Uniform convergence of $\varphi_n(x)=\int_{-\infty}^{\infty}\frac{\sin(n(y-x))}{n(y-x)}f(y)dy$

Question

Let $f\in L^1(\mathbb{R})$ and $n\in\mathbb{Z}_{+}$, we define $$\varphi_n(x):=\int_{-\infty}^{\infty}\frac{\sin(n(y-x))}{n(y-x)}f(y)dy\ .\ \ (x\in\mathbb{R})$$ (However, if $y=x$, it is interpreted as $\frac{\sin(n(y-x))}{n(y-x)}=1$.)

Proof that $\{\varphi_n\}$ converges uniformly to $0$.

Answer 1

1^st proof. Let $\operatorname{Leb}$ denote the Lebesgue measure. For any $\epsilon > 0$,

• there exists $\delta > 0$ such that $\operatorname{Leb}(E) \leq \delta$ implies $\int_{E} |f(x)| \, \mathrm{d}x < \epsilon/2$, and

• there exists $N \geq 1$ such that $ \frac{2}{n\delta}\int_{-\infty}^{\infty} |f(x)| \, \mathrm{d}x < \epsilon/2$ for all $n \geq N$.

Then for any $n \geq N$ and $x \in \mathbb{R}$,

\begin{align*} \left| \varphi_n(x) \right| &\leq \left| \int_{|y-x|<\delta/2} \frac{\sin(n(y-x))}{n(y-x)} f(y) \, \mathrm{d}y \right| + \left| \int_{|y-x|>\delta/2} \frac{\sin(n(y-x))}{n(y-x)} f(y) \, \mathrm{d}y \right| \\ &\leq \int_{|y-x|<\delta/2} \left| f(y) \right| \, \mathrm{d}y + \int_{|y-x|>\delta/2} \frac{2}{n\delta} \left| f(y) \right| \, \mathrm{d}y \\ & < (\epsilon/2) + (\epsilon/2) = \epsilon. \end{align*}

Therefore $\varphi_n \to 0$ uniformly as $n\to\infty$.

---

2^nd proof. Write

$$ S(x) = \int_{-\infty}^{x} \frac{\sin t}{t} \, \mathrm{d} t. $$

We know that $S(x)$ is bounded on $\mathbb{R}$. Let $M > 0$ denote a bound of $S$, and write $C^1_c(\mathbb{R})$ for the set of all compactly supported $C^1$-functions on $\mathbb{R}$. Then for any $g \in C^1_c(\mathbb{R})$,

\begin{align*} \left| \varphi_n(x) \right| &\leq \left| \int_{-\infty}^{\infty} \frac{\sin(n(y-x))}{n(y-x)} (f(y) - g(y)) \, \mathrm{d}y \right| + \left| \int_{-\infty}^{\infty} \frac{\sin(n(y-x))}{n(y-x)} g(y) \, \mathrm{d}y \right| \\ &\leq \| f - g \|_{L^1} + \left| \frac{1}{n}\int_{-\infty}^{\infty} S(n(y-x)) g'(y) \, \mathrm{d}y \right| \\ &\leq \| f - g \|_{L^1} + \frac{M}{n} \int_{-\infty}^{\infty} |g'(y)| \, \mathrm{d}y. \end{align*}

So, taking supremum over $x \in \mathbb{R}$ and taking limsup as $n\to\infty$,

$$ \limsup_{n\to\infty} \left( \sup_{x\in\mathbb{R}} \left| \varphi_n(x) \right| \right) \leq \| f - g\|_{L^1}. $$

However, since the left-hand side is independent of $g$ and $C^1_c(\mathbb{R})$ is dense in $L^1(\mathbb{R})$, letting $g \to f$ in $L^1$ proves the desired claim.