Prove that if $\sum_{k=0}^{\infty}a_k$ converges, $\sum_{k=0}^{\infty}{a_k}{x^k}$ converges uniformly on $[0, 1]$.

I already know that $\sum_{k=0}^{\infty}{a_k}{x^k}$ converges pointwise on $[0, 1]$
by either Abel's or Dirichlet's Test for pointwise convergence.
Now, can I do this proof without Dirichlet's Test for uniform convergence?
I don't want to use it because, in the text,
this question came before Dirichlet's Test for uniform convergence.

Since we don't know that $\sum_{k=0}^{\infty}a_k$ converges absolutely,
we can't say that for all $x \in [0, 1]$, $|{a_k}{x^k}| \le |a_k|$ and apply the Weiertrass M-test.

Also, since we don't know if $\sum_{k=0}^{\infty}a_k$ is an alternating series,
we can't say that that since $\sum_{k=0}^{\infty}{a_k}{x^k}$ converges pointwise,
let $\sum_{k=0}^{\infty}{a_k}{x^k} = A(x)$ and observe that for all $x \in [0, 1]$, $$|A(x) - \sum_{k=0}^{n}{a_k}{x^k}| \le |a_{n+1}|x^{n+1} \le |a_{n+1}| \rightarrow 0$$ as $n$ goes to infinity.

A clue is sufficient. I really want to think this problem through. TY!

  • $\begingroup$ @Arthur A series without a variable can't converge uniformly. It needs to be a series of functions for the notion to make sense. Also, the $|a_k|$ are bounded, but that only gives pointwise convergence unless I'm missing something obvious. $\endgroup$ – Zach L. Jul 13 '13 at 6:47
  • $\begingroup$ @ZachL. I'm confusing it with absolute convergence, sorry. $\endgroup$ – Arthur Jul 13 '13 at 6:50
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    $\begingroup$ Another attempt, then. Set $A_k = \sum_{i = 0}^k a_i$ and $f_k(x) = \sum_{i = 0}^k a_ix^i$ for notational convinience. Given an $\epsilon > 0$, choose $N$ so that $|A_\infty - A_n| < \epsilon$ for all $n> N$. Such an $N$ exists by convergence of $A_\infty$. Now if we can prove that $|f_\infty(x) - f_k(x)| \leq |A_\infty - A_k|$ for all $x$ and $k$, we're done. If you write out the sum on each side, it shouldn't be too hard. $\endgroup$ – Arthur Jul 13 '13 at 7:01
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    $\begingroup$ You could use Abel's lemma, I believe. $\endgroup$ – David Mitra Jul 13 '13 at 10:25
  • $\begingroup$ See this for a proof using Abel's lemma. $\endgroup$ – David Mitra Jul 13 '13 at 13:17

For $n \in \mathbb{N}$, define

$$R_n = \sum_{k = n}^{\infty} a_k$$

and $M_n = \sup \{ \lvert R_k\rvert : k \geqslant n\}$. Then $M_n$ decreases to $0$, and using $a_k = R_k - R_{k+1}$ a summation by parts yields

\begin{align} \sum_{k = m}^n a_k x^k &= \sum_{k = m}^n (R_k - R_{k+1})x^k \\ &= R_m x^{m-1} + \sum_{k = m}^n R_k(x^k - x^{k-1}) - R_{n+1}x^n \tag{$\ast$} \end{align}

for $1 \leqslant m \leqslant n$. For $0 \leqslant x \leqslant 1$, we obtain

\begin{align} \Biggl\lvert \sum_{k = m}^n a_k x^k\Biggr\rvert &\leqslant \lvert R_m\rvert + (1-x)\sum_{k = m}^n \lvert R_k\rvert x^{k-1} + \lvert R_{n+1}\rvert \\ &\leqslant \Biggl( 2 + (1-x)\sum_{k = m}^n x^{k-1}\Biggr) M_m \\ &= (2 + x^{m-1} - x^n)M_m \\ &\leqslant 3 M_m \end{align}

uniformly in $x\in [0,1]$.

If we don't restrict to $x \in [0,1]$ and allow arbitrary complex $z$ with $\lvert z\rvert < 1$, taking the absolute value of each term in $(\ast)$ yields the estimate

\begin{align} \Biggl\lvert \sum_{k = m}^n a_k z^k\Biggr\rvert &\leqslant \Biggl( 2 + \lvert 1-z\rvert \sum_{k = m}^n \lvert z\rvert^{k-1}\Biggr) M_m \\ &\leqslant \biggl( 2 + \frac{\lvert 1-z\rvert}{1-\lvert z\rvert}\biggr) M_m. \end{align}

Thus, for every $K \geqslant 1$ (and $1 \leqslant m \leqslant n$) we have

$$\Biggl\lvert \sum_{k = m}^n a_k z^k\Biggr\rvert \leqslant (2 + K)M_m$$

and hence uniform convergence on the (closed) Stolz region $S_K = \{ z : \lvert 1-z\rvert \leqslant K(1 - \lvert z\rvert)\}$. The interval $[0,1]$ is the special case $S_1$.

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  • $\begingroup$ Urgh, I started to write the same :-) You was faster +1. $\endgroup$ – vesszabo Jul 13 '13 at 12:53
  • $\begingroup$ I know that feeling. It's not too bad when I just started, but when I'm about half through, it's really annoying for a moment or two. $\endgroup$ – Daniel Fischer Jul 13 '13 at 12:57
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    $\begingroup$ But I don't mind it. Next time I will be faster ;-) $\endgroup$ – vesszabo Jul 13 '13 at 13:02
  • $\begingroup$ Hello All, I thought that to use the Weierstrass test, $|a_{k}(x)| \le M$ for all $x$. So either $\sum a_{k}$ converges absolutely or there is another series of positive terms that converges. $\endgroup$ – Andy Tam Jul 15 '13 at 19:55
  • $\begingroup$ @AndyTam Consider $a_k\cdot r^k$ for an $r < 1$. Then Weierstraß gives uniform convergence on $[0,\, s]$ for all $s < r$. $\endgroup$ – Daniel Fischer Jul 15 '13 at 19:57

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