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I was going through some questions on pointwise and uniform convergence. Got stuck in one of those which says:

Let $g_n(x) = \sin^2(x+\frac{1}{n})$ be defined on $[0,\infty).$

and $f_n(x) = \int_0^xg_n(t)\,dt.$

I am supposed to discuss about its uniform-convergence of $(f_n).$

The terms are really looking complicated to try it by the definition. Should I first show that $(g_n)$ is uniformly convergent? How am I supposed to do even that?

Help, please.

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  • $\begingroup$ $[0,\infty]?$ Why not $[0,\infty)?$ $\endgroup$ – zhw. Jun 4 '20 at 19:36
  • $\begingroup$ Sorry...It was an inadvertent error in typing..edited it $\endgroup$ – Gitika Jun 4 '20 at 19:38
  • $\begingroup$ Did you try using $\sin^2(x) = \frac{1}{2}(1 - \cos(2x))$? $\endgroup$ – Bungo Jun 4 '20 at 19:46
  • $\begingroup$ See my edits for proper MathJax usage. You shouldn't keep alternating in and out of MathJax within a single expression. Just stay in MathJax until the whole expression is done. Also, note proper punctuation. A period's purpose is not to separate two sentences; it is to end a sentence. Thus it should not be omitted in "Help, please." and a space after the period should precede the first letter of the next sentence. That is standard in all European languages. $\endgroup$ – Michael Hardy Jun 4 '20 at 20:05
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You have

$$\begin{aligned}g_n(x)&=\sin^2\left(x + \frac{1}{n}\right) = \frac{1}{2}\left(1- \cos\left(2(x + \frac{1}{n})\right)\right)\\ &=\frac{1}{2}\left(1 - \cos 2x \cos\frac{1}{n} + \sin 2x \sin \frac{1}{n}\right). \end{aligned}$$

Therefore

$$f_n(x)= \frac{1}{2}\left(x - \frac{1}{2}\cos\frac{1}{n}\sin 2x-\frac{1}{2}\sin\frac{1}{n}\left(\cos 2x -1\right)\right).$$

From there, you can prove that $\{f_n\}$ converges uniformly to

$$f(x) = \frac{x}{2} - \frac{1}{4} \sin 2x$$

as $$\begin{aligned}\left\vert f_n(x) - f(x) \right\vert &= \frac{1}{4}\left\vert \left(1 - \cos\frac{1}{n} \right)\sin 2x + \sin\frac{1}{n}\left(\cos 2x -1\right)\right\vert\\ &\le \frac{1}{4}\left(\left\vert \left(1 - \cos\frac{1}{n} \right)\sin 2x\right\vert + \left\vert\sin\frac{1}{n}\left(\cos 2x -1\right)\right\vert\right)\\ &\le \frac{1}{4}\left(\left\vert 1 - \cos\frac{1}{n} \right\vert + 2\left\vert\sin\frac{1}{n}\right\vert\right)\\ \end{aligned}$$

and the RHS of above inequality converges to zero independently of $x$.

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  • $\begingroup$ I see that happening point-wise..How to show it does so uniformly? $\endgroup$ – Gitika Jun 4 '20 at 19:57
  • $\begingroup$ Added a couple of lines. $\endgroup$ – mathcounterexamples.net Jun 4 '20 at 20:09
  • $\begingroup$ Thank you..thank you $\endgroup$ – Gitika Jun 4 '20 at 20:20
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This doesn't have much to do with trig identities, etc. We have a bounded continuous function $h$ on $[0,\infty)$ (for example $h(t)=\sin^2(t)),$ and we define

$$f_n(x) = \int_{0}^{x} h(t+1/n)\,dt = \int_{1/n}^{x+1/n} h(s)\,ds.$$

Suppose $N\le m < n.$ Then

$$\tag 1 f_m(x) - f_n(x) = \int_{1/n}^{1/m} h(s)\,ds - \int_{x+1/n}^{x+1/m} h(s)\,ds.$$

Let $M$ be the bound on $|h|.$ Then $(1)$ is dominated in absolute value by $2M(1/m-1/n)\le 2M/N.$ This shows $f_n$ is uniformly Cauchy on $[0,\infty),$ hence is uniformly convergent there.

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