Uniform Convergence of sequence of functions $\int_{0}^{x} \sin^2(t+\frac{1}{n}) dt$

Question

Given a sequence $f_n(x) = \int_{0}^{x} \sin^2(t+\frac{1}{n})\ dt$. Now for checking the uniformity $$|f_n(x)-g(x)|\leq\int_{0}^{x}\left|\,\sin^2\left(t+\frac{1}{n}\right)-\sin^2(t)\right|\ dt$$

where, $g(x)$ is the uniform limit of $\sin^2(t+\frac{1}{n})$ as $n \rightarrow \infty$

Now, $|\sin^2(t+\frac{1}{n})-\sin^2(t)|\leq \frac{2}{n}$

$|f_n(x)-g(x)|\leq \frac{2x}{n}$

The above inequality implies the convergence is not uniform in $[0,\infty)$ but answer given to me states it is uniform over this interval. So the question which one is correct?

Answer 1

You have

\begin{align*} f_n(x) - g(x) &= \int_{\frac{1}{n}}^{x+\frac{1}{n}} \sin^2 t \, dt - \int_{0}^{x} \sin^2 t \, dt \\ &= \int_{x}^{x+\frac{1}{n}} \sin^2 t \, dt - \int_{0}^{\frac{1}{n}} \sin^2 t \, dt \end{align*}

and hence

$$ |f_n(x) - g(x)| \leq \frac{2}{n}. $$

Answer 2

1) I think, you got $g$ as the pointwise limit of $f$. If you know, that $g$ is the uniform limit of $f$, then there is nothing to prove anymore...

2) The inequality doesn't prove anything. To prove the uniformity of the convergence the inequality is not sufficient, but it doesn't disprove it either! To prove that $f_n$ doesn't converge uniformly to $g$, you have to find $x_n\in[0,\infty)$ such that $|f_n(x_n)-g(x_n)|\not\to 0$.