8-18

Prove the parallelogram law for L^{2}:
\|f+g\|^{2}+\|f-g\|^{2}=2\|f\|^{2}+2\|g\|^{2} 
Is this true for L^{p} when p≠2? The geometric interpretation is that the sum of the squares of the lengths of the diagonals of a parallelogram equals the sum of the squares of the edge lengths. 
Solution:
\|f+g\|^{2}_{2}=\int_{E}^{} |f+g|^{2} = \int_{E}^{} \left ( f+g \right ) ^{2},
\|f-g\|^{2}_{2}=\int_{E}^{} |f-g|^{2} = \int_{E}^{} \left ( f-g \right ) ^{2},
so, we have 
\begin{aligned}
\|f+g\|^{2}_{2}+\|f-g\|^{2}_{2} &= \int_{E}^{} \left ( f+g \right ) ^{2}+ \int_{E}^{} \left ( f-g \right ) ^{2} \\
&=\int_{E}^{} \left ( f+g \right ) ^{2}+\left ( f-g \right ) ^{2}\\
&=\int_{E}^{} 2f^{2}+2g^{2} \\
&=2\int_{E}^{} f^{2}+ 2\int_{E}^{} g^{2} \\
&=2(\int_{E}^{} f^{2})^{\frac{1}{2} \times 2}+ 2(\int_{E}^{} g^{2})^{\frac{1}{2} \times 2}\\
&=2\|f\|^{2}_{2}+ 2\|g\|^{2}_{2}
 \end{aligned}
The proof is done now for p=2.
Now we claim it's not true for p≠2. 
Assume WLOG that E is finite (if it is not, then replace the set E with a finite subset). Let us write E=E_{1} \cup E_{2}, where E_{1},E_{2} are non-overlapping measurable subsets of E so that |E2| = |E1| = |E|/2. We define f=\chi_{E_{1}}, g=\chi_{E_{2}}.
Then f+g =\chi_{E_{1}}+\chi_{E_{2}}=\chi_{E} and f-g =\chi_{E_{1}}-\chi_{E_{2}}= \left\{\begin{matrix} 
 1 \ ,x \in E_{1}\\ 
 -1 \ ,x \in E_{2}\\
\end{matrix}\right.. Note that |f+g| =|f-g|=\chi_{E}.
When p = \infty, we check that
\|f\|_{\infty}=\|g\|_{\infty}=\|f+g\|_{\infty}=\|f-g\|_{\infty}= ess \ \sup_{E} |f| =1.
Obviously, the identity doesn't hold.
Let us assume now that p is finite. According to the definition of L_{p} norm, we have 
 \|f+g\|_{p}^{2}=\|f-g\|_{p}^{2}=|E|^{2 / p},\|f\|_{p}^{2}=\left|E_{1}\right|^{2 / p}=|\frac{E}{2}|^{2 / p} ,\|g\|_{p}^{2} =\left|E_{2}\right|^{2 / p}=|\frac{E}{2}|^{2 / p},
Therefore,
 2\|f\|_{p}^{2}+2\|g\|_{p}^{2}=2\left(\left|E_{1}\right|^{2 / p}+\left|E_{2}\right|^{2 / p}\right)=4|\frac{E}{2}|^{2 / p} 
and 
\|f+g\|_{p}^{2}+\|f-g\|_{p}^{2}=2|E|^{2 / p}
They are equal if and only if 4 \times (\frac{1}{2})^{2 / p}=2, which means p=2.
Hence this formula is not true for L^{p} when p≠2, only when p=2 can it be true.
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