Let $\Omega \subset \mathbb{C}$ and $f_n: \Omega\to \mathbb{C}$ is a
Cauchy sequence in the supremum norm $||.||$. Check the uniform
convergence of $( f_n)$
$$(f_n) \text{ Cauchy in } (C, || .||) \\
\rightarrow \forall \epsilon > 0\;\; \exists N_0 \in\mathbb{N} \; \ni |f_n(z) -f_m(z)| < \epsilon \;\;\forall n,m \geq N_0 \text{ and } \forall\; z\in \Omega \;\;\ldots(1) \\
\rightarrow \forall \epsilon > 0\;\; \text{ and fixed } z\in \Omega\;\; \exists N_0 \in\mathbb{N} \; \ni |f_n(z) -f_m(z)| < \epsilon \;\;\forall n,m \geq N_0 \\
\rightarrow (f_n(z)) \text{ is Cauchy
in } (\mathbb{C}, |.|) \; \forall z\in \Omega \\
(\mathbb{C}, |.|)\text{ is complete and so converges pointwise to say } f(z) \; \forall z\in \Omega \\ \text{This means that } \forall z\in\Omega \lim_{n\to\infty} f_n(z) = f(z) \\
\text{Letting } m\to \infty \text{ in } (1) \text{ we get } \\
\rightarrow \forall \epsilon > 0\;\; \exists N_0 \in\mathbb{N} \; \ni |f_n(z) -f(z)| < \epsilon \;\;\forall n\geq N_0 \text{ and } \forall\; z\in \Omega \\ \rightarrow || f_n – f|| \to 0 \\
\rightarrow (f_n) \text{ converges uniformly to } f$$
Is my proof correct?
Best Answer
There's a problem when you say the, letting $m\to\infty$, you get $|f_n(z)-f(z)|<\varepsilon$. All you can deduce is that $|f_n(z)-f(z)|\leqslant\varepsilon$. So, take a number $\varepsilon'\in(0,\varepsilon)$. Using your argument you get that $|f_n(z)-f(z)|\leqslant\varepsilon$ and that therefore $\|f_n-f\|\leqslant\varepsilon'<\varepsilon$.
Otherwise, it looks fine.