# Uniform convergence of a sequence of functions which is integral of another sequence

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?

• $[0,\infty]?$ Why not $[0,\infty)?$
– zhw.
Jun 4, 2020 at 19:36
• Sorry...It was an inadvertent error in typing..edited it Jun 4, 2020 at 19:38
• Did you try using $\sin^2(x) = \frac{1}{2}(1 - \cos(2x))$?
– user169852
Jun 4, 2020 at 19:46
• 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. Jun 4, 2020 at 20:05

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 $$x$$.

• I see that happening point-wise..How to show it does so uniformly? Jun 4, 2020 at 19:57
• Added a couple of lines. Jun 4, 2020 at 20:09
• Thank you..thank you Jun 4, 2020 at 20:20

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.