# Prove that $\lim_{n \to \infty} \int_{-\infty}^{\infty} \sin(nt) f(t) d t = 0$.

I am trying to prove that $\lim_{n \to \infty} \int_{-\infty}^{\infty} \sin(nt) f(t) d t = 0$ for every Lebesgue integrable function $f$ on $\mathbb{R}$. My first thoughts were to use Dominated Convergence Theorem but I realised that there is no pointwise limit of the sequence of functions $f_n = \sin(nt) f(t)$. I do not know how to proceed.

Any help would be appreciated. Thanks!

(This is the content of the Riemann-Lebesgue Lemma.)

Sketch of proof:

1. Fix $\epsilon > 0$, there exists $g$ that is continuously differentiable and has compact support, such that $\|f - g\|_{L^1} < \epsilon / 2$.
2. $\int \sin(nt) g(t) dt = \frac{1}{n} \int \cos(nt) g'(t) dt \to_{n\to\infty} 0$.
3. Hence there exists $N > 0$ such that for every $n > N$, $|\int \sin(nt) f(t) dt| < \epsilon$.
• The first assumption is too strong and it requires a proof. – Danny Pak-Keung Chan Mar 22 '18 at 22:33
• I don't understand your comment, the first step is a fact, not an assumption; that it requires a proof is given since I am only outlining the proof for the reader. – Willie Wong Mar 23 '18 at 16:01
• I mean the proof of Step 1 is more difficult than the proof of Riemann-Lebesgue Lemma. Reader may wonder why Step 1 is true. – Danny Pak-Keung Chan Mar 23 '18 at 21:05
• This will make your argument not self-contained. (as it assumes a result that is hard to proof). If I remember correctly, $g$ is obtained by convoluting $f$ with a $C^\infty$ approximate identity. The approximate identity is a net (in this concrete case, it can be chosen as a sequence) of non-negative, $C^\infty$ functions whose supports are compact and are shrinking to the singleton $\{0\}$ and each of them has integral equals to 1. – Danny Pak-Keung Chan Mar 23 '18 at 21:10

1. It is true for function of the form $1_{[a,b]}$, where $-\infty<a<b<\infty$ by direct computation.

2. By linearity, it is true for all step functions.

3. Step functions is $||\cdot||_1$-dense in $L^1$.

People may question about the validity of step 3. But this ultimately boils down to the regularity of Lebesgue measure: For any measurable set $A$ with $m(A)<\infty$, and $\varepsilon>0$ we can find an open set $U$ such that $A\subseteq U$ and $m(U)-m(A)< \varepsilon$. Write $U$ as a countably union of disjoint open intervals. Since the sum of the length of all intervals is $m(U)$ which is finite, we can keep finitely many of these intervals and use them to approximate $U$. The sum of the characteristic functions of these intervals will approximate the characteristic function of $A$, in $||\cdot||_1$ norm.

To facilitate readers about the existence of the so-called approximated identity, I prove the following.

For any $\delta>0$, there exists a function $\phi:\mathbb{R}\rightarrow[0,\infty)$ such that:

1. $\phi$ is $C^{\infty}$, i.e., For each $k\in\mathbb{N}$, the $k$-th order derivative $\phi^{(k)}$ exists.

2. $\phi(x)=0$ whenever $x\in\mathbb{R}\setminus(-\delta,\delta)$.

3. $\int_{-\infty}^{\infty}\phi(x)dx=1$.

Sketch of proof:

Let $\psi:\mathbb{R}\rightarrow\mathbb{R}$ by $$\psi(x)=\begin{cases} e^{-\frac{1}{x}}, & \mbox{ if }x>0\\ 0, & \mbox{ if }x\leq0 \end{cases}.$$ It can be proved that $\psi$ is $C^{\infty}$ (The most difficult part is probably proving $\psi^{(k)}(0)=0$ for each $k\in\mathbb{N}$. This can be done by induction by working out the form of $\psi^{(k)}(x)$, for $x>0$. The form is $\frac{P(x)}{x^{n(k)}}e^{-\frac{1}{x}}$, where $P(x)$ is a polynomial in $x$ and $n(k)$ is a positive integer that depends on $k$ only. In fact, this is a modification of Cauchy's famous counter-example that $C^{\infty}$does not imply analytic.). Note that $\psi(x)$ has the property that $\psi(x)=0$ for all $x\leq0$ and $0<\psi(x)<1$ for all $x>0$. Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by $f(x)=\psi(x+\delta)\psi(-x+\delta)$, then $f$ is $C^{\infty},$ non-negative, and $f(x)=0$ whenever $x\in\mathbb{R}\setminus(-\delta,\delta)$ but $0<f(x)<1$ whenever $x\in(-\delta,\delta)$. Observe that $f$ is bounded and hence $0<c:=\int_{-\infty}^{\infty}f(x)dx<\infty.$

Finally, define $\phi(x)=\frac{1}{c}\cdot f(x)$.

For each $\delta=\frac{1}{n}$, denote such a function by $\phi_n$. Then the sequence $(\phi_n)$ has the property that for any integrable function $f$, $f\star\phi_n$ is $C^\infty$, integrable, and $||f\star\phi_n -f||_1\rightarrow 0$ as $n\rightarrow \infty$.

The result follows from the Riemann-Lebesgue theorem:

If $f$ is an integrable function in the sense of Lebesgue then $$\lim_{n\to\infty}\int f(t)e^{int}\,dt=0$$ It is easy to show that when $f(t):=\chi(t)_{(a,b)}$ on the open interval $(a,b)$ then $$\lim_n\Big|\int f(t)e^{i nt}\,dt\Big|=\lim_n\Big|\int^b_ae^{int}\,dt\Big|=\lim_n\Big|\frac{1}{in}(e^{inb}-e^{ina})\Big|\leqslant\lim_n\frac{2}{n}=0$$ hence it follows that for any such characteristic function $$\lim_n\int f(t)e^{int}\,dt=0$$ This in turn implies that any step function $g$ (a linear combination of characteristic functions) the same vanishing limit holds since integration is a linear operator. Now take any Lebesgue integrable function $f$. Then for each $\varepsilon>0$ there exists a step function $g$ such that $$\int|f(t)-g(t)|\,dt<\varepsilon$$ Therefore we have $$\Big|\int f(t)e^{int}\,dt\Big|\leqslant \Big|\int(f(t)-g(t))e^{int}\,dt\Big|+\Big|\int g(t)e^{int}\,dt\Big|\\\leqslant \int|f(t)-g(t)|\,dt+\Big|\int g(t)e^{int}\,dt\Big|<\varepsilon+\varepsilon=2\varepsilon$$ since for every $\varepsilon>0$ there is some $N$ such that whenever $n\geqslant N$ then $|\int g(t)e^{int}\,dt|<\varepsilon$. Therefore whenever $f$ is Lebesgue integrable then $$\lim_n\int f(t)e^{int}\,dt=0$$ This requires both the real part and the imaginary part to vanish in the limit i.e. $$\lim_n\int f(t)\cos(nt)\,dt=\lim_n\int f(t)\sin(nt)\,dt=0$$