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How to show $u_n(x)=\dfrac{\sin^2(nx)}{n\sin(x)}$ converges uniformly

Attempt :

Let $\varepsilon >0$, and this interval $[\varepsilon, \pi-\varepsilon]$

Then $\forall x\in[\varepsilon, \pi-\varepsilon], \quad \dfrac{\sin^2(nx)}{n\sin(x)}\le \dfrac{1}{n\varepsilon}$ hence $u_n(x)\overset{\text{unif.}}{\longrightarrow}0$ on $[\varepsilon, \pi-\varepsilon]$

We consider now $C(t):=\dfrac{\sin t}{t}$

Then we have $u_n(x)=\sin(nx)C(nx)$, and since $|C(nx)|\le1$

We have $|\sin(nx)C(nx)|\le|\sin(nx)|\le |nx|$

We can conclude that : $|x|<\dfrac{\varepsilon}{n}\implies |u_n(x)|<\varepsilon$ and

$u_n(x)\overset{\text{unif.}}{\longrightarrow}0$ on $\left[0, \dfrac{\varepsilon}{n}\right)$

Now I 've got a gap for $x\in \left[\dfrac{\varepsilon}{n},\varepsilon\right)$, which $u_n(x)$ doesn't seem to converge uniformly, May you confirm please?

Stu
  • 1,750

1 Answers1

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Uniform convergence fails on $(0,\pi).$ Suppose otherwise. It should be clear that the pointwise limit of the $u_n$'s is $0$ on this interval. Thus $u_n \to 0$ uniformly on $(0,\pi).$ That implies $\sup_{(0,\pi)} u_n \to 0.$ However, letting $x_n=1/n,$ we see

$$\sup_{(0,\pi)} u_n \ge \frac{\sin^2 (nx_n)}{n\sin x_n} = \frac{\sin^2 (1)}{n\sin (1/n)} \to \sin^2 (1)>0.$$ That's a contradiction, proving the result.

zhw.
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