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$f$ and $g$ are measurable functions on a measure space $(\mathbb R, \mathcal{B}, m)$ where $m$ is Lebesgue measure. $f(x)$ satisfies $$\int_\mathbb R f^2(x) dm(x) < \infty$$ Where, $$f_k(x) = \sqrt{k} f(kx).$$

Prove that if $g(x)$ satisfies $$\int_\mathbb R g^2 dm < \infty$$ then $$\lim\limits_{k \rightarrow \infty} \int_\mathbb R f_k(x) g(x) dm(x) = 0.$$

I can prove that $ f_k(x) g(x) $ is integrable, but I didn't figure out how to get the final result.

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  • $\begingroup$ @Jack: Yes, Cauchy inequality makes sure that $f_k(x) g(x) $ is integrable. How can I prove the limit is zero? $\endgroup$ – David Nov 15 '17 at 15:29
  • $\begingroup$ @GuyFsone: No, that's all. I thought the way to prove dominated convergence theorem could work. $\endgroup$ – David Nov 15 '17 at 15:47
  • $\begingroup$ I have solved it by approximation of integral by continuous and compactly supported functions it is correct as well $\endgroup$ – Guy Fsone Nov 15 '17 at 16:32
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    $\begingroup$ Strongly related $\endgroup$ – Giuseppe Negro Nov 16 '17 at 18:11
  • $\begingroup$ @GiuseppeNegro: Thank you! I didn't find this question before. $\endgroup$ – David Nov 17 '17 at 6:16
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Lemma: Continuous function of compact support are dense in $L^2(dm)$. Hence,

For, $\varepsilon>0$ we know that there exist a $f_\varepsilon,g_\varepsilon\in C_c(\Bbb R)$ a continuous function of compact support such that, $$ \|g-g_\varepsilon\|_2\le \varepsilon \|f\|_2^{-1} $$ and $$ \|f-f_\varepsilon\|_2\le \varepsilon \|g_\varepsilon\|_2^{-1} $$

Then,using Cauchy Schwartz inequalty we have, \begin{align} \left|\int_\mathbb R f_k(x) g(x) dm(x)\right| &= \left|\int_\mathbb R f_k(x) [g(x)-g_\varepsilon(x)] dm(x)+\int_\mathbb R f_k(x) g_\varepsilon(x) dm(x)\right|\\ &\le \|g-g_\varepsilon\|_2 \left( \int_\mathbb R k f^2(kx)dm(x)\right)^{1/2} +\left|\int_\mathbb R \sqrt{k}f(kx) g_\varepsilon(x) dm(x)\right| \\& \overset{\color{red}{u=kx}}{=} \|g-g_\varepsilon\|_2 \|f\|_2 + \left|\int_\mathbb R\frac{1}{ \sqrt{k}}f(x) g_\varepsilon(\frac{x}{k}) dm(x)\right|\\ & \le \varepsilon + \frac{1}{ \sqrt{k}}\left|\int_\mathbb Rf(x) g_\varepsilon(\frac{x}{k}) dm(x)\right| \end{align}

Similarly we have, $$ \left|\int_\mathbb Rf(x) g_\varepsilon(\frac{x}{k}) dm(x)\right| \le \varepsilon+ \left|\int_\mathbb Rf_\varepsilon(x) g_\varepsilon(\frac{x}{k}) dm(x)\right| \\= \varepsilon +\left|\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(\frac{x}{k}) dm(x)\right|$$

Since, $f_\varepsilon$ $h_\varepsilon$ are continuous functions with compact support, we get $$\color{blue}{\lim_{k\to \infty}\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(\frac{x}{k}) dm(x)=\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(0) dm(x)}$$

Whence, $$ \lim_{k\to \infty} \left|\int_\mathbb R f_k(x) g(x) dm(x)\right| \le\lim_{k\to \infty} \left[\varepsilon+\frac{\varepsilon}{ \sqrt{k}}+ \frac{1}{ \sqrt{k}}\left|\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(\frac{x}{k}) dm(x)\right| \right]=\varepsilon $$

That is for all $\varepsilon>0,$ we have, $$ \limsup_{k\to \infty} \left|\int_\mathbb R f_k(x) g(x) dm(x)\right| \le \varepsilon $$ letting $\varepsilon\to 0$ yields the require result.

$$ \color{red}{\lim_{k\to \infty} \int_\mathbb R f_k(x) g(x) dm(x)=0.} $$

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  • $\begingroup$ @peterag you are right Ok I forgotten to remove the squares aftermath $\endgroup$ – Guy Fsone Nov 16 '17 at 13:50

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