The Proposition states if $(f_n)_{n\in\mathbb{N}}$ is a secuence of functions $f_n:D\rightarrow \mathbb{C}$ with $\sum_{n=1}^{\infty} ||f_n||_D < +\infty$ then the series $\sum_{n=1}^{\infty}f_n$ converges absolutely uniform and pointwise.

The series converges uniformly (or uniformous?)if and only if the secuence of partial sums do so. The series of partialsums is $(\sum_{i=1}^{n}f_n)_{n\in\mathbb{N}}$

A sequence of functions converge if and only if

$\exists_{f:D\rightarrow\mathbb{C}}\forall_{\epsilon>0}\exists_{n(\epsilon)\in\mathbb{N}}\forall_{n\geq n(\epsilon)}\forall_{z\in D}:|f_n(z)-f(z)|<\epsilon$

Which would mean

$\exists_{f:D\rightarrow\mathbb{C}}\forall_{\epsilon>0}\exists_{n(\epsilon)\in\mathbb{N}}\forall_{n\geq n(\epsilon)}\forall_{z\in D}:|\sum_{i=1}^{n}f_n(z)-f(z)|<\epsilon$

My Question what would be the $f$ in this Proposition?

  • 1
    $\begingroup$ Perhaps you are having a translation difficulty (English is an immensely irregular language). "Uniformous" is not a word in English, although logically it could be. The correct word is "uniformly". $\endgroup$ – DanielWainfleet Jan 10 at 9:28

For each $z$ the series $ \sum\limits_{k=1}^{\infty} f_k(z)$ is absolutely convergent. $f(z)$ is the sum of this series. Note that $| \sum\limits_{k=1}^{n} f_k(z) - \sum\limits_{k=1}^{m} f_k(z) |<\epsilon$ for all $z \in D$ for all $n,m \geq n_0$ implies $| \sum\limits_{k=1}^{n} f_k(z) -f(z) |\leq \epsilon$ for all $z \in D$ for all $n \geq n_0$.


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.