose $(a_n)$ and $(b_n)$ are in $L^2$.
$$
\left( \sum_{n=1}^N |a_n|^2 \right) \left( \sum_{n=1}^N |b_n|^2 \right) =
\sum_{n=1}^N \sum_{m=1}^N |a_n b_m|^2 \geq \sum_{n=1}^N |a_n b_n|^2
$$
so for all $N$
$$
\sum_{n=1}^N |a_n b_n|^2 \leq \left( \sum_{n=1}^{\infty} |a_n|^2 \right) \left( \sum_{n=1}^{\infty} |b_n|^2 \right)
$$
with the right member of the inequality independent of $N$,
and therefore the sum of the $|a_n b_n|^2$ converges and
$$
\sum_{n=1}^{\infty} |a_n b_n|^2 \leq \left( \sum_{n=1}^{\infty} |a_n|^2 \right) \left( \sum_{n=1}^{\infty} |b_n|^2 \right).
$$
