1
$\begingroup$

For a series with $\sum u_n'(x)$ not uniformly convergent, and If $f '(x) = \lim_{n\to\infty} f_n'(x) $
where $f(x)=\lim_{n\to\infty} f_n(x) $ and $ f_n(x) $ $=u_1+u_2+ . . . +u_n$

Then the series $\sum u_n(x)$ can be differentiated term by term, is this condition true for any series?

$\endgroup$
1
$\begingroup$

If $\sum_{k}f_k(x)=f(x)$, even if it's not uniformly convergent on $]a,b[$, it will be uniformly convergent on $[c,d]$ for all $a<c<d<b$. Therefore, if $\sum_{k}f_k'(x)=f'(x)$ on $]a,b[$, and $x\notin\{a,b\}$, then you can always differentiate $\sum_{k}f_k(x)$ term by term.

$\endgroup$
  • $\begingroup$ The doubt is both the series with $ f_n(x) = \frac {\log(1+n^4x^2)}{2n^2}$ and the series with $ f_n(x)=\dfrac{nx}{1+n^2x^2} $ satisfies this condition. But series with $ f_n(x)=\dfrac{nx}{1+n^2x^2} $ cannot be differntiated term by term at $x=0$ , WHY? $\endgroup$ – ravenclaw May 22 '16 at 1:47
  • $\begingroup$ You need some additional conditions for locally uniform convergence of the series. Let $g \colon \mathbb{R}\to \mathbb{R}$ a sufficiently smooth function with $\operatorname{supp} g = [0,1]$, and for $n \in \mathbb{N}\setminus \{0\}$ let $g_n(x) = g\bigl(n(n+1)x - \frac{1}{n}\bigr)$. Set $f_1 = g_1$ and $f_n = g_n - g_{n-1}$ for $n \geqslant 2$. Then $\sum_{n = 1}^\infty f_n(x) = 0$ for all $x$, and $\sum_{n = 1}^\infty f_n'(x) = 0$ for all $x$. But neither convergence is uniform on any neighbourhood of $0$. $\endgroup$ – Daniel Fischer May 22 '16 at 10:06
1
$\begingroup$

This is a stronger version (uniformly integrability condition) of what you need:

If a sequence of absolutely continuous functions {$f_n$} converges pointwise to some $f$ and if the sequence of derivatives {$f_n’$} converges almost everywhere to some $g$ and if {$f_n’$} is uniformly integrable then $\lim\limits_{n\mapsto \infty} f_n’ = g= f’$ almost everywhere. Where the derivative of $f$ is $f’$. If the convergence is pointwise and $ g $ is continuous then $ f'$ = $ g $ everywhere.

Proof : by FTC $f_n(x) – f_n(a) = \int_a^x f_n’ dx$

By Vitali convergence theorem : $\lim\limits_{n\mapsto \infty}\int_a^x f_n’ dx = \int_a^x g dx$

Therefore $\lim\limits_{n\mapsto \infty}( f_n(x) – f_n(a))= \int_a^x g dx$

$f(x)-f(a) = \int_a^x g dx$

$f(x)’=g$ almost everywhere

If the convergence is pointwise and $ g $ is continuous then $ f'$ = $ g $ everywhere.

$\endgroup$
  • $\begingroup$ My doubt is both the series with $ f_n(x) = \frac {\log(1+n^4x^2)}{2n^2}$ and the series with $ f_n(x)=\dfrac{nx}{1+n^2x^2} $ satisfies this condition. But series with $ f_n(x)=\dfrac{nx}{1+n^2x^2} $ cannot be differntiated term by term at $x=0$ , WHY? $\endgroup$ – ravenclaw May 22 '16 at 1:46
  • $\begingroup$ Have you proved uniform integrability of {$f_n$}? $\endgroup$ – ibnAbu May 22 '16 at 1:51
  • $\begingroup$ I didn't see any problem. why did you say it cannot be differentaited term by term at x=0? $\endgroup$ – ibnAbu May 22 '16 at 2:02

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.