# counterexample to parallelogram law

• April 25th 2010, 10:05 AM
vinnie100
counterexample 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??
• April 25th 2010, 10:11 AM
edit: stupid mistake

Those two do not contradict the parallelogram law. Try the indicators on [0,1/2] and [1/2,1].
• April 25th 2010, 12:28 PM
vinnie100
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
• April 25th 2010, 12:45 PM
Um, I don't really understand... ;[

The parallelogram law states that $2||f||^2 + 2||g||^2 - ||f+g||^2 = ||f-g||^2$. Now $||f||=||g|| = 1/2$ and $||f+g||=1$. Therefore the LHS is 0, so the RHS is too. But $||f-g|| = 1$, a contradiction. Therefore the parallelogram law does not hold in L1([0,1]).
• April 25th 2010, 12:56 PM
vinnie100
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.
• April 25th 2010, 01:03 PM
Let $f:=1_{[0,1/2)},\;g:=1_{[1/2,1]}$.
$|f-g| = |1_{[0,1/2)} - 1_{[1/2,1]}|$. Since the indicators are on disjoint sets, the absolute value of their sum is the sum of their absolute values vis. $|1_{[0,1/2)} - 1_{[1/2,1]}| = |1_{[0,1/2)}| + |-1_{[1/2,1]}| = |1_{[0,1/2)}| + |1_{[1/2,1]}| = 1$. Integrating this gives 1.