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.
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 $2\|f\|_1^2+2\|g\|_1^2=2{1\over 4}+2{1\over 4}=1$.