show that the following sequence function converges uniformly to 0 on the given set $\left\{\frac{\sin nx}{nx}\right\}$ on $[\alpha,\infty)$ ...

Question

... where $\alpha>0$.

question: show that the following sequence function converges uniformly to $0$ on the given set $\displaystyle\left\{\frac{\sin nx}{nx}\right\}$ on $[\alpha,\infty)$ where $\alpha>0$.

my thoughts: let $\displaystyle f_n(x)= \frac{\sin nx}{nx}$ so $\displaystyle\left|f_n(x) - 0\right|= \frac{\sin nx}{n}< x <\epsilon$.

$\displaystyle\left|\frac{\sin nx}{nx}\right| < 1$ for all $n > N$

Am I on the right track?

Answer 1

Because $\alpha>0$, we can always find number $N$ such that $N\alpha>1/\varepsilon$ for any $\varepsilon>0$: \begin{align} \alpha >0 \implies \forall\, \varepsilon>0 \;\exists\,N\in\mathbb N: \dfrac{1}{N\alpha}<\varepsilon\; \forall \, x\in\left[\alpha, +\infty\right) \end{align}

Since $x\in\left[\alpha, +\infty\right)$ we know that $x\geq \alpha$, and therefore $Nx \geq N\alpha>1/\varepsilon$: \begin{align} \forall \, \varepsilon>0,\ \forall \, x\in\left[\alpha, +\infty\right) \quad \exists\,N\in\mathbb N: \dfrac{1}{Nx}<\varepsilon \end{align}

On the other hand, $\sin x$ is a bounded function, so that \begin{align} \forall \, n\in\mathbb N,\ \forall \, x\in\left[\alpha, +\infty\right) \quad -1\leq \sin nx\leq 1 \iff \left\lvert\sin nx\right\rvert<1 \end{align}

But then \begin{align} \forall \, \varepsilon>0,\ \forall \, x\in\left[\alpha, +\infty\right) \quad \exists \, N\in\mathbb N: \quad\forall n\geq N\quad\left\lvert\dfrac{\sin nx}{nx}\right\rvert\leq\dfrac{1}{Nx}<\varepsilon \end{align}

Thus we have shown that the sequence $\;\left\lbrace\dfrac{\sin nx}{nx}\right\rbrace$ converges uniformly to $0$:

\begin{align} \bbox[1ex, border:1.5pt solid #e10000]{\dfrac{\sin nx}{nx} \rightrightarrows 0} \end{align}