Does $\{f_n\}$ converge uniformly on $(0,1) $ and $(0,\infty)?$

Question

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 show that $\{f_n\}$ converges uniformly on $(0,1)$. Does $\{f_n\}$ converge uniformly on $(0,\infty)?$.

$f(x)=\lim_{n\to \infty}f_n(x)=\lim_{n\to \infty}\int _0^x g_n(t) dt=\int _0^x\lim_{n\to \infty}\sin ^2(t+\frac {1}{n})=\int_0^x \sin^2 t$ [Since $\int$ and $\sin^2x$ are continuous functions].

Hence $f(x)=\int _0^x \sin ^2 t dt$.

Now $\sup_{(0,\infty)} |f_n(x)-f(x)|=\sup_{(0,\infty)}|\int _0^x \{\sin ^2(t+\frac{1}{n})-\sin^2 t\}|\le \sup_{(0,\infty)} \int _0^x |\{\sin ^2(t+\frac{1}{n})-\sin^2 t\}|$

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

Hence $\sup_{(0,\infty)} |f_n(x)-f(x)|\le \sup_{(0,\infty)} 2\frac{x}{n}$

Hence if $x\in (0,1)\implies \sup_{(0,\infty)} |f_n(x)-f(x)|\le \frac{2}{n}\to 0$

Hence $f_n$ converges uniformly on $(0,1)$.

But $f_n$ does not converge uniformly on $(0,\infty)$ because I can choose $x$ quite large such that $\frac{x}{n}>1$.

Please validate.

Answer 1

Alternative way. In order to avoid integration, note that $$f_n(x)-f(x)= \int_{1/n}^{x+1/n} \sin^2(t) \, dt-\int_0^x \sin^2(t) \, dt= \int_{x}^{x+1/n} \sin^2(t) \, dt-\int_0^{1/n} \sin^2(t) \, dt$$ Hence for $x\geq 0$, $$|f_n(x)-f(x)|\leq \int_{x}^{x+1/n} 1 \, dt+\int_0^{1/n} 1 \, dt= \frac{2}{n} $$ which implies that $(f_n)_n$ converges uniformly to $f$ in $[0,+\infty)$. The same argument works if we replace $\sin(x)$ with any bounded continuous function in $[0,+\infty)$.

Answer 2

Note that

$$\int_0^x \sin^2(t +1/n) \, dt = \frac{x}{2} - \frac{1}{4}\sin(2x+2/n) + \frac{1}{4} \sin (2/n),$$

and

$$\int_0^x \sin^2(t) \, dt = \frac{x}{2} - \frac{1}{4}\sin(2x) .$$

Hence,

$$|f_n(x) - f(x)| \leqslant \frac{1}{4}\left|\sin(2x+2/n) - \sin(2x) \right|+ \frac{1}{4} |\sin(2/n)| \leqslant \frac{1}{2n} + \frac{1}{4}|\sin(2/n)|$$

We have uniform convergence on $(0,\infty)$.