Define an inner product in the normed linear space Is it possible to define an inner product $(\cdot,\cdot)$ in the normed linear space $(L^1(R),\|\cdot\|_{L^1})$ such that $(u,u)=\|u\|_{L^1}^2$?
I don't really understand what it is asking for.
I think that it is possible but I don't know how to do it.
 A: You may want to try to use (and prove) that 
a normed linear space is an inner product space iff the norm ||.|| satisfies 
the parallelogram equality
$$||f + g||^2 + ||f-g||^2 = 2 ( ||f||^2 + ||g||^2 )$$
A: As you may know, the Lebesgue space $L^1(\mathbb{R})$ with the usual norm is a Banach space. You may even now the definition of Hilbert space. The question means precisely: is there some inner product giving the $L^1$-norm? Is it possible to endow $L^1(\mathbb{R})$ with an Hilbert space structure whose norm is the $L^1$-norm? The answer is no: $L^2(\mathbb{R})$ is the only Hilbert space among $L^p$ spaces. 
We want to find a pair of functions $f,g\in L^2(\mathbb{R})$ violating the Parallelogram Law:
$$
\|f+g\|^2_1+\|f-g\|^2_1=2\|f\|^2_1+2\|g\|^2_1,
$$ 
where $\|\,\cdot\,\|_1$ denotes the $L^1$-norm.
Consider the real intervals $I=[0,{1\over 2}]$ and $J=[{1\over 2},1]$, choose $f=\mathbb{1}_I$ and $g=\mathbb{1}_J$. Then we have: 
$$
\int |f|\,\mathrm{d}x=\int_0^{{1\over 2}}\mathrm{d}x={1\over 2},
$$
so that $\|f\|_1^2=\left(\int |f|\,\mathrm{d}x\right)^2={1\over 4}$; similarly, $\|g\|_1^2={1\over 4}$.
On the other hand,
$$
\int |f+g|\,\mathrm{d}x=\int_0^1\mathrm{d}x=1,
$$
while 
$$
\int |f-g|\,\mathrm{d}x=\int_0^{{1\over 2}}|f-g|\,\mathrm{d}x+\int_{{1\over 2}}^1|f-g|\,\mathrm{d}x
$$
$$
=\int_0^{{1\over 2}}|f|\,\mathrm{d}x+\int_{{1\over 2}}^1|-g|\,\mathrm{d}x=\int_0^{{1\over 2}}|f|\,\mathrm{d}x+\int_{{1\over 2}}^1|g|\,\mathrm{d}x={1\over 2}+{1\over 2},
$$
therefore $\|f+g\|_1^2+\|f-g\|_1^2=1+1=2$, while $2\|f\|_1^2+2\|g\|_1^2=2{1\over 4}+2{1\over 4}=1$.
