Abel's Test for Uniform Convergence

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Theorem

Let $\sequence {\map {a_n} z}$ and $\sequence {\map {b_n} z}$ be sequences of complex functions on a compact set $K$.

Let $\sequence {\map {a_n} z}$ be such that:

$\sequence {\map {a_n} z}$ is bounded in $K$

$\ds \sum \size {\map {a_n} z - \map {a_{n + 1} } z}$ is convergent with a sum which is bounded in $K$

$\ds \sum \map {b_n} z$ is uniformly convergent in $K$.

Then $\ds \sum \map {a_n} z \map {b_n} z$ is uniformly convergent on $K$.

Proof

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First we modify the statement of Abel's Lemma

Lemma

Suppose $\sum b_k$ converges.

Let $B_k = b_k + b_{k + 1} + b_{k + 2} + \cdots$

Then:

$a_n b_n + \cdots + a_{n + k} b_{n + k} = B_n a_n + B_{n + 1} \paren {a_{n + 1} - a_n} + \cdots + B_{n + k} \paren {a_{n + k} - a_{n + k - 1} } - B_{n + k + 1} a_{n + k}$

Proof of lemma

\(\ds a_n b_n + \cdots + a_{n + k} b_{n + k}\)

\(=\)

\(\ds a_n \paren {B_n - B_{n + 1} } + \cdots + a_{n + k} \paren {B_{n + k} - B_{n + k + 1} }\)

\(\ds \)

\(=\)

\(\ds B_n a_n + B_{n + 1} \paren {a_{n + 1} - a_n} + \cdots + B_{n + k} \paren {a_{n + k} - a_{n + k - 1} } - B_{n + k + 1} a_{n + k}\)

$\Box$

Proof of theorem

We show that $\ds \sum_{j\mathop = n}^{n + k} \map {a_j} z \map {b_j} z$ is uniformly small if $n$ is large enough.

Using the notation of the lemma, since $\map {B_n} z \to 0$ uniformly, let $n$ be so large that $\size {\map {B_N} z} \le \epsilon$ in $K$ for all $N \ge n$.

Since $\sequence {\map {a_n} z}$ is uniformly bounded in $K$, let $\size {\map {a_n} z} \le M$ for all $z \in K$.

Since $\ds \sum \size {\map {a_n} z - \map {a_{n + 1} } z}$ is convergent with a sum which is bounded in $K$, let $n$ be so large that:

$\forall k \in \N: \size {a_{n + 1} - a_n} + \cdots + \size {a_{n + k} - a_{n + k - 1} } \le \epsilon$ for all $z \in K$

then we have:

$\forall z \in K: \size {B_{n + 1} \paren {a_{n + 1} - a_n} } + \cdots + \size {B_{n + k} \paren {a_{n + k} - a_{n + k - 1} } } \le M \epsilon$

Therefore:

\(\ds \size {a_n b_n + \cdots + a_{n + k} b_{n + k} }\)

\(=\)

\(\ds \size {B_n a_n + B_{n + 1} \paren {a_{n + 1} - a_n} + \cdots + B_{n + k} \paren {a_{n + k} - a_{n + k - 1} } + B_{n + k + 1} a_{n + k} }\)

Lemma

\(\ds \)

\(\le\)

\(\ds \size {B_n a_n} + \size {B_{n + 1} \paren {a_{n + 1} - a_n} } + \cdots + \size {B_{n + k} \paren {a_{n + k} - a_{n + k - 1} } } + \size {B_{n + k + 1} a_{n + k} }\)

Triangle Inequality for Complex Numbers

\(\ds \)

\(\le\)

\(\ds M \epsilon + M \epsilon + M \epsilon\)

\(\ds \)

\(=\)

\(\ds 3 M \epsilon\)

Therefore, $\ds \sum \map {a_n} z \map {b_n} z$ is uniformly convergent on $K$.

$\blacksquare$

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This needs considerable tedious hard slog to complete it. In particular: by my understanding, we are only so far as having proved it for monotone sequences, but a question remains as to what that actually means in this context. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Source of Name

This entry was named for Niels Henrik Abel.

Sources

• 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Abel's test: 2. Abel's test for uniform convergence.