1
$\begingroup$

For example: Suppose I have the following series:

$$\sum_{k=0}^{\infty}e^{-k}\sin(kt)$$

The Weierstrass-M-Test shows that the series is uniformly convergent on $\mathbb R$. Does this imply differentiability and continuity on $\mathbb R$ aswell?

$\endgroup$
  • 2
    $\begingroup$ Continuity yes, differentiability no. The uniform convergence of the derivatives gives you differentiability. $\endgroup$ – Daniel Fischer Jun 6 '15 at 14:56
  • 1
    $\begingroup$ In fact, the $n$-th derivatives all converge uniformly for any $n$, so the limit is smooth. $\endgroup$ – JHance Jun 6 '15 at 15:05
2
$\begingroup$

Yes, since each finite sum is continuous, then the uniform convergence of the series implies the continuity of the limit on any compact of $\mathbb{R}$, thus the continuity of the limit sum on $\mathbb{R}$.

For the differentiability, you can check that the series $$ \left|\sum_{k=0}^{\infty}k\:e^{-k}\cos(kt)\right|\leq\left|\sum_{k=0}^{\infty}k\:e^{-k}\right|=\frac{e}{(1-e)^2}<+\infty $$ is (normally) uniformly convergent on $\mathbb{R}$, giving the desired the differentiability of the limit sum.

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

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.