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I know that $\int f(x) \overline{g(x)} dx$ is an inner product on $L^2$. But is it one on $L^1$? I think it isn't, but I am have had difficulty figuring out which defining property is violated.

Thanks in advance for any pointers!

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Think about a real space and about $x^a$ for some appropriate $a$. –  Jonas Teuwen Jan 21 '11 at 15:15

3 Answers 3

More explicitely: Take $L^p([0,1])$ and consider $f(x) = 1/\sqrt{x}$. Then $f\in L^1$ but $\int_0^1 f(x)^2 dx$ is infinte.

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Take $f$ a function in $L^1$ but not in $L^2$. We had examples recently-they blow up too much near a point. Then $\int f(x) \overline{f(x)} dx$ is not defined.

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Suppose that $g$ is a measurable function on $[0,1]$ such that $\int |fg|<\infty$ for all $f\in L^1[0,1]$. Then $g$ is essentially bounded.

If $g$ is not essentially bounded and $E_n=\{x\in[0,1]:n\leq |g(x)|\lt n+1\}$ for each positive integer $n$, then there is a subsequence $E_{n_1},E_{n_2},\ldots$ with $m(E_{n_k})\gt 0$ for all $k$. Let $\displaystyle{f=\sum_{k=1}^\infty}\frac{1}{k^2m(E_{n_k})}\chi_{E_{n_k}}$, so $f$ is in $L^1$, and $\displaystyle{\int |gf| \geq \sum_{k=1}^\infty\frac{n_k}{k^2}\geq\sum_{k=1}^\infty}\frac{1}{k}=\infty$.

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@Jonas Meyer: Nice, but a more interesting question to me would be: Why is $L^2$ the only Hilbert space (and the rest are isometrically isomorphic)? I know you can use that to show that $L^2$ must be the only one but is there a direct argument why we cannot define an inner product on $L^p$ $p > 1, p \neq 2$ that makes the space into a Hilbert space? –  Jonas Teuwen Jan 22 '11 at 14:40
    
Doesn't it have to do with the parallelogram law? It can be shown that the parallelogram property doesn't hold for $p \ne 2$ I believe. I guess that only tells you that $L^p$ can't be a Hilbert Space using the usual $L^p$ norm though. –  Brian Jan 22 '11 at 18:40

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