Proving that the metric space $(\mathcal{H}(\mathbb{D}), d)$ is complete using Complex Analysis

Call the vector space of holomorphic functions on the unit disk $\mathcal{H}(\mathbb{D})$. Define for $f, g \in \mathcal{H}(\mathbb{D})$: $$d(f,g) = \sum^{\infty}_{k=1}2^{-k}\min\left(1, \sup_{|z|\leq1-\frac{1}{k}}|f(z)-g(z)|\right).$$ I think it's quite straightforward to prove that this defines a metric on $\mathcal{H}(\mathbb{D})$ (only my proof for the triangle inequality might be not entirely correct, but that another question in itself).

Now I want to prove that this metric space $(\mathcal{H}(\mathbb{D}), d)$ is complete. For this I first wanted to prove that a sequence $(f_n)_n$ in $\mathcal{H}(\mathbb{D})$ converges in $(\mathcal{H}(\mathbb{D}), d)$ if and only if it converges uniformly on compact subsets of $\mathbb{D}$. I tried proving the first implication by noticing that since $\forall \epsilon > 0$ there exists a $N$ such that for all $n \geq N$ it holds that $d(f_n,f) < \epsilon$. Therefore the sum in $d(f_n,f)$ must be small in each term, so for the first few terms it must hold that $$\min\left(1, \sup_{|z|\leq1-\frac{1}{k}}|f(z)-g(z)|\right) = \sup_{|z|\leq1-\frac{1}{k}}|f(z)-g(z)|.$$ But does this hold? And how I proceed with the rest of the proof?

For the converse statement, it must hold in any compact subset $K$ of $\mathbb{D}$ that $\forall \epsilon > 0$ there exists a $N$ such that $\forall n \geq N$ $\textbf{and}$ $\forall z \in K$ it holds that $|f_n(z),f(z)| < \epsilon$. But I'm now confused on how to proceed with the proof because in $d(f,g)$ the supremum is taken over all $z \leq 1 - 1/k$, not just the ones in some compact set $K$.

Any help is appreciated!

Solution 1:

For one direction one can split the sum in two parts: One finite sum and a “small“ remainder: $$d(f_n,f) \le \sum^{N}_{k=1}2^{-k}\min\left(1, \sup_{|z|\leq1-\frac{1}{k}}|f_n(z)-f(z)|\right) + \frac{1}{2^N} $$ for every positive integer $N$. If $f_n \to f$ uniformly on compact sets of $\Bbb D$ then, given $\epsilon > 0$, you can first choose $N$ so large that $1/2^N < \epsilon/2$, and then $n > N$ so large such that all of the finitely many terms in the sum are $< \epsilon/(2 N )$. It follows that $d(f_n, f) \to 0$.

For the other direction, if $d(f_n, f) \to 0$ and $K \subset \Bbb D$ is compact then $K \subset \{ z : |z| < 1-1/k \}$ for some $k$, and $$ 2^{-k}\min\left(1, \sup_{|z|\leq1-\frac{1}{k}}|f_n(z)-f(z)|\right) \le d(f_n, f) \, . $$ Given $0 < \epsilon < 1$, $d(f_n, f) < \epsilon/2^k$ for $n \ge n_0$ then implies $$ \sup_{z \in K}|f_n(z)-f(z)| \le \sup_{|z|\leq1-\frac{1}{k}}|f_n(z)-f(z)| < \epsilon $$ so that $f_n \to f$ uniformly on $K$.

In order to prove that $(\mathcal{H}(\mathbb{D}), d)$ is complete one can proceed as follows: Let $(f_n)$ be a Cauchy sequence in $(\mathcal{H}(\mathbb{D}), d)$.

Similar as in the second part above it follows that for every compact set $K \subset \Bbb D$, $(f_n)$ restricted to $K$ is a Cauchy sequence with respect to the supremums norm. It is a standard argument to show that $(f_n)$ is uniformly convergent on $K$, and that the limit function $f$ is holomorphic.

It then follows from the first part above that $f_n \to f$ in $(\mathcal{H}(\mathbb{D}), d)$.

One can also demonstrate $d(f_n, f) \to 0$ directly: If $\epsilon > 0$ then for all $N$ and sufficiently large $n, m \ge n_0$ $$ \sum^{N}_{k=1}2^{-k}\min\left(1, \sup_{|z|\leq1-\frac{1}{k}}|f_n(z)-f_m(z)|\right) \le d(f_n, f_m) < \epsilon \, . $$ Letting $m \to \infty$ first and then $N \to \infty$ it follows that $d(f_n, f) < \epsilon$ for $n \ge n_0$.