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Can you tell me if this is correct? Thanks.

Claim: The following inner product is well-defined, i.e. finite for all $f,g \in L^2$: $$ (f,g) = \int_X f \overline{g} d \mu$$

Proof: We may assume $f,g$ are real-valued since we can otherwise write them as sums of real and imaginary part. We can also assume that they are $\geq 0$ since otherwise we can write $f = f^+ - f^-$ for $f^+ , f^- \geq 0$. By assumption, $f,g \geq 0$ are measurable hence there are sequences of simple functions $s_n \to f, t_n \to g$ converging pointwise and $s_n \leq s_{n+1}, t_n \leq t_{n+1}$ and hence $s_n t_n \to fg$ pointwise and $s_n t_n$ is non-decreasing. Simple functions form an inner product space with respect to $\langle .,.\rangle$ so Cauchy Schwarz holds. Then

$$\begin{align} \int_X fg d \mu = \int_X \lim_{n \to \infty} s_n t_n d \mu & \stackrel{\text{monotone conv.}}{=}\lim_{n \to \infty} \int_X s_n t_n d \mu \\ & = \lim_{n \to \infty} \int_X |s_n| | t_n| d \mu \\ & \stackrel{\text{Cauchy Schwarz}}{\leq} \lim_{n \to \infty} \|s_n \|_2 \|t_n \|_2 \\ & \stackrel{\text{continuity of norm}}{=} \|f \|_2 \|g \|_2 < \infty \end{align}$$

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I'm quite sure it's correct but I need to be sure. Sorry. And sorry for the align gone wrong, I don't know how to fix it. –  Matt N. Jul 16 '12 at 14:56
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Yes, looks right. You proved Hölder only for simple functions? I wonder, why you can't just say that $|(f,g)| \le \int |f||g| \le \|f\|_2\|g\|_2$ by Hölder ... –  martini Jul 16 '12 at 14:56
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Hm ... for simple functions I think you can get your needed inequality from Cauchy-Schwarz on $\mathbb R^n$ ... $s_n = \sum_k \alpha_k \chi_{A_k}$, $t_n = \sum_\ell \beta_\ell \chi_{B_\ell}$, then $\int_X s_nt_n = \sum_{k,\ell} \alpha_k\beta_\ell \mu(A_k\cap B_\ell) \le \left(\sum_{k,\ell} \alpha_k^2\mu(A_k\cap B_\ell)\right)^{1/2}\left(\sum_{k,\ell} \beta_\ell^2\mu(A_k\cap B_\ell)\right)^{1/2}$. –  martini Jul 16 '12 at 15:00
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Just notice that $$ |f\bar{g}| \leq \frac{1}{2}(|f|^2 + |g|^2) $$ by AM-GM inequality. This is sufficient to prove that $f \bar{g}$ is integrable. –  sos440 Jul 16 '12 at 15:01
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There is another point to be made: remember that elements of $L^2$ are not actually functions, but equivalence classes of functions. I would argue that you should also show that if $f\sim f'$ and $g\sim g'$, then $\langle f,g\rangle = \langle f',g'\rangle$. –  Arturo Magidin Jul 16 '12 at 18:18
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