This is a counterexample to the parallelogram law to show that the L1 norm is not induced by an inner product in [0,1]:

f(x) = 1

g(x) = 1

|f|=|g|=1 and |f+g|=|f-g|=2

Why is |f-g|=2??

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- Apr 25th 2010, 10:05 AMvinnie100counterexample to parallelogram law
This is a counterexample to the parallelogram law to show that the L1 norm is not induced by an inner product in [0,1]:

f(x) = 1

g(x) = 1

|f|=|g|=1 and |f+g|=|f-g|=2

Why is |f-g|=2?? - Apr 25th 2010, 10:11 AMmaddas
edit: stupid mistake

Those two do not contradict the parallelogram law. Try the indicators on [0,1/2] and [1/2,1]. - Apr 25th 2010, 12:28 PMvinnie100
f(x) = 1 on [0,1/2] and f(x) = 0 on [1/2,1]

g(x) = 1 on [1/2,1] and f(x) = 0 on [0,1/2]

|f|=|g|=1/2 and |f+g|=|f-g|=1

But again then why is |f-g|=1??

I am confused:( Hope you can help M. Thanks

- Apr 25th 2010, 12:45 PMmaddas
Um, I don't really understand... ;[

The parallelogram law states that $\displaystyle 2||f||^2 + 2||g||^2 - ||f+g||^2 = ||f-g||^2$. Now $\displaystyle ||f||=||g|| = 1/2$ and $\displaystyle ||f+g||=1$. Therefore the LHS is 0, so the RHS is too. But $\displaystyle ||f-g|| = 1$, a contradiction. Therefore the parallelogram law does not hold in L1([0,1]). - Apr 25th 2010, 12:56 PMvinnie100
Everything makes sense there except why is http://www.mathhelpforum.com/math-he...a1d64ac1-1.gif

?

Surely that would be 0?

Thanks ever so much for your help. - Apr 25th 2010, 01:03 PMmaddas
Let $\displaystyle f:=1_{[0,1/2)},\;g:=1_{[1/2,1]}$.

$\displaystyle |f-g| = |1_{[0,1/2)} - 1_{[1/2,1]}|$. Since the indicators are on disjoint sets, the absolute value of their sum is the sum of their absolute values vis. $\displaystyle |1_{[0,1/2)} - 1_{[1/2,1]}| = |1_{[0,1/2)}| + |-1_{[1/2,1]}| = |1_{[0,1/2)}| + |1_{[1/2,1]}| = 1$. Integrating this gives 1. - Apr 25th 2010, 01:06 PMvinnie100
Maddas - you are absolutely awesome! This makes perfect sense. I have an exam after 2 days and has lowered my stress a bit! Thank you!!:)