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Let $n>0$ and define the function $(f_n)$ as $f_n(x)=\sin(nx)$. Show that $T_{f_n} \underset{\mathcal{D'}}{\rightarrow} T_0$. As a hint, I was told to use the fact that for distributions one can interchange differentiation and limit.

What I did: The idea is that if I calculate the primitive $(F_n)$ of $(f_n)$ and show that $(F_n)$ converges uniformly on any compact set to $0$, then I could somehow use differentiation and interchange it with the limit.

$\displaystyle \int \sin(nx)dx=-\frac{\cos(nx)}{n}$

$\displaystyle \sup_{x \in [-\pi,\pi] }|-\frac{\cos(nx)}{n}| \leq \frac{1}{n} \leq \epsilon$

if $\epsilon \geq \frac{1}{n}$

This means $T_{F_n} \underset{\mathcal{D'}}{\rightarrow} T_0$

If $L=\frac{d}{dx}$ , then $L^*=\frac{d}{dx}$

$$\lim_{n \rightarrow \infty} \langle T_{f_n},\phi \rangle=\lim_{n \rightarrow \infty} \langle \frac{d}{dx}T_{F_n},\phi \rangle \underset{(1)}{=} \frac{d}{dx}\langle \lim_{n \rightarrow \infty}T_{F_n},\phi \rangle$$ I am a little bit confused on how to write that down. I don't even know how $\frac{d}{dx} \langle T,\phi \rangle$ is defined in case of Distributions.

I only know that if $L: \mathcal{D}(\Omega_1) \rightarrow \mathcal{D}(\Omega_2)$ is a linear and sequentially continuous function and further if $L^*$ exists and is also sequentially continuous with $\langle L(\phi),\psi=\langle \phi, L^* \psi \rangle$ for all $\phi \in \mathcal{D'}(\Omega_1), \psi \in \mathcal{D'}(\Omega_2) $ Then the operator $L$ is defined as $\langle L(T),\psi \rangle:=\langle T, L^* (\psi)\rangle$

Hope someone could show me how to write it down correctly (and if my approach is correct).

ACB
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John.W
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  • Your Q could benefit from some added punctuation to your train of reasoning (comma, period, implication arrow as intended), and TeX placement of a backslash [ \ ] before common proper non-variable names (operators, bounds, etc.; e.g. $\sin$, $\sup$) to make them appear more properly (i.e., not italicized like something generic). For the particular case of derivative "d" one way to format it is via \mathrm{d} (which can be quick-sourced by right-clicking the LaTeX portion of the title of this top Q). – user946772 Dec 31 '21 at 03:43
  • How about evaluating against a given test function, then integrating by parts before letting $n\rightarrow +\infty$? – Stefan Lafon Dec 31 '21 at 03:56
  • @StefanLafon Do you mean like this $lim_{n \rightarrow \infty} \int_{\Omega} f_n(x) \phi(x)dx=lim_{n \rightarrow \infty} \int_{\Omega} \frac{d}{dx} F_n(x) \phi(x)dx= \frac{d}{dx} \int_{\Omega} lim_{n \rightarrow \infty} F_n(x) \phi(x)dx$. Is there some way to write this down with the Distribution Notation? – John.W Dec 31 '21 at 04:29
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    $\int \phi (x)f_n(x)dx \to 0$ as $n \to \infty$ for every test function $\phi$ by Riemann Lebesgue Lemma. – Kavi Rama Murthy Dec 31 '21 at 04:56
  • @KaviRamaMurthy But why? As mentioned in the question, I should use the fact that I can change limit and derivative. – John.W Dec 31 '21 at 19:15

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By definition, if $\varphi$ is a test function, $$\langle T_{f_n}, \varphi\rangle = \int_{\mathbb R} f_n(x)\varphi(x)dx=\int_{\mathbb R} \sin(nx)\varphi(x)dx$$ If you integrate by parts, noting that the test function evaluated at $\pm \infty$ is $0$: $$\langle T_{f_n}, \varphi\rangle = \int_{\mathbb R} \frac{\cos(nx)}n\varphi^\prime(x)dx = \frac 1 n\int_I \cos(nx)\varphi^\prime(x)dx$$ where $I$ is the support of $\varphi$. Finally, you can conclude by noticing that $$\left|\langle T_{f_n}, \varphi\rangle \right|\leq \frac{\|\varphi^\prime\|_\infty |I|}{n}\xrightarrow[n\rightarrow+\infty]{}0$$

As noted in the comments, this is known as the Riemann-Lebesgue lemma.

Stefan Lafon
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