# 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.

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!

• Do you have an hypothesis on the regularity of the limit? – charmd Nov 28 '20 at 11:53
• There is a reason they included the assumption that $|S_n'| < M$. – Paul Sinclair Nov 28 '20 at 15:18
• Something like this might help: $$|S_{n+p}(x) - S_n(x)| = \left|\int_a^x(S'_{n+p}(t) - S'_n(t))dt\right|\leq\int_a^x|S'_{n+p}(t) - S'_n(t)|dt\leq...$$ – dezdichado Nov 28 '20 at 18:19
• 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. – 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.