convergence of sequence of functions on an infinite interval

Question

For $n\ge 1$, Let $$g_n(x)=\sin^2(x+\frac{1}{n}), x\in[0,\infty)$$ and $$f_n(x)=\ \int_{0}^{x}g_n(t)dt$$. Then

1) $\{f_n\}$ converges pointwise to a funtion $f$ on $[0,\infty)$ but does not converge uniformly on $[0,\infty)$

2) $\{f_n\}$ does not converge pointwise to any function on $[0,\infty)$

3) $\{f_n\}$ converges uniformly on $[0,1]$

4) $\{f_n\}$ converges uniformly on $[0,\infty)$.

i found the pointwise limit of $g_n(x)$ then I don't know how to solve.

$\begingroup$ what did you get for the pointwise limit? $\endgroup$ – Adam Hughes Jun 22 '15 at 19:03 — Adam Hughes, Jun 22 '15 at 19:03

RRL · Accepted Answer · 2015-06-22 19:55:25Z

2

Hints:

$$f_n(x) = \int_0^x\sin^2\left(t+\frac1{n}\right)\,dt=\int_0^x \frac1{2}\left[1-\cos\left(2t+\frac{2}{n}\right)\right]\,dt$$

answered Jun 22 '15 at 19:40

RRL

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$\begingroup$ can you tell me, which one is the right answer? $\endgroup$ – aryan Jun 23 '15 at 3:36
$\begingroup$ Sure. After integrating we see that the limit is $f(x) = x/2 - (1/4)\sin(2x)$. Did you get that part? $\endgroup$ – RRL Jun 23 '15 at 3:51
1

$\begingroup$ Then $|f_n(x) - f(x)| = (1/4)| \sin(2x + 2/n) - \sin(2x) - \sin(2/n)| \leqslant (1/4)|\sin(2x + 2/n) - \sin(2x)| + (1/4)|\sin(2/n)| = O(1/n)$ for $x \in [0,\infty)$. $\endgroup$ – RRL Jun 23 '15 at 3:55
$\begingroup$ Limit function is continuous, but it doesn't imply that sequence is uniformly convergent, so what we should do next? $\endgroup$ – aryan Jun 23 '15 at 3:55
$\begingroup$ So it appears the sequence converges uniformly on $[0,\infty)$. $\endgroup$ – RRL Jun 23 '15 at 3:57

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