Convergence of Sequences to $0$ and $\delta$ in Distribution

Question

Let $\lambda>0$, $\lambda_n \to +\infty$, and set $$f_n(x) = \sin (\lambda_n x) \ \ \ g_n(x) =\frac{\sin(\lambda_n x)}{\pi x}$$

1. Show that $f_n \to 0$ in $\mathcal{D}(\mathbb{R})$ as $n \to \infty$.
2. Show that $g_n \to \delta$ in $\mathcal{D}(\mathbb{R})$ as $n \to \infty$.

(You may use the fact that the improper integral $\int_{-\infty}^\infty \frac{\sin x}{x} dx = \pi$.)

This is a problem from an old applied analysis qualifying exam. I know have part 1 (see solutions below), but am still working on part 2. I have seen some "solutions" to 2, but all involve a questionable passage of the limit, or a vague reference to the Mean Value Theorem for Integrals that doesn't seem to fit...

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Here are two methods for the first question:

Problem 1, Method 1 - Riemann Lebesgue Lemma

Choose $\phi \in \mathcal{D}(\mathbb{R})$ with $supp\phi \in [-M,M]$ for some $M>0$. Notice that as $n \to \infty, \lambda_n \to \infty$. Furthermore, $\phi$ is clearly integrable on $[-M,M]$. Thus, by the Riemann Lebesgue Lemma, we have $$ \lim_{n \to \infty} f_n\phi = \lim_{n \to \infty}\int_{\mathbb{R}} \sin(\lambda_n x) \phi(x) dx = \lim_{n \to \infty}\int_{-M}^M \sin(\lambda_n x) \phi(x) dx =0 $$

Problem 1, Method 2 - Integration by Parts

Choose $\phi \in \mathcal{D}$ with $supp \in [-M,M]$. Then, using integration by parts with $u=\phi(x)$, we obtain: $$f_n \phi = \int_{\mathbb{R}} \sin (\lambda_n x) \phi(x)dx = \phi'(x)\frac{1}{\lambda_n}\sin (\lambda_n x) \rvert_{-\infty}^\infty+\frac{1}{\lambda_n}\int_{-\infty}^\infty \cos (\lambda_n x) \phi'(x) dx $$ $$= \frac{1}{\lambda_n}\int_{-\infty}^\infty \cos (\lambda_n x) \phi'(x) dx = \frac{1}{\lambda_n}\int_{-M}^M \cos (\lambda_n x) \phi'(x) dx $$ where the second to last step is because $\phi \in \mathcal{D}(\mathbb{R})$. Furthermore, as $\phi \in \mathcal{D}(\mathbb{R})$, we also know that $||\phi'||_\infty < \infty$, and we then have $$|f_n\phi| = \big|\frac{1}{\lambda_n}\int_{-M}^M \cos (\lambda_n x) \phi'(x) dx\big | \leq \frac{1}{\lambda_n}\int_{-M}^M |\phi'(x) |dx \leq \frac{1}{\lambda_n}\int_{-M}^M ||\phi' ||_\infty dx = \frac{1}{\lambda_n} 2 M ||\phi'||_\infty \to 0 \ \text{as} \ n \to \infty.$$

Answer 1

Having gotten the first part out of the way via integration by parts, let's focus on part 2.

We note that for every $\eta > 0$ and every $\varphi \in \mathcal{D}(\mathbb{R})$ we have

\begin{align} \Biggl\lvert \int_{\mathbb{R}} g_n(x)\varphi(x)\,dx - \varphi(0)\Biggr\rvert &= \Biggl\lvert \int_{\mathbb{R}} g_n(x)\bigl(\varphi(x) - \varphi(0)\bigr)\,dx\Biggr\rvert \\ &\leqslant\int_{-\eta}^{\eta} \lvert g_n(x)\rvert\cdot\lvert \varphi(x) - \varphi(0)\rvert\,dx + \Biggl\lvert\int_{\lvert x\rvert > \eta} g_n(x)\varphi(x)\,dx\Biggr\rvert + \Biggl\lvert \int_{\lvert x\rvert > \eta} g_n(x)\varphi(0)\,dx\Biggr\rvert. \end{align}

We estimate the three integrals in turn. By the mean value theorem, we have $\lvert\varphi(x) - \varphi(0)\rvert \leqslant K\lvert x\rvert$, where $K = \max \{ \lvert \varphi'(t)\rvert : t \in \mathbb{R}\}$, and hence

$$\int_{-\eta}^{\eta} \lvert g_n(x)\rvert\cdot \lvert \varphi(x) - \varphi(0)\rvert\,dx \leqslant \frac{K}{\pi}\int_{-\eta}^{\eta}\lvert\sin (\lambda_n x)\rvert\,dx \leqslant \frac{2K}{\pi}\cdot \eta.$$

For the second integral, choose $M$ so that $\operatorname{supp} \varphi \subset [-M,M]$. Integration by parts yields

$$\int_{\eta}^M g_n(x)\varphi(x)\,dx = \frac{\cos (\lambda_n\eta)\varphi(\eta)}{\lambda_n\pi\eta} + \frac{1}{\lambda_n\pi}\int_{\eta}^M \cos(\lambda_n x)\frac{x\varphi'(x) - \varphi(x)}{x^2}\,dx,$$

and a similar expression for the integral over $[-M,-\eta]$. By the boundedness of all involved functions on $[-M,M]\setminus (-\eta,\eta)$, we conclude that the second integral is bounded by $C\cdot \lambda_n^{-1}$ for a suitable $C$, and every fixed $\eta > 0$. The third integral is also bounded (for fixed $\eta > 0$) by $\tilde{C}\cdot \lambda_n^{-1}$, as can be seen by the substitution $y = \lambda_n x$ and an integration by parts.

From this, it follows that

$$\limsup_{n\to\infty} \Biggl\lvert \int_{\mathbb{R}} g_n(x)\varphi(x)\,dx - \varphi(0)\Biggr\rvert \leqslant \frac{2K}{\pi}\cdot \eta,$$

for every fixed $\eta > 0$. But the left hand side is independent of $\eta$, so

$$\limsup_{n\to\infty} \Biggl\lvert \int_{\mathbb{R}} g_n(x)\varphi(x)\,dx - \varphi(0)\Biggr\rvert \leqslant 0,$$

that is,

$$\lim_{n\to\infty} \int_{\mathbb{R}} g_n(x)\varphi(x)\,dx = \varphi(0),$$

or $g_n \to \delta$ in $\mathcal{D}'(\mathbb{R})$.