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For $n\geq 1$ and 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. {$f_n(x)$} converges point wise to a function on $[0,\infty)$ , but does not converge uniformly on $[0,\infty)$
  2. {$f_n(x)$} does not converge point wise to any function on $[0,\infty)$
  3. {$f_n(x)$} converges uniformly to a function on $[0,1]$
  4. {$f_n(x)$} converges uniformly to a function on $[0,\infty)$

What I answered is {$f_n(x)$} converges uniformly to a function on $[0,\infty)$ because the limit converges to $\sin^2(x)$ which is bounded continuous function

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    $\begingroup$ convergence of $f_n(x)$ in $[0,\infty)$ assures its convergence in $[0,1]$ too. $\endgroup$ – Nitin Uniyal Jun 20 '17 at 14:38
  • $\begingroup$ You have $x$ as the upper limit and also as variable of integration. Also, the limit doesn't converge, it simply is $\sin^2(x).$ $\endgroup$ – zhw. Jun 20 '17 at 15:13
  • $\begingroup$ that means what?? $\endgroup$ – user229886 Jun 20 '17 at 15:22
  • $\begingroup$ My last sentence above is just about word usage: Limits don't converge. They either exist or don't exist. $\endgroup$ – zhw. Jun 20 '17 at 16:48
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Right answer, wrong reason. Yes, the limit function is bounded and continuous, but that does not imply the convergence is uniform. For example, $f_n(x) = x/n$ converges pointwise to $0$ on $[0,\infty),$ but the convergence is not uniform there ($\sup_{[0,\infty)} |f_n| = \infty$ for all $n$).

Hint to see that $\int_0^x \sin^2(t+1/n)\,dt$ converges uniformly to $\int_0^x \sin^2(t)\,dt$ on $[0,\infty):$

$$\int_0^x \sin^2(t+1/n)\,dt = \int_{1/n}^{x+1/n} \sin^2(t)\,dt.$$

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  • $\begingroup$ The continuity of $\sin^2$ in fact plays no part - you could replace $\sin^2$ with any bounded function with the same conclusion. $\endgroup$ – Jason Jun 20 '17 at 16:25
  • $\begingroup$ Or in fact, any function which is $L^p$ for some $p>1$. $\endgroup$ – Jason Jun 20 '17 at 16:28
  • $\begingroup$ @MasterX Why do you ask about $\int_0^\infty?$ Just compute $ \int_{1/n}^{x+1/n} \sin^2(t)\,dt - \int_{0}^{x} \sin^2(t)\,dt.$ $\endgroup$ – zhw. Jun 20 '17 at 16:51
  • $\begingroup$ ooh I was thinking about we can directly yield it from $\int_{0}^{x} sin^2(t+\frac{1}{n} )$ but why did you used $ \int_{\frac{1}{n} }^{x+ 1/n} sin^2(t+ \frac{1}{n}} $ $\endgroup$ – user229886 Jun 20 '17 at 17:11

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