How to show the parallelogram law holds in $L^p$ (or $l^p$) if and only if $p=2$?

Question

Parallelogram law: $$||x+y||^2+||x-y||^2=2||x||^2+2||y||^2.$$ How to show this statement? I tried to prove the $l^p$ case. But it seems not easy to manipulate these norms since they have power and summation signs.

Answer 1

Take $f=I_{(0,1/2)},g=I_{(1/2,1)}$ and compute all the 4 terms. You will see that parallelogram law holds iff $2^{1-2/p}=1$ which means $p=2$.

Answer 2

Consider two simple functions:

$f(x) = I_{[a, (a + b)/2]}$

$g(x) = I_{[(a + b)/2, b]}$

We can see that the parallelogram law won’t hold for $p \ne 2$:

$\| f - g \|_p^2 + \| f + g \|_p^2$

$= (\int_{[a, b]} |f(x)-g(x)|^p dx)^{2/p} + (\int_{[a, b]} |f(x)+g(x)|^p dx)^{2/p}$

$= (\int_{[a, b]} 1 dx)^{2/p} + (\int_{[a, b]} 1 dx)^{2/p}$

$= 2(b-a)^{2/p}$

$\ne 2^{-\frac{2}{p} + 1}2(b-a)^{2/p}$

$= 2^{-\frac{2}{p} + 2}(b-a)^{2/p}$

$= 2((\frac{b-a}{2})^{2/p} + (\frac{b-a}{2})^{2/p})$

$= 2((\int_{[a, b]} f(x) dx)^{2/p} + (\int_{[a, b]} g(x) dx)^{2/p})$

$= 2((\int_{[a, b]} |f(x)|^p dx)^{2/p} + (\int_{[a, b]} |g(x)|^p dx)^{2/p})$

$= \|f\|_p^2 + \|g\|_p^2$

Indeed this equality will only occur for $1 = 2^{-2/p}\cdot 2 \iff p = 2$.