Property 3: Triangle Inequality
For any non-negative real numbers \(a,b\) we have \(\frac{a}{1+a}+\frac{b}{1+b} \geq \frac{a+b}{1+a+b}\), since the inequality can be rearranged to \((1+a)(1+b)(a+b) \geq (1+a+b)^2\) which is true.
Let \(a = \|f-h\|_n\) and \(b = \|g-h\|_n\). Then we have
$$ \frac{\|f-h\|_{n}}{1+\|f-h\|_{n}}+\frac{\|g-h\|_{n}}{1+\|g-h\|_{n}} \geq \frac{\|f-h\|_{n}+\|g-h\|_{n}}{1+\|f-h\|_{n}+\|g-h\|_{n}} \geq \frac{\|f-g\|_{n}}{1+\|f-g\|_{n}}$$
where the last inequality follows from the triangle inequality for the norm. Now we can sum over all terms in \(k\):
$$\rho(f,h) + \rho(g,h) = \sum_{n=1}^{\infty} \frac{1}{2^{n-1}} \left( \frac{\|f-h\|_{n}}{1+\|f-h\|_{n}} + \frac{\|g-h\|_{n}}{1+\|g-h\|_{n}} \right) \geq \sum_{n=1}^{\infty} \frac{1}{2^{n-1}} \frac{\|f-g\|_{n}}{1+\|f-g\|_{n}} = \rho(f,g) $$.
Thus, Property 3 holds.
Since all properties hold, \(\rho\) is indeed a metric on \(C[0, \infty)\).
##Part (b): Convergence in terms of uniform convergence##
Given that \(\lim_{k \rightarrow \infty} f_k = f\) under the metric \(\rho\), we have to show that \(\lim_{k \rightarrow \infty} f_k(x) = f(x)\) uniformly on every finite subinterval of \([0, \infty)\).
Suppose \([a,b]\) is a finite subinterval of \([0,\infty)\). For any given \(\epsilon > 0\), choose \(K \in \mathbb{N}\) such that \(2^{-(K-1)} < \frac{\epsilon}{3}\).
Due to the convergence of \(f_k\) to \(f\) in the metric space defined by \(\rho\), we know that there must exist some \(N\) such that for all \(k \geq N\),
$$\rho(f_k, f) = \sum_{n=1}^{\infty} \frac{1}{2^{n-1}} \rho_n(f_k, f) < 2^{-(K-1)}.$$
Now, choose \(n \geq K+1\) such that \(n \geq b\). Then for any \(x \in [a,b]\) and \(k \geq N\), we have
$$|f_k(x)-f(x)| \leq \|f_k - f\|_n \leq (1+ \|f_k - f\|_n) \rho_n(f_k, f) \leq (1+\|f_k - f\|_n)\frac{1}{2^{n-1}}\rho(f_k, f) < \frac{\epsilon}{3}(1+\|f_k - f\|_n)$$.
Now, looking at the first \(K-1\) terms in the sum defining \(\rho(f_k, f)\), we have
$$\sum_{n=1}^{K-1} \frac{1}{2^{n-1}} \rho_n(f_k, f) \geq \frac{1}{2^b} \rho_{b}(f_k, f).$$
But, we also have
$$\sum_{n=1}^{K-1} \frac{1}{2^{n-1}} \rho_n(f_k, f) < 2^{-(K-1)} < \frac{\epsilon}{3}.$$
Hence,
$$ \frac{1}{2^{b}}\rho_b(f_k, f) < \frac{\epsilon}{3}, \Rightarrow \rho_b(f_k, f) < \frac{\epsilon}{3}(1+\|f_k-f\|_b) \Rightarrow \|f_k-f\|_b < \epsilon$$.
Thus, \(|f_k(x)-f(x)| < \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon\) for every \(x \in [a,b]\), which implies uniform convergence on every finite subinterval.
Conversely, if the functions \(f_k\) converge uniformly to \(f\) on every finite subinterval of \([0,\infty)\), suppose \(\epsilon > 0\) and choose any \(M \in \mathbb{N}\) such that \(2^{-(M-1)} < \frac{\epsilon}{2}\). We can find some \(N\) such that for all \(n \leq M\) and \(k \geq N\), we have \(\|f_k - f\|_n < \frac{\epsilon}{2}(1+\|f_k - f\|_n)\). This implies that \(\rho_n(f_k, f) < 2^{-(n-1)} \frac{\epsilon}{2}\) for all \(k \geq N\) and \(n \leq M\).
Hence, for all \(k \geq N\), we have
$$\rho(f_k, f) = \sum_{n=1}^{\infty} \rho_n(f_k, f) \leq \sum_{n=1}^{M} \rho_n(f_k, f) + \sum_{n=M+1}^{\infty} \rho_n(f_k, f) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$
Thus, \(\lim_{k\to\infty} f_k=f\) in the metric space \((C[0, \infty), \rho)\).
##Part (c): Completeness of \((C[0, \infty), \rho)\)##
To show that \((C[0, \infty), \rho)\) is complete, we need to prove that every Cauchy sequence converges in this metric space.
Let \(\{f_k\}\) be a Cauchy sequence in \((C[0, \infty), \rho)\). For any \(\epsilon > 0\), we can find some \(N\) such that for all \(m, n \geq N\), \(\rho(f_m, f_n) < \epsilon\). This implies that the sequence \(\{f_k\}\) is Cauchy under the metric \(\rho_n\) for any \(n \in \mathbb{N}\).
Now, consider a fixed \(n\) and let \(M= \max(\|f_n\|_n, 1)\). By the Heine-Cantor theorem, any continuous function on the interval \([0,n]\) is uniformly continuous, especially \(f_1, f_2, \dots\) on this interval.
As \(\{f_k\}\) is a Cauchy sequence in the metric \(\rho_n\), it is uniformly Cauchy on \([0, n]\), meaning for any \(x \in [0, n]\), \(\{f_k(x)\}\) is a Cauchy sequence of real numbers. Thus, for every \(x \in [0, n]\), there exists a limit \(\lim_{k\to\infty} f_k(x) = f(x)\).
We can now construct a function \(f: [0, \infty) \to \mathbb{R}\), by defining \(f(x) = \lim_{k \to \infty} f_k(x)\) for every \(x \in [0, \infty)\). As each \(f_k\) is continuous on \([0, \infty)\) and the convergence is uniform on every finite interval, \(f\) is continuous on \([0, \infty)\), and thus \(f \in C[0, \infty)\).
Since the convergence of \(\{f_k\}\) to \(f\) is uniform on every finite subinterval, we have shown in Part (b) that this implies \(\lim_{k \to \infty} f_k = f\) in the metric space \((C[0, \infty), \rho)\).
Thus, every Cauchy sequence \(\{f_k\}\) converges to a limit in \(C[0, \infty)\), and the metric space \((C[0, \infty), \rho)\) is complete.
