# Convergence of the following sequence of functions.

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

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

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

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

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

I have found that $$f_n (x) = \frac 1 2 \left (x - \sin x \cos \left (x + \frac 2 n \right ) \right)$$ which converges to the function $$f$$ on $$[0, \infty)$$ defined by $$f(x) = \frac 1 4 ( 2x - \sin {2x}), x \in [0, \infty)$$. But I am not quite sure about whether this convergence is uniform or not. Please help me in this regard.

Thank you very much.

• Yeah you are right. I have edited my body. Oct 12, 2018 at 19:05
• What is the definition of uniform convergence, and how is it different from point-wise convergence. Oct 12, 2018 at 19:28

hint

$$\sin^2(t+\frac 1n)=\frac{1-\cos(2t+\frac 2n)}{2}$$

$$f_n(x)=\int_0^x\sin^2(t+\frac 1n)dt=\frac{1}{2}\Big[t-\frac{\sin(2t+\frac 2n)}{2}\Bigr]_0^x$$

$$=\frac x2-\frac 14\sin(2x+\frac 2n)+\frac 14\sin(\frac 2n)$$

The pointwise limit is $$f(x)=\frac x2-\frac{\sin(2x)}{4}$$

For the uniform convergence, use MVT to get

$$|\sin(2x+\frac 2n)-\sin(2x)|\le \frac 2n$$

and

$$|f_n(x)-f(x)|\le \frac 1n.$$

The convergence is uniform at $$[0,\infty)$$.

• I have done that. First see my body and then answer. Oct 12, 2018 at 19:09
• @Dbchatto67 Your $f_n(x)$ is not correct. Oct 12, 2018 at 19:10
• Have you evaluated your integral between the limits $0$ and $x$? Oct 12, 2018 at 19:11
• In place of $t$ we have $t + \frac 1 n$. Oct 12, 2018 at 19:11
• Now calculate and simplify. Oct 12, 2018 at 19:13