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??
Follow Math Help Forum on Facebook and Google+
edit: stupid mistake Those two do not contradict the parallelogram law. Try the indicators on [0,1/2] and [1/2,1].
Last edited by maddas; April 25th 2010 at 11:23 AM.
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
Um, I don't really understand... ;[ The parallelogram law states that . Now and . Therefore the LHS is 0, so the RHS is too. But , a contradiction. Therefore the parallelogram law does not hold in L1([0,1]).
Everything makes sense there except why is ? Surely that would be 0? Thanks ever so much for your help.
Let . . Since the indicators are on disjoint sets, the absolute value of their sum is the sum of their absolute values vis. . Integrating this gives 1.
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!!
View Tag Cloud