# If $\sum_{n=0}^\infty|a_n|<\infty$, prove that $\sum_{n=0}^\infty a_nx^n$ converges uniformly for $x\in[0,1]$.

If $\sum_{n=0}^\infty|a_n|<\infty$, prove that $\sum_{n=0}^\infty a_nx^n$ converges uniformly for $x\in[0,1]$.

Here's my attempt: Let $x\in[0,1]$. Then,

\begin{align}\sum_{n=0}^\infty a_nx^n & \le\sum_{n=0}^\infty|a_nx^n|\\ & = \sum_{n=0}^\infty|a_n||x^n|\\ &\le \sum_{n=0}^\infty|a_n||1|^n \\ & = \sum_{n=0}^\infty|a_n| <\infty, \tag{*}\end{align} which by the Weierstrass $M$-test, we have that $\displaystyle \sum_{n=0}^\infty a_nx^n$ converges uniformly on $[0,1]$.

Here's my question: Have I correctly used the Weierstrass $M$-test? The conditions of the Weierstrass $M$-test are:

Let $\sum_{k=1}^\infty u_k$ be a series of real-valued functions on $E$. If there exists positive $M_1,M_2,\ldots$ with $\sum_{k=0}^\infty M_k <\infty$ such that $\sum_{k=0}^\infty u_k \ll \sum_{k=0}^\infty M_k$, then $\sum_{k=0}^\infty u_k$ converges uniformly on $E$.

I don't necessarily know that the $|a_n|$ are necessarily positive. However, I do know that they are non-negative. Thus, can I not use the Weierstrass $M$-test here? And I'm not entirely sure that I showed the dominated part entirely, but I've shown that the series $\displaystyle \sum_{n=0}^\infty a_nx^n$ converges for all $x\in[0,1]$. I suppose I just have a hard time distinguishing between pointwise convergence and uniform convergence.

• I don't know if you know this theorem, but I think you should : if a series of real-valued functions converges normally, then it converges uniformly. But here $||x\to a_n x^n||_{\infty, [0;1]} = |a_n|$ – Max Mar 14 '17 at 20:19

You should not use any theorem. For $|x| \le 1$ : $$f_k(x) = \sum_{n=0}^k a_n x^n, \qquad f(x) = \sum_{n=0}^\infty a_n x^n$$ Since $\sum_{n=0}^\infty |a_n| < \infty$, the series for $f(x)$ converges absolutely and is well-defined.

Also, again for $|x| \le 1$ : $$|f(x)-f_k(x)| =| \sum_{n=k+1}^\infty a_n x^n| \le \sum_{n=k+1}^\infty |a_n x^n| \le \sum_{n=k+1}^\infty |a_n |$$

where $\lim_{k \to \infty}\sum_{n=k+1}^\infty |a_n | =0$.

Q.e.d. this is the definition of the uniform convergence.

• In other words, don't use Weierstrass M, reprove Weierstrass M whenever you want to use it. – zhw. Mar 14 '17 at 22:23
• @zhw. Sarcasm ? :D Well yes, at least the first few months you are studying real/complex/functional analysis (I wanted to show to the OP there is nothing difficult here, just need to look at the good things) – reuns Mar 14 '17 at 22:26
• Sarcasm, maybe, but that's exactly what you did here. Also, you did use a theorem or two along the way. – zhw. Mar 14 '17 at 22:54

You aren't really using WM correctly. What you wrote was, essentially, if $M_n\ge 0$ for each $n,$ with $\sum_n M_n < \infty,$ and if $\sum |f_n(x)| \le \sum_n M_n$ on $E,$ then $\sum_n f_n$ converges uniformly on $E.$ That is false (and indeed it is a mistake I see on MSE often, even by skilled people of high reputation).

WM actually says: If for each $n$ we have $|f_n(x)|\le M_n$ for $x \in E,$ and if $\sum_n M_n <\infty,$ then $\sum_n f_n$ converges uniformly on $E.$ In your problem, you would take $E=[0,1],$ $f_n(x) = a_nx^n,$ and $M_n= |a_n|.$ That should lead to a correct proof for you.