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Welcome to MSE! I need assistance in demonstrating the uniform convergence of the sequence $\frac{\sin nx}{n\sqrt{x}} \to 0$. I've attempted to approach the problem by trying a bounding technique, but my attempts have not been successful. Additionally, I've explored the possibility of finding a suitable sequence to show that it doesn't converge to zero, but my efforts in this direction have been unsuccessful as well.

Could someone help me understand where I might be going wrong in my attempts? Perhaps there's a different approach or insight that I'm missing. Your guidance and suggestions would be immensely valuable. Thank you!

mathselite
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Jan 06 '24 at 09:39
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    Hint: write this as $\frac{\sin(nx)}{nx} \sqrt{x} < \sqrt{x}$. It is of course also bounded above by $\frac{1}{n \sqrt{x}}$. – Toby M Jan 06 '24 at 09:52
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    In what domain do you want to check uniform convergence? Without this information, the question makes no sense. – Mark Jan 06 '24 at 10:11
  • The domain is $\mathbb{R}^+$? – mathselite Jan 06 '24 at 10:23

1 Answers1

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Clearly, the sequence converges uniformly on any set that is bounded away from $0$. For example, if $x \in [a,\infty)$ where $a>0$, then uniform convergence holds since

$$\sup_{x \in [a,\infty)}\left|\frac{\sin nx}{n\sqrt{x}} \right|\leqslant \frac{1}{n\sqrt{a}}\underset{n \to \infty}\longrightarrow 0$$

To prove uniform convergence on all of $\mathbb{R}^+$ consider $x \in (0,a]$. If the sequence failed to converge uniformly then there would exist $\epsilon_0 > 0$ and a sequence $x_n \in (0,a]$ such that for all $n$

$$\left|\frac{\sin nx_n}{n\sqrt{x_n}} \right|\geqslant \epsilon_0$$

The sequence $(x_n)$ may or may not have a limit but it is bounded and $l=\liminf_{n \to \infty}x_n \geqslant 0.$

If $l =0$, then there is a subsequence $x_{n_k}$ such that $x_{n_k} \to 0$ and for all positive integrers $k$

$$\left|\frac{\sin n_kx_{n_k}}{n_k\sqrt{x_{n_k}}} \right|= \left|\frac{\sin n_kx_{n_k}}{n_kx_{n_k}} \right|\sqrt{x_{n_k}}\geqslant \epsilon_0$$

However, this is impossible because $\frac{\sin x}{x}$ is bounded for all $x \in \mathbb{R}$ and $\sqrt{x_{n_k}} \to 0$ as $ k \to \infty$.

On the other hand, if $l >0$, then $x_n > l/2$ eventually which means that there exists $N$ such that for all $n \geqslant N$ we have both

$$\left|\frac{\sin nx_{n}}{n\sqrt{x_{n}}} \right|\geqslant \epsilon_0,$$

and

$$\tag{*}\left|\frac{\sin nx_{n}}{n\sqrt{x_{n}}} \right| \leqslant \frac{1}{n\sqrt{l/2}}$$

which, again, is impossible since the RHS of (*) tends to $0$ as $n \to \infty$.

Using a proof by contradiction we have shown that the convergence of the sequence in question is uniform for all $x \in \mathbb{R}^+$.

RRL
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