# Dirichlet's Test for Uniform Convergence

From ProofWiki

## Contents

## Theorem

Suppose:

- The sequence of partial sums of $\displaystyle \sum_{n \mathop = 1}^\infty a_n (x)$ is bounded on $D$.

- ${b_n(x)}$ is monotonic for each $x\in D$.

- $b_n(x)\to 0$ converges uniformly on $D$.

Then:

- $\displaystyle \sum_{n \mathop = 1}^{\infty}a_n(x)b_n(x)$ converges uniformly on $D$.

## Proof

Suppose $b_n(x)\ge b_{n+1}(x)$ for each $x \in D$.

All we need to show is that $\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{b_n(x)-b_{n+1}(x)}\right\vert$ converges uniformly on $D$.

To do this we show that the Cauchy Criterion holds.

Assign $\epsilon < 0$, then $\exists N \in \N$ such that:

- $\displaystyle \forall x \in D, \forall n \ge N: \left\vert{b_n(x)}\right\vert < \frac \epsilon 2$

If $x\in D$ and $n > m \ge N$ then,

\(\displaystyle \sum_{k \mathop = m+1}^n \left \vert {b_k(x)-b_{k+1}(x)} \right \vert\) | \(=\) | \(\displaystyle \sum_{k \mathop = m+1}^n(b_k(x)-b_{k+1}(x))\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle b_{m+1}(x)-b_{n+1}(x)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left \vert {b_{m+1}(x)-b_{n+1}(x)} \right \vert\) | |||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \left \vert {b_{m+1}(x)+b_{n+1}(x)} \right \vert\) | |||||||||||

\(\displaystyle \) | \(<\) | \(\displaystyle \frac \epsilon 2 + \frac \epsilon 2\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \epsilon\) |

$\blacksquare$

## Source of Name

This entry was named for Johann Lejeune Dirichlet.

## Sources

- Bruce Watson:
*Real Analysis Course Notes*, Memorial University (2007)