0
$\begingroup$

The sequence $\displaystyle{f_n(x)= \left(\frac{\sin x}{x}\right)^{\frac{1}{n}},\, x\in [0,π] .}$ Find whether it converges pointwise, uniformly or not?

Since $\displaystyle{\lim_{n \to \infty} \left(\frac{\sin x}{x}\right)^{\frac{1}{n}} = 1 \,\forall x \in [0,π]}$, it is pointwise convergent. I know it's not uniform convergent but how to approach for that?

|cite|improve this question
$\endgroup$
4
$\begingroup$

For $x=\pi$ the pointwise limit is not $1$, it is $0$. This also proves that the convergence is not uniform because the pointwise limit is not a continuous function.

[I am interpreting $f_n(0)$ as $1$ for all $n$].

|cite|improve this answer
$\endgroup$
0
$\begingroup$

If the interval includes pi, then there is no pointwise convergence. Otherwise there is and to prove that it is not uniform show that $$\forall n \exists x \in (0,π)\ f_n(x)<1/2$$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ For $x=\pi$, $ f_n(x)=0$ for all $n$, so pointwise limit does exist. $\endgroup$ – Kabo Murphy Sep 17 at 12:33

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.