2
$\begingroup$

Find out whether the following sequence is uniformly convergent on $E$. $$ f_n(x)=\ln\left(\sin x+\frac{1}{n}\right),\ \ x\in E=\left(0;\frac{\pi}{6}\right) $$

Here is what I did: $$ \left\{ \begin{aligned} &f(x)=\lim_{n\rightarrow\infty}\ln\left(\sin x+\frac{1}{n}\right)=\ln(\sin x)\\ &0<\sin x<\frac{1}{2} \end{aligned} \right.\Rightarrow \ln(\sin x)\ \exists\Rightarrow\\ \Rightarrow f_n(x) \text{ converges to } f(x)=\ln(\sin x)\\ t_n=|f_n(x)-f(x)|=\left|\ln\left(1+\frac{1}{n\sin x}\right)\right| $$ Then I thought that $$\forall n\ \ \exists x: t_n > 0.\ \ \text{Thus, } \lim_{n\rightarrow\infty}\sup_{x\in E}t_n\ne0\Rightarrow f_n(x)\ \ \text{does not converge uniformly on E.} $$ Is my solution correct? And if so, please, tell me how to properly prove that $\forall n\ \ \exists x:\ \lim_{n\rightarrow\infty}\sup_{x\in E}t_n\ne0.$

Thank you in advance!

$\endgroup$
  • 1
    $\begingroup$ If we take $x=\frac1{n^2}$ then $t_n$ is asymptotic with $\ln(1+n)$ which definitely doesn't go to $0$. $\endgroup$ – kingW3 Oct 11 '19 at 22:11
2
$\begingroup$

Note that $\lim_{n \to \infty}n\, \sin \frac{1}{n} = 1$.

Take $x_n = \frac{1}{n} \in (0,\pi/6)$ and find that as $n \to \infty$,

$$ \sup_{x \in (0,\pi/6)}|f_n(x) - f(x) | \geqslant \left|\ln\left(1+\frac{1}{n\sin x_n}\right)\right| \to |\ln(2)| \neq 0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.