# convergence of sequence of functions on an infinite interval

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.

• what did you get for the pointwise limit? Jun 22, 2015 at 19:03

Hints:

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

Then consider bounding $|f_n(x)-f(x)|$ using the inequality $|\sin a - \sin b|\leqslant |a-b|.$

• can you tell me, which one is the right answer? Jun 23, 2015 at 3:36
• Sure. After integrating we see that the limit is $f(x) = x/2 - (1/4)\sin(2x)$. Did you get that part?
– RRL
Jun 23, 2015 at 3:51
• 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 $x \in [0,\infty)$.
– RRL
Jun 23, 2015 at 3:55
• Limit function is continuous, but it doesn't imply that sequence is uniformly convergent, so what we should do next? Jun 23, 2015 at 3:55
• So it appears the sequence converges uniformly on $[0,\infty)$.
– RRL
Jun 23, 2015 at 3:57