2

I am to computing this limit:

$$\lim_{n\to+\infty}\int_{\mathbb{R}} \sin(nx)h(x) \,, dx$$ where $h(x)$ stays in $L^1(\mathbb{R})$.

Is it correct this method? I choose a sequence of functions $h_n(x)\in C_c^\infty$ which converges to $h$ in $L^1$ sense and I have

$$\int_{\mathbb{R}} \sin(nx)h_n(x) \, dx = \left.-\frac{\cos(nx)} n h_n(x)\right|_{-\infty}^{+\infty} + \int_{\mathbb{R}} \frac{\cos(nx)}{n}h_n'(x) \, dx$$

so

$$\left|\int_{\mathbb{R}} \sin(nx)h_n(x) \, dx\right|\le \frac Cn\to0.$$

In this way I have

$$\|\sin(nx)h(x)\|_{L^1}\le\| \sin(nx)h(x)-\sin(nx)h_n(x)\|_{L^1}+\| \sin(nx) h_n(x) \|_{L^1} \to 0$$

Michael Hardy
  • 1
aleio1
  • 995
  • What if $h$ is not differentiable ? – Atmos Jan 19 '18 at 16:55
  • $h$ doesn't need to be differentiable. If I have not written wrong, I performed integration by parts on $h_n$ which are infinitely differentiable compactly supported functions. – aleio1 Jan 19 '18 at 17:03
  • 1
    Your proof is fine. $L_1$ functions can be approximated "inside the integral" by $C^\infty$ function. – pisco Jan 19 '18 at 17:32

0 Answers0