# A question about a proof of the Dirichlet test for uniform convergence

I don't understand the proof of Wikipedia. D:

For the Dirichlet test for uniform convergence, I saw a proof in WikiProofs. But I did not understand the proof, they say that all we need to show is that the other series converges uniformly; why is that?

-
1. These should be three separate questions. 2. For each, you should show what you've tried and precisely where you are stuck. 3. If these are homework problems, you should tag them as such. –  Austin Mohr Nov 17 '11 at 4:44
@Austin: +1 for this self-consistent admonition :-) –  joriki Nov 17 '11 at 8:07

The Dirichlet criterion states in detail:

Let $a_k: D \to \mathbb{C}$ be functions for which there exists a constant $M$ such that for all $n$ and all $x \in D$ we have $$\left|\sum_{k=1}^n a_k(x)\right| \leq M \tag{1}$$ and let $b_k: D \to [0,\infty)$ be functions that converge pointwise monotonically and uniformly to $0$, which means

• for all $x\in D$ and all $k$ we have $$b_k (x) \geq b_{k+1}(x) \geq 0\tag{2}$$
• for all $\varepsilon \gt 0$ there is $K$ such that for all $x \in D$ and all $k\geq K$ we have $$b_k(x) \lt \varepsilon. \tag{3}$$

Then the series $\sum_{k=1}^\infty a_k(x) b_k(x)$ converges uniformly on $D$.

The main idea of the proof is to use the so-called Abel summation trick or summation by parts. To avoid need- and useless clutter in the formulas, let me drop the $x$ from the notation.

1. Warm-up/preparation

Define $A_\ell = \sum_{k=1}^\ell a_k$ and note that $(1)$ tells us that $|A_\ell| \leq M$. Then $a_k = A_{k}-A_{k-1}$ and \begin{align*} \sum_{k=1}^n \phantom{|} a_k b_k &= A_1 b_1 + (A_2 - A_1)b_2 + \cdots + (A_n - A_{n-1})b_n \\ & = A_1 (b_1 - b_2) + A_2 (b_2 - b_3) + \cdots + A_{n-1}(b_{n-1}-b_n) + A_n b_n \\ & = \sum_{k=1}^{n-1} \phantom{|} A_k\,(b_{k}-b_{k+1}) + A_n \,b_n. \end{align*} For $n \lt m$ we conclude from this and $\sum_{n+1}^m = \sum_{1}^m - \sum_{1}^n$ that $$\tag{4} \sum_{k=n+1}^{m} \phantom{|}a_k b_k = \sum_{k=n}^{m-1} \phantom{|} A_k\,(b_{k}-b_{k+1}) + (A_m\,b_m - A_n\,b_n).$$

2. Conclusion

Recall from $(1)$ that $|A_\ell| \leq M$ and from $(2)$ that $b_k - b_{k+1} \geq 0$ as well as $b_m,b_n \geq 0$ so that the last equality $(4)$ lets us estimate \begin{align*} \left|\sum_{k=n+1}^{m} \phantom{|}a_k b_k\right| &\leq \sum_{k=n}^{m-1} \phantom{|} |A_k|\,(b_{k}-b_{k+1}) + (|A_m|\,b_m + |A_n|\,b_n) \\ &\leq M \cdot \sum_{k=n}^{m-1} \phantom{|}(b_{k}-b_{k+1}) + (M \,b_m + M \,b_n) = 2M \,b_n. \end{align*} By $(3)$ there is for every $\varepsilon \gt 0$ a $K$ such that for all $n \geq K$ we have $b_n \lt \varepsilon / 2M$ and thus we conclude for $m \gt n \geq K$ that $$\left|\sum_{k=n+1}^{m} \phantom{|}a_k b_k\right| \leq 2M \,b_n \lt \varepsilon$$ which tells us that the series $\sum_{k=1}^\infty a_k(x)\,b_k(x)$ converges uniformly on $D$ by the Cauchy criterion.

Notice that in essence we exploited the telescoping property of the series $\sum_{k=n+1}^{m} (b_{k} - b_{k+1})$ in our penultimate estimate (which gives us its uniform convergence), as asserted on the ProofWiki page you linked to. However, phrasing it the way it is done there seems rather misleading to me, so I wouldn't have understood that proof either if I hadn't known about the Abel summation trick beforehand.

-
Excelent Answer!!!!!!!!! –  August Nov 17 '11 at 16:09

You want to prove that the sequence of partial sums $$\begin{eqnarray} S_N(x) = \sum_{n=0}^{N} a_n(x) b_n(x) \end{eqnarray}$$ converges uniformly as $N \rightarrow \infty$. Observe that $$\begin{eqnarray} S_N(x) &=& \sum_{n=0}^{N} a_n(x) b_n(x) = \sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) - \sum_{k=0}^{n-1} a_k(x) \right) b_n(x) \\ &=& \sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) b_n(x) - \sum_{n=0}^{N} \left( \sum_{k=0}^{n-1} a_k(x) \right) b_n(x) \\ &=& \sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) b_n(x) - \sum_{n=0}^{N+1} \left( \sum_{k=0}^{n-1} a_k(x) \right) b_n(x) + \left( \sum_{k=0}^{N} a_k(x) \right) b_{N+1}(x) \\ &=& \sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) b_n(x) - \sum_{n=0}^{N} \left( \sum_{k=0}^{n} a_k(x) \right) b_{n+1}(x) + \left( \sum_{k=0}^{N} a_k(x) \right) b_{N+1}(x) \\ &=& \sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) b_n(x) - \sum_{n=0}^{N} \left( \sum_{k=0}^{n} a_k(x) \right) b_{n+1}(x) + \left( \sum_{k=0}^{N} a_k(x) \right) b_{N+1}(x) \\ &=& \sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) (b_n(x) - b_{n+1}(x)) + \left( \sum_{k=0}^{N} a_k(x) \right) b_{N+1}(x). \end{eqnarray}$$ (This manipulation is called partial summation or summation by parts.) So it will be enough to show that both $\sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) (b_n(x) - b_{n+1}(x))$ and $\left( \sum_{k=0}^{N} a_k(x) \right) b_{N+1}(x)$ converge uniformly as $N \rightarrow \infty$

Since $\left( \sum_{k=0}^{N} a_k(x) \right)$ is bounded and since $b_{n}(x)$ converges to $0$ uniformly, it follows that $\left( \sum_{k=0}^{N} a_k(x) \right) b_{N+1}(x)$ converges to $0$ uniformly.

I leave it to you to show that the boundedness of $\left( \sum_{k=0}^{n} a_k(x) \right)$ and the uniform convergence of $\sum_{n=0}^{N} |b_n(x) - b_{n+1}(x)|$ implies that $\sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) (b_n(x) - b_{n+1}(x))$ converges uniformly.

Remark: Instead of showing that $\sum_{n=0}^{N} |b_n(x) - b_{n+1}(x)|$ converges uniformly using the Cauchy Criterion and then that the boundedness of $\left( \sum_{k=0}^{n} a_k(x) \right)$ and the uniform convergence of $\sum_{n=0}^{N} |b_n(x) - b_{n+1}(x)|$ implies that $\sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) (b_n(x) - b_{n+1}(x))$ converges uniformly, I suggest just showing directly that $\sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) (b_n(x) - b_{n+1}(x))$ converges uniformly using the Cauchy Criterion. The argument is very similar to the one used to prove that $\sum_{n=0}^{N} |b_n(x) - b_{n+1}(x)|$ converges uniformly. Just add in the fact that $\left| \sum_{k=0}^{n} a_k(x) \right| \leq M$.

-
No, I refuse to observe that! –  Alexei Averchenko Nov 17 '11 at 7:51