Prove some facts about a sequence of functions?

Question

I have this simple sequence of functions: $$f_{n}(x)=\frac{\sin(nx)}{nx^{\frac{3}{2}}}$$

I want to see if I can correctly:

1. Prove that $f_{n}\in\ L^{1}(0,+\infty)$ for every $n\in\mathbb N$.
2. Prove that $f_{n}\rightarrow0$ as $n\rightarrow \infty$ pointwise in $(0,+\infty)$
3. Prove that $\int_{0}^{+\infty}f_{n}(x)dx \rightarrow0$ as $n \rightarrow \infty$.

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So, starting from point one: I see that $f_{n}\in C^{0}(0+\infty)$.

I have to study $|f_{n}|$ in $U(0^{+})$ and in $U(+\infty)$.

In $U(0^{+})$,

$|f_{n}|\approx \frac{nx}{nx^{\frac{3}{2}}}\approx x^{-\frac{3}{2}+1}\approx x^{-\frac{1}{2}}\approx\frac{1}{\sqrt{x}}$

I think this prove that $f_{n}\in L^{1}(U(0^{+}))$ because it gives an integrable function.

In $U(+\infty)$ I think that $|f{n}|\rightarrow0$ (I just computet the limit). So, this should be fine with the hypothesis.

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Point two

$f_{n}$ it's a bounded function and for $n\rightarrow\infty$ is easy to see that $f_{n}\rightarrow0$ , $n \rightarrow \infty$ and poinwise on $(0,+\infty)$.

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Point three

As I have said above "by eyes" I would say that the hypothesis is right but I hav esome difficult to put it on more detailled "math language". Could someone help me and help me on my work?

Answer 1

In order to prove that a function $f$ is an element of $L^1(0,\infty)$, one must prove the following: $$\int_0^{\infty} |f(x)| \;\mathrm{d}x < \infty$$ So, in your case, observe that we have two ways to bound the sine: $$|\sin(nx)| \leq nx \qquad \text{and} \qquad |\sin(nx)| \leq 1$$ Observe that on $(0,\frac1n)$, the left bound is better, whereas on $(\frac1n,\infty)$, the right one is tighter. Hence, by splitting up the integral, we obtain for all $n \in \mathbb{N}$: \begin{align*} \int_0^{\infty} |f_n(x)| \;\mathrm{d}x &= \int_0^{\infty} \left| \frac{\sin(nx)}{nx^{3/2}}\right| \;\mathrm{d}x \leq \int_0^{\frac1n} \frac{nx}{nx^{3/2}} \;\mathrm{d} x + \int_{\frac1n}^{\infty} \frac{1}{nx^{3/2}} \;\mathrm{d}x \\ &= \int_0^{\frac1n} \frac{1}{\sqrt{x}} \;\mathrm{d}x + \frac{1}{n} \int_{\frac1n}^{\infty} \frac{1}{x\sqrt{x}} \;\mathrm{d}x \\ &= \left[2\sqrt{x}\right]_0^{\frac1n} + \frac{1}{n}\left[-\frac{2}{\sqrt{x}}\right]_{\frac1n}^{\infty} = 2\sqrt{\frac1n} + \frac{2\sqrt{n}}{n} \\ &= \frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}} = \frac{4}{\sqrt{n}} \end{align*} So, for every $n \in \mathbb{N}$, we find that the integral is finite, as $4/\sqrt{n} < \infty$. This proves claim one. And moreover, as $4/\sqrt{n} \to 0$ as $n \to \infty$, we have proven claim three as well.

Now, can you do the remaining second claim in a rigorous way? Hint, choose $x \in (0,\infty)$ arbitrarily, and go from there.

Answer 2

• Note that $|\sin(nx)|\leq \min\{xn,1\}$ for all $x\geq 0$. Thus $$\int_0^\infty\left|\frac{\sin (nx)}{nx^{\frac{3}{2}}}\right|dx\leq\int_0^1x^{-\frac{1}{2}}dx+\frac{1}{n}\int_1^\infty x^{-\frac{3}{2}}dx=2+\frac{2}{n}<\infty.$$ This proves $f_n\in L^1(0,\infty)$.
• It's trivial that $|f_n(x)|\to 0$ as $n\to\infty$.
• As shown in the first claim, we have $$|f_n(x)|\leq \frac{nx}{nx^{\frac{3}{2}}}\chi_{[0,1]}(x)+\frac{1}{nx^{\frac{3}{2}}}\chi_{[1,\infty)}(x)\leq x^{-\frac{1}{2}}\chi_{[0,1]}(x)+x^{-\frac{3}{2}}\chi_{[1,\infty)}(x),$$ so $f_n$ is uniformly bounded by an integrable function. The last claim now follows right away by the dominated convergence theorem.