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.

  • $\begingroup$ Yeah you are right. I have edited my body. $\endgroup$
    – little o
    Oct 12, 2018 at 19:05
  • $\begingroup$ What is the definition of uniform convergence, and how is it different from point-wise convergence. $\endgroup$
    – Doug M
    Oct 12, 2018 at 19:28

1 Answer 1



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

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

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

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

For the uniform convergence, use MVT to get

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


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

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

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

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