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For, $n\ge 1$ let, $g_n(x)=\sin^2(x+1/n)$ , $x\in [0,\infty)$ and $f_n(x)=\int_0^xg_n(t)\,dt.$

Then, which is(/are) correct ?

(A) $\{f_n(x)\} $ converges pointwise to a function $f$ on $[0,\infty)$ but does not converge uniformly on $[0,\infty)$.

(B) $\{f_n(x)\} $ does not converge pointwise to any function on $[0,\infty)$.

(C) $\{f_n(x)\} $ converges uniformly on $[0,1]$.

(D) $\{f_n(x)\} $ converges uniformly on $[0,\infty)$.

I found that, $$f_n(x)=\frac{1}{2}\left[x-\frac{1}{2}\sin(x+1/n)+\frac{1}{2}\sin(1/n)\right].$$

So, $f(x)=\lim_{n}f_n(x)=\frac{x}{2}-\frac{1}{4}\sin x$.

Now, $|f_n(x)-f(x)|\le 3/4$. So, by M-test, $f_n(x)$ is not uniformly convergent, but point-wise convergent. So option (A) is correct..Am I right??

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  • $\begingroup$ Your output for $f_n(x)$ includes a non-essential inaccuracies, I think. But in any case the speed of convergence $f_n(x)$ to $f(x)$ does not depend on $x$. Therefore, the convergence is uniform on $[0, \infty)$. $\endgroup$ – grizzly Jul 7 '15 at 7:06
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We have

$$f_n(x) = \int_0^x\sin^2\left(t+\frac1{n}\right)\,dt=\int_0^x \frac1{2}\left[1-\cos\left(2t+\frac{2}{n}\right)\right]\,dt = \\ \frac{x}{2} - \frac1{4}\sin\left(2x+ \frac{2}{n}\right) + \frac1{4}\sin\left(\frac{2}{n}\right).$$

Hence,

$$\lim_{n \to \infty} f_n(x) = f(x) = \frac{x}{2} - \frac1{4}\sin\left(2x\right).$$

Using $|\sin a - \sin b| \leqslant |a-b|$,

$$|f_n(x) - f(x)| = \frac1{4}\left| \sin\left(2x+ \frac{2}{n}\right)-\sin\left(2x\right)-\sin\left(\frac{2}{n}\right)\right| \\ \leqslant \frac1{4}\left| \sin\left(2x+ \frac{2}{n}\right)-\sin\left(2x\right)\right|+\frac1{4}\left|\sin\left(\frac{2}{n}\right)\right| \leqslant \frac1{n}.$$

The sequence converges uniformly on $[0,\infty)$.

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Since $<f_{n}^{'}(x)>=<g_{n}(x)>$ if you prove that $<f_{n}^{'}(x)>$ is unformly convergent then $<f_{n}(x)>$ will be uniformly convergent. But it is clear that $<f_{n}^{'}(x)>=<g_{n}(x)>$ is uniformly convergent on $[0,\infty)$ as below

$$|sin^{2}(x+\frac{1}{n})-sin^{2}(x)|\leq 2|\frac{1}{n}|\rightarrow 0 ~as~n\rightarrow 0.$$

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Observe that

$$\int_0^x \sin^2(t+1/n)\, dt - \int_0^x \sin^2(t)\, dt = \int_{1/n}^{x+1/n} \sin^2(t)\, dt - \int_0^x \sin^2(t)\, dt$$ $$ = \int_{x}^{x+1/n} \sin^2(t)\, dt- \int_{0}^{1/n} \sin^2(t)\, dt.$$

In absolute value the last expression is less the sum of the two integrals, which is less than $1\cdot (1/n) + 1\cdot (1/n) = 2/n.$ This proves $f_n(x) \to \int_0^x \sin^2(t)\, dt$ uniformly on $[0,\infty).$

The only properties of $\sin^2 (t)$ used: Boundedness, and integrability on bounded intervals.

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