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Let $\phi \in C^1_c(\mathbb R)$. Prove that $ \lim_{n \to +\infty} \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx = \pi\phi(0). $

Unfortunately, I didn't manage to give a complete proof. First of all, I fixed $\varepsilon>0$. Then there exists a $\delta >0$ s.t. $ \vert x \vert < \delta \Rightarrow \vert \phi(x)-\phi(0) \vert < \frac{\varepsilon}{\pi}. $ Now, I would use the well-known fact that $ \int_\mathbb R \frac{\sin x}{x} \, dx = \pi. $ On the other hand, by substitution rule, we have also $ \int_\mathbb R \frac{\sin(nx)}{x} \, dx = \int_\mathbb R \frac{\sin x}{x} \, dx = \pi. $ Indeed, I would like to estimate the quantity $ \begin{split} & \left\vert \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx - \pi \phi(0) \right\vert = \\ & = \left\vert \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx - \phi(0)\int_\mathbb R \frac{\sin{(nx)}}{x}dx \right\vert \le \\ & \le \int_\mathbb R \left\vert \frac{\sin(nx)}{x}\right\vert \cdot \left\vert \phi(x)-\phi(0) \right\vert dx \end{split} $ but the problem is that $x \mapsto \frac{\sin(nx)}{x}$ is not absolutely integrable over $\mathbb R$. Another big problem is that I don't see how to use the hypothesis $\phi$ has compact support.

I think that I should use dominated convergence theorem, but I've never done exercises about this theorem. Would you please help me? Thank you very much indeed.

4 Answers 4

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Note that $ \small \int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\phi(x)dx-\pi\phi(0)= \int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\phi(x)dx-\phi(0)\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}dx= \int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx $ Denote $ \small I_{m}(n):=\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx\qquad I(n):=\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx $ We claim that $I_m(n)$ converges to $I(n)$ uniformly by $n\in\mathbb{N}$ when $m\to\infty$. Indeed, since $\phi$ is compactly supported $ \small \lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|I_m(n)-I(n)\right|= \lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx\right|= $ $ \small \lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(nx)}{x}(-\phi(0))dx\right|= \lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\phi(0)\int\limits_{\mathbb{R}\setminus[-\pi mn,\pi mn]}\frac{\sin(y)}{y}dy\right|= $ $ \small |\phi(0)|\lim\limits_{m\to\infty}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(y)}{y}dy\right|=0 $ Since convergence is uniform by $n\in\mathbb{N}$ we can say $ \small \lim\limits_{n\to\infty} I(n)= \lim\limits_{n\to\infty} \lim\limits_{m\to\infty} I_m(n)= \lim\limits_{m\to\infty}\lim\limits_{n\to\infty} I_m(n)= \lim\limits_{m\to\infty}\lim\limits_{n\to\infty}\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx $ Since $\varphi\in C_c^1(\mathbb{R})$, then the function $x^{-1}(\varphi(x)-\varphi(0))$ is in $L^1([-\pi m,\pi m])$ for all $m\in\mathbb{N}$. Then by Riemann–Lebesgue lemma $ \small \lim\limits_{n\to\infty}\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx=0 $ so $\small\lim\limits_{n\to\infty}I(n)=0$. This is exactly what we wanted to prove.

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    @Romeo No, I wont!2012-07-26
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A quick proof of this can be given using the Riemann-Lebesgue lemma, which is covered in Rudin and a number of other texts. Write your limit as $\lim_{n \rightarrow \infty} \int_\mathbb R \sin(nx)\frac{\phi(x) - \phi(0)\chi_{[-1,1]}(x)}{x}\, dx + \lim_{n \rightarrow \infty} \int_\mathbb R \sin(nx)\frac{\phi(0)\chi_{[-1,1]}(x)}{x}\, dx $ Here $\chi_{[-1,1]}(x)$ denotes the characteristic function of $[-1,1]$. Since $\phi(x) \in C_c^1({\mathbb R})$, the function ${\displaystyle \frac{\phi(x) - \phi(0)\chi_{[-1,1]}(x)}{x}}$ is a bounded function with compact support; the only place you have to worry about is $x = 0$ and you can use the mean value theorem for example to show it's bounded near $x = 0$. Since the function is bounded function with compact support it is in $L^1$, which is enough to apply the Riemann-Lebesgue lemma and say the first term goes to zero. As for the second term, after changing variables to $y = nx$ we may rewrite it as $\lim_{n \rightarrow \infty} \int_\mathbb R \sin(y)\frac{\phi(0)\chi_{[-n,n]}(y)}{y}\, dy $ $= \phi(0)\lim_{n \rightarrow \infty} \int_{-n}^n \frac{\sin(y)}{y}\, dy $ $= \phi(0)\int_\mathbb R \frac{\sin(y)}{y}\, dy $ $= \pi \phi(0)$ So this will be the overall limit.

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    If you assume it, then the argument for the convergence on $0$ may not follow by mean value theorem, but just using boundedness of the ratio $\frac{\phi(x)-\phi(0)}x$.2012-07-26
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We can assume WLOG that $\phi\in C^3_c(\Bbb R)$, since this subset is dense for the supremum norm and $\frac{\sin x}x$ is integrable over compact subsets. We have $\phi(x)-\phi(0)=x\phi'(0)+x\int_0^1(1-s)\phi''(sx),$ hence, if the support of $\phi$ is contained in $[-R,R]$ \begin{align} \small \int_{-\infty}^{+\infty}\frac{\sin(nx)}x\phi(x)dx&=\small\phi(0)\int_{-R}^R\frac{\sin(nx)}xdx+\phi'(0)\int_{-R}^R\sin(nx)dx+\int_{—R}^R\sin(nx)\int_0^1(1-s)\phi''(sx)dsdx\\ &=\small\phi(0)\int_{-nR}^{nR}\frac{\sin t}tdt-\left[\frac{\cos(nx)}n\int_0^1(1-s)\phi''(sx)ds\right]_{-R}^R+\int_{—R}^R\frac{\cos(nx)}n\int_0^1 s(1-s)\phi'''(sx)dsdx \end{align} We have $\lim_{n\to+\infty}\int_{-nR}^{nR}\frac{\sin t}tdt=\int_{-\infty}^{+\infty}\frac{\sin t}tdt;$ $\left|\left[\frac{\cos(nx)}n\int_0^1(1-s)\phi''(sx)ds\right]_{-R}^R\right|\leq \frac 2n\sup_{t\in \Bbb R}|\phi''(t)|,$ and $\left|\int_{—R}^R\frac{\cos(nx)}n\int_0^1 s(1-s)\phi'''(sx)dsdx\right|\leq \frac{2R}n\sup_{t\in \Bbb R}|\phi'''(t)|$

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Assume that $\phi(x)$ is supported in $|x|< L$. Since $\phi$ is differentiable, $\frac{\phi(x)-\phi(0)}{x}$ is bounded and therefore integrable on $|x|. $ \begin{align} \int_{-\infty}^\infty\frac{\sin(nx)}{x}\phi(x)\,\mathrm{d}x &=\pi\,\phi(0)+\int_{-\infty}^\infty\sin(nx)\frac{\phi(x)-\phi(0)}{x}\,\mathrm{d}x\\ &=\pi\,\phi(0)+\color{#C00000}{\int_{-L}^L\sin(nx)\frac{\phi(x)-\phi(0)}{x}\,\mathrm{d}x}\\ &-2\,\phi(0)\color{#00A000}{\int_{nL}^\infty\frac{\sin(x)}{x}\,\mathrm{d}x}\\ \end{align} $ As $n\to\infty$, the red integral vanishes by the Riemann-Lebesgue Lemma and the green integral vanishes because The Dirichlet Integral converges. This leaves us with $ \lim_{n\to\infty}\int_{-\infty}^\infty\frac{\sin(nx)}{x}\phi(x)\,\mathrm{d}x=\pi\,\phi(0) $