Is the parallelogram equality satisfied in $l^1$?

Question

I can't show that the parallelogram equality is satisfied / or is not satisfied in $l^1$.

If $(v_n), (w_n) \in l^1$, then we have $$ || (v_n) + (w_n) ||_1^2 + || (v_n) - (w_n) ||_1^2 = || (v_n + w_n) ||_1^2 + || (v_n - w_n) ||_1^2 = $$ $$= (\underset{n=1}{\overset{\infty}{\sum}} | v_n + w_n |)^2 + (\underset{n=1}{\overset{\infty}{\sum}} | v_n - w_n |)^2 \leq 2 [ \underset{n=1}{\overset{\infty}{\sum}} (| v_n | + | w_n | )]^2 = 2 (\underset{n=1}{\overset{\infty}{\sum}} | v_n | + \underset{n=1}{\overset{\infty}{\sum}} | w_n | )^2 = 2 (|| (v_n) ||_1 + || (w_n) ||_1 )^2 = 2 || (v_n) ||_1^2 + 2 || (w_n) ||_1^2 + 4 || (v_n) ||_1 || (w_n) ||_1 $$ Therefore $$ || (v_n) + (w_n) ||_1^2 + || (v_n) - (w_n) ||_1^2 \leq 2 || (v_n) ||_1^2 + 2 || (w_n) ||_1^2 + 4 || (v_n) ||_1 || (w_n) ||_1 $$

And then?

Thank you!

Answer 1

The parallelogramm identity does not hold in $\ell_1$. Just consider

$$(v_n) := (1,0,0,\ldots) \qquad \quad (w_n) := (0,1,0,\ldots).$$

Then

$$\|(v_n)+(w_n)\|_1^2 + \|(v_n)-(w_n)\|_1^2 = 2^2+2^2 = 8$$

whereas

$$\|(v_n)\|_1^2 + \|(w_n)\|_1^2 = 1^2+1^2 = 2.$$

Remark: One can even show that the parallelogramm identity implies a Hilbert space structure. And it is widely know that $\ell^1$ is not a Hilbert space.

Answer 2

$\ell^1$ is teh space of absolutely summable sequences. The space of bounded sequences is $\ell_\infty$.

The only $\ell^p$ inner product space is $\ell^2$ (and parallelogram law holds iff inner product space).

Try to come up with a counter example.