Demonstrate that $\lim_{n \rightarrow \infty} \int_{\mathbb{R}} f(x)\sin(nx) ~ dx = 0 $ for $f \in L^{1}(\mathbb{R}).$

We will first demonstrate that this holds for step functions. Let, $\phi = \mathcal{X}_{(a,b)}$ where $a < b$. Now, $$\lim_{n \rightarrow \infty} \int_{\mathbb{R}}\phi \sin(nx) ~ dx = \lim_{n \rightarrow \infty} \int^{b}_{a}\sin(nx) dx = 0.$$ The last statement can be obtained via direct computation of the integral. Let, $\varphi = \sum^{N}_{j=1}c_j\mathcal{X}_{I_j}$ where $I_j$ is an interval for $j \in [1,N]$. Using additivity of integral, we obtain: $$\lim_{n \rightarrow \infty} \int_{\mathbb{R}}\varphi \sin(nx)~dx = 0. $$
We know that step functions are dense in $L^1(\mathbb{R})$. Hence, there exists a sequence of step functions $\{\varphi_k\}_{k \ge 1}$ such that: $$\lim_{k \rightarrow \infty} \left\lVert f - \varphi_k\right\rVert_{1} = 0.$$ Hence, $$|\int_{\mathbb{R}}f\sin(nx)~ dx | \le |\int_{\mathbb{R}}f\sin(nx)~ dx - \int_{\mathbb{R}}\varphi_k\sin(nx)~ dx | + |\int_{\mathbb{R}}\varphi_k\sin(nx)~ dx |.$$ Then, $$ |\int_{\mathbb{R}}f\sin(nx)~ dx | \le \int_{\mathbb{R}}|f(x)-\varphi_k(x)| dx +|\int_{\mathbb{R}}\varphi_k\sin(nx)~ dx |.$$ First take the limit as $n \rightarrow \infty$ with respect to the previous inequality and we obtain: $$ \lim_{n \rightarrow \infty}|\int_{\mathbb{R}}f\sin(nx)~ dx | \le \int_{\mathbb{R}}|f(x)-\varphi_k(x)| dx.$$ Then we take the limit as $k \rightarrow \infty$ with respect to the previous inequality and then we obtain: $$ \lim_{k \rightarrow \infty} \lim_{n \rightarrow \infty}|\int_{\mathbb{R}}f\sin(nx)~ dx | \le 0 \implies \lim_{n \rightarrow \infty}|\int_{\mathbb{R}}f\sin(nx)~ dx | \le 0.$$

Does my proof look good? I'm wondering if the last implication in my proof is valid? Thanks.

Solution 1:

Basically correct. Perhaps, you may use $\varepsilon>0$ and $\limsup$ to eliminate ambiguity.

For any $\varepsilon >0$, there exists a step function $\varphi$ such that $\|f-\varphi\|<\varepsilon$. Then, $$ \left |\int_{\mathbb{R}}f \sin (nx)\, dx \right |\leq \ \|f-\varphi\|+\left|\int_{\mathbb{R}}\varphi\sin(nx)\, dx \right | < \varepsilon + \left|\int_{\mathbb{R}}\varphi\sin(nx)\, dx \right | $$

Taking $n \to \infty$ implies $$ \limsup_{n\to \infty}\left |\int_{\mathbb{R}}f \sin (nx)\, dx \right | \leq \varepsilon $$

Since $\varepsilon>0$ can be arbitrary number, we conclude

$$ \limsup_{n\to \infty}\left |\int_{\mathbb{R}}f \sin (nx)\, dx \right | =0. $$