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??
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]).
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.