Consider the sequence of functions defined by the following $$ f_n(x) = x^n \sin \left(\frac{1} {x^n} \right), 0\lt x \lt \frac {1} {\sqrt[n] \pi}, f_n(x)=0,$$ elsewhere in $\mathbb R$. What can we say about its convergence? I think the sequence converges pointwise but not uniformly.
I have made an attempt I now write as an answer not to make the question too long. I'm not completely sure about it. Please check it and feel free to suggest hints or other ways to solve this problem. Thank you.
On
The "interesting" part of the function is the open interval where it is not defined as the 0 function. For every positive integer we have a different interval, although they all are subsets of $(0,1)$. However, for every x and n we can say $$ - x^n \le x^n sin \left(\frac{1} {x^n} \right) \le x^n $$ so the sequence converges pointwise to the 0 function everywhere in $\mathbb R$ by the squeeze theorem. Now I want to prove it doesn't converge uniformly, though. For any fixed $n$, I consider the point $\frac {1} {\sqrt[n] \frac {5\pi}{2} } $ which is in the interval $(0, \frac {1} {\sqrt[n] \pi}) $ because $\pi \lt \frac {5\pi}{2}$ and so their n-th root. $$ \sup _ {x \in (0, \frac {1} {\sqrt[n] \pi})} | f_n (x)| \ge f_n\left(\frac {1} {\sqrt[n] \frac {5\pi}{2} } \right ) = \frac {2} {5\pi} $$ thus $$ lim _ {n \to \infty} \sup _ {x \in (0, \frac {1} {\sqrt[n] \pi})} | f_n (x)| \neq 0$$
Note that the function $g(u)=u\sin {1\over u}$ has a maximum in $u_0$ such that $u_0$ is the largest positive solution to $$\tan{1\over u}={1\over u}$$which means that $$ \pi<{1\over u_0}<{3\pi\over 2}\implies {2\over 3\pi}<{u_0}<{1\over \pi} $$ hence the function $f_n(x)=x^n\sin{1\over x^n}$ always has a fixed-value extremum at $x=\sqrt[n]{u_0}$ which lies within $(0,{1\over \sqrt[n]{\pi}})$, therefore meaning that the convergence is not uniform.