proves and disproves about inner product spaces

Question

$l_p=\{[{a_n}]_{n=1}^ {\infty}|\sum_{n=1}^{\infty}|a_n|^p < \infty \}$

with the norm $||a_n||_p = (\sum_{n=1}^{\infty}|a_n|^p)^\frac{1}{p} $

prove or disprove:

1. $L_2\subset L_1$

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

2. 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

3. 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.

Answer 1

1. 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 $L^1$.

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

3. 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

$$ \sum_{n=1}^{\infty} |a_n b_n|^2 \leq \left( \sum_{n=1}^{\infty} |a_n|^2 \right) \left( \sum_{n=1}^{\infty} |b_n|^2 \right). $$