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.
Best Answer
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 $[0,\infty)$.