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...

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 $$|f_n\phi| = \big|\frac{1}{\lambda_n}\int_{-M}^M \cos (\lambda_n x) \phi'(x) dx\big | \leq \frac{1}{\lambda_n}\int_{-M}^M |\phi'(x) |dx \leq \frac{1}{\lambda_n}\int_{-M}^M ||\phi' ||_\infty dx = \frac{1}{\lambda_n} 2 M ||\phi'||_\infty \to 0 \ \text{as} \ n \to \infty.$$

  • $\begingroup$ For 1, try an integration by parts rather than a substitution. Also look up or recall the Riemann-Lebesgue lemma. $\endgroup$ Jan 5, 2017 at 18:42
  • $\begingroup$ Ohhhh. As $\phi$ integrable on its support $[-M,M]$, would it be enough then to simply say $\int_{\mathbb{R}} \sin \lambda_n x\phi(x)dx \to 0$ as $n \to \infty$ by the Riemann-Lebesgue Lemma? And we wouldn't even need to use integration by parts? $\endgroup$
    – Chriz26
    Jan 5, 2017 at 18:57
  • $\begingroup$ Or rather $\int_{\mathbb{R}} \sin (\lambda_n x) \phi(x)dx = \int_{-M}^M \sin (\lambda_n x) \phi(x)dx \to 0$ as $n \to \infty$. $\endgroup$
    – Chriz26
    Jan 5, 2017 at 19:06
  • $\begingroup$ Yes, one could appeal to the Riemann-Lebesgue lemma. But that is (typically at least) proved using integration by parts for differentiable functions, and then by an approximation argument for general functions. So avoiding the integration by parts by appealing to the Riemann-Lebesgue lemma is … actually using more work. $\endgroup$ Jan 5, 2017 at 19:11
  • $\begingroup$ Makes sense! Thank you! I have part 1 now! $\endgroup$
    – Chriz26
    Jan 5, 2017 at 20:17

1 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 $\mathcal{D}'(\mathbb{R})$.

  • $\begingroup$ Just clarifying that it's fine that $C$ and $\tilde{C}$ depend on $\eta$ as they are controlled by the $\frac{1}{\lambda_n}$...but it's crucial that $K$ be independent of $\eta$, correct? $\endgroup$
    – Chriz26
    Jan 7, 2017 at 0:44
  • $\begingroup$ Correct. Although we could permit a dependence of $K$ on $\eta$ as long as $K(\eta)\cdot \eta \to 0$. But it's crucial that $K\cdot \eta \to 0$, while the dependence of $C,\tilde{C}$ on $\eta$ is irrelevant because we let $n\to\infty$ for a fixed $\eta$. $\endgroup$ Jan 7, 2017 at 11:01

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