Real Analysis, Folland Problem 6.1.12 $L^p$ spaces

Question

Problem 6.1.12 - If $p\neq 2$, the $L^p$ norm does not arise from an inner product on $L^p$, except in trivial cases when $\dim(L^p) \leq 1$. (Show that the parallelogram law fails.)

Attempted proof (as suggested by Qiyu Wen) - Let $f = 1_{A}$ and $g = 1_{B}$ where $A$ and $B$ are disjoint sets with positive measure then, $$2\left(\int_{X}|1_{A}|^p d\mu\right)^{2/p} + 2\left(\int_{X}|1_{B}|^p d\mu\right)^{2/p} = 2\left[\int_{X} |1_{A} + 1_{B}|^pd\mu \right]^{2/p} $$ Should we let $X$ be equal to something where we will violate the parallelogram law? Not sure how to proceed further.

Answer 1

You were on the right track notice if $A,B$ are disjoint sets with non-zero finite measure then: $$ [\int_X |1_A|^p d\mu]^{1/p} = [\int_X 1_A d\mu]^{1/p} = \mu(A)^{1/p}$$

$$ [\int_X |1_B|^p d\mu]^{1/p} = [\int_X 1_B d\mu]^{1/p} = \mu(B)^{1/p}$$

$$ [\int_X |1_A+1_B|^p d\mu]^{1/p} = [\int_X (1_A+ 1_B) d\mu]^{1/p} = (\mu(A)+\mu(B))^{1/p}$$

$$ [\int_X |1_A-1_B|^p d\mu]^{1/p} = [\int_X (1_A+ 1_B) d\mu]^{1/p} = (\mu(A)+\mu(B))^{1/p}$$

So as you have above we need to satisfy the parallelogram law we need to show:

$$2\left(\int_{X}|1_{A}|^p d\mu\right)^{2/p} + 2\left(\int_{X}|1_{B}|^p d\mu\right)^{2/p} = \left[\int_{X} |1_{A} + 1_{B}|^pd\mu \right]^{2/p} + \left[\int_{X} |1_{A} - 1_{B}|^pd\mu \right]^{2/p} $$

Subbing in above we have: $$2(\mu(A))^{2/p}+2(\mu(B))^{2/p} = (\mu(A)+\mu(B))^{2/p}+(\mu(A)+\mu(B))^{2/p}$$

This holds only when $p=2$.

So we have shown provided we are dealing $L^p(X,M,\mu)$ where $X$ has two disjoint sets with non-zero finite measure the $L^p$ norm does not arrise from the inner product if $p \neq 2$.

Now here is where $dim(L^p)$ comes in. Notice if the dimension is $0$ then this thing is always trivially true as the only element is $0 \in L^p$ so that in our parallelogram law everything turns to $0$. It is also true if the dimension is $1$ all you a have to worry about is $0$ and the scalar multiples of one function say $f$. Simply check the identity holds. Now if the dimension is larger than 1 that gives me at least two linearly independent functions in $L^p$ so I finally have two disjoint sets with non-zero finite measure so all the above holds.

Answer 2

You need to show that $$ 2\left(\int_X|f|^p\,d\mu\right)^{2/p} + 2\left(\int_X|g|^p\,d\mu\right)^{2/p} = \left(\int_X|f+g|^p\,d\mu\right)^{2/p} + \left(\int_X|f-g|^p\,d\mu\right)^{2/p} $$ does not hold for all $f,g \in L^p(\mu)$ if your $\mu$ is reasonable enough.