# Term-by-Term Differentiation and UNIFORM CONVERGENCE: True Relation

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?

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.

• 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? – ravenclaw May 22 '16 at 1:47
• 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 $0$. – Daniel Fischer May 22 '16 at 10:06

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.

• 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? – ravenclaw May 22 '16 at 1:46
• Have you proved uniform integrability of {$f_n$}? – ibnAbu May 22 '16 at 1:51
• I didn't see any problem. why did you say it cannot be differentaited term by term at x=0? – ibnAbu May 22 '16 at 2:02