1

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.

lethecullen
  • 43

2 Answers2

0

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

$$||f + g||^2 + ||f-g||^2 = 2 ( ||f||^2 + ||g||^2 )$$

Bjørn Kjos-Hanssen
  • 2,915
0

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

EM90
  • 479
  • 2
  • 10
  • For the case of $L^1([0,1])$, you can see this discussion: https://math.stackexchange.com/questions/621934/l10-1-is-not-a-hilbert-space. – EM90 Oct 08 '17 at 07:16
  • Is it possible that you can give me more details? I checked the case for $L^1([0,1])$, but I'm still confused about it. I also tried to use the parallelogram equality, but I'm still struggling with it:( – lethecullen Oct 08 '17 at 23:43
  • Notice that $\mathbb{1}{[0,{1\over 2}]}$ and $\mathbb{1}{[{1\over 2},1]}$ are integrable over $\mathbb{R}$. I tried to add details to the answer. Please, let me know if it is more clear now (or if you find some mistake!). – EM90 Oct 09 '17 at 08:16
  • Thank you! I was confused by the domain R. I thought that the function has to be defined on R. – lethecullen Oct 09 '17 at 16:14
  • Yeah, well: both $\mathbb{1}{[0,{1\over 2}]}$ and $\mathbb{1}{[{1\over 2},1]}$ can be defined on the whole $\mathbb{R}$. Or on the smaller space $[0,1]$. The point is that both our functions are null ouside of a compact set! – EM90 Oct 10 '17 at 17:12