Dirichlet's Test for Uniform Convergence

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Theorem

Let $D$ be a set.

Let $\struct {V, \norm {\,\cdot\,} }$ be a normed vector space.

Let $a_i, b_i$ be mappings from $D \to M$.


Let the following conditions be satisfied:

$(1): \quad$ The sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty \map {a_n} x$ be bounded on $D$
$(2): \quad \sequence {\map {b_n} x}$ be monotonic for each $x \in D$
$(3): \quad \map {b_n} x \to 0$ converge uniformly on $D$.


Then:

$\ds \sum_{n \mathop = 1}^\infty \map {a_n} x \map {b_n} x$ converges uniformly on $D$.


Proof

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

All we need to show is that:

$\ds \sum_{n \mathop = 1}^\infty \size {\map {b_n} x - \map {b_{n + 1} } x}$

converges uniformly on $D$.


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To do this we show that the Cauchy criterion holds.

Assign $\epsilon < 0$.

Then by definition of uniform convergence:

$\exists N \in \N: \forall x \in D: \forall n \ge N: \size {\map {b_n} x} < \dfrac \epsilon 2$

Let $x \in D$ and $n > m \ge N$.

Then:

\(\ds \sum_{k \mathop = m + 1}^n \size {\map {b_k} x - \map {b_{k + 1} } x}\) \(=\) \(\ds \sum_{k \mathop = m + 1}^n \paren {\map {b_k} x - \map {b_{k + 1} } x}\)
\(\ds \) \(=\) \(\ds \map {b_{m + 1} } x - \map {b_{n + 1} } x\)
\(\ds \) \(=\) \(\ds \size {\map {b_{m + 1} } x - \map {b_{n + 1} } x}\)
\(\ds \) \(\le\) \(\ds \size {\map {b_{m + 1} } x + \map {b_{n + 1} } x}\)
\(\ds \) \(<\) \(\ds \frac \epsilon 2 + \frac \epsilon 2\)
\(\ds \) \(=\) \(\ds \epsilon\)

$\blacksquare$


Source of Name

This entry was named for Johann Peter Gustav Lejeune Dirichlet.


Sources

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