Let series of functions $\sum_{n=1}^\infty a_n(x)$ converges on $[a, b]$, its partial sum are $S_n(x)$. If there exsits a constant M such that $$ \forall x\in [a,b], \forall n\geq 1,|S'_n(x)|\leq M, $$ then $\sum_{n=1}^\infty a_n(x)$ converges uniformly.

I tried to prove the problem by Cauchy criterion. $$ S_{n+p}(x)-S_n(x)=S_{n+p}'(\xi_1)(x-a)+S_{n+p}(a)-S_{n}'(\xi_2)(x-a)-S_{n}(a) $$ But it seems we are unable to to control $|S'_{n+p}-S'_n|$, so I amstuck here.

Is my way wrong, if so, please suggest a correct way. Appreciate any help!

  • $\begingroup$ Do you have an hypothesis on the regularity of the limit? $\endgroup$ – charmd Nov 28 '20 at 11:53
  • $\begingroup$ There is a reason they included the assumption that $|S_n'| < M$. $\endgroup$ – Paul Sinclair Nov 28 '20 at 15:18
  • $\begingroup$ Something like this might help: $$|S_{n+p}(x) - S_n(x)| = \left|\int_a^x(S'_{n+p}(t) - S'_n(t))dt\right|\leq\int_a^x|S'_{n+p}(t) - S'_n(t)|dt\leq...$$ $\endgroup$ – dezdichado Nov 28 '20 at 18:19
  • $\begingroup$ Actually, I just realized that the bounded limit implies $f_n$ are Lipshitz and it's well-known that if a sequence of Lipshitz functions on a compact set converges pointwise, then it must also converges uniformly. I believe it's the standard $\frac{\epsilon}{3}$ argument for the $|S(x) - S_n(x)|$. Besides, I am sure this has been answered before on this site if you search it using the phrasing I used. $\endgroup$ – dezdichado Nov 28 '20 at 18:27

We have $S_n(x) \to S(x)$ pointwise and $|S_n(x) - S_n(y)| \leqslant |S_n'(\xi)||x-y| \leqslant M|x-y|$ for all $x,y \in [a,b]$ and all $n \in \mathbb{N}$.

For all $\epsilon > 0$ there exists $\delta > 0$ such that if $|x-y| < \delta$, then for all $n \in\mathbb{N}$ we have $|S_n(x) - S_n(y)| < \epsilon/3$. Taking the limit as $n \to \infty$ it follows that $|S(x) - S(y)| \leqslant \epsilon/3$ as well.

Since $[a,b]$ is compact there exist finitely many points $x_1, \ldots x_m$ such that $[a,b] \subset \cup_jI(x_j,\delta),$ where $I(x_j,\delta)$ is the open interval with center $x_j$ and radius $\delta$.

By pointwise convergence, there exists $N_j \in \mathbb{N}$ such that if $n \geqslant N_j$ then $|S_n(x_j) - S(x_j)|< \epsilon/3.$ Let $N = \max_{1 \leqslant j \leqslant m}N_j.$

Given any $x \in [a,b]$ there exists $j$ such that $x \in I(x_j,\delta)$. This implies $|x - x_j| < \delta$ and if $n \geqslant N$ we have

$$|S_n(x) - S(x)| \leqslant |S_n(x) - S_n(x_j)|+|S_n(x_j) - S(x_j)|+|S(x_j) - S(x)| \\\leqslant \frac{\epsilon}{3} + \frac{\epsilon}{3}+ \frac{\epsilon}{3} = \epsilon,$$

proving uniform convergence.


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.