In Lang – Complex Analysis – 1999 there is following theorem:
Theorem. If a sequence $(f_n)$ of complex functions on $S$ is Cauchy, then it converges uniformly.
Now, since I do no want to the same thing many times (real analysis, complex analysis, metric space) I would like to know what is the best generalization of the theorem above, for instance to a sequence of maps $(f_n:X\to Y)$ between two metric spaces $(X,d)$ any $(Y,d')$.
What are the conditions on $S$ and on the two metric spaces in order for the sequence to converge and to converge uniformly?
Q: How do I generalize the theorem above to metric spaces?
Best Answer
Let $X$ be a set and $(Y, d')$ be a complete metric space.
Let $(f_n)_{n\geq 1}$ be a sequence of functions $f_n: X\rightarrow Y$ such that for every $\varepsilon>0$ there exists $N\in \mathbb{N}$ such that for all $n,m \geq N$ holds $$ \sup_{x\in X} d'(f_n(x), f_m(x)) <\varepsilon. $$ Then holds for $x\in X$ $$ d'(f_n(x), f_m(x))\leq \sup_{y\in X} d'(f_n(y), f_m(y)) <\varepsilon. $$ Hence, $(f_n(x))_{n\geq 1}$ is a Cauchy sequence in $(Y, d')$ and as $(Y, d')$ is complete there exists $$f(x):= \lim_{n\rightarrow \infty} f_n(x).$$ Now we show that $$ \lim_{n\rightarrow \infty} \sup_{x\in X} d'(f_n(x), f(x)) =0.$$ Pick $N$ such that $$ \sup_{y\in X} d'(f_n(y), f_m(y)) <\varepsilon/2. $$ Now let $x\in X$. Then there exists $M\geq N$ such that for all $ n\geq M$ holds $$ d'(f_n(x), f(x))<\varepsilon/2.$$ Thus, we get for $n\geq N$ $$ d'(f_n(x), f(x)) \leq d'(f_n(x), f_M(x)) + d'(f_M(x), f(x)) < \varepsilon/2 + \varepsilon/2 =\varepsilon.$$ As this holds for all $x\in X$ we get for $n\geq N$ $$\sup_{x\in X} d'(f_n(x), f(x)) < \varepsilon.$$ Hence, $(f_n)_{n\geq 1}$ converges unifornly to $f$.