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This question has been posed in multiple forms on here, but there is always one bit of information lacking. Suppose one considers the space $C$ of continuous functions $f:\mathopen [ 0,+\infty \mathclose [ \rightarrow Y$ where $Y$ is a complete metric space with metric $\rho \leqslant 1$. Define $d_n(f,g)=\sup \left \{\rho(f(x),g(x))| \ x\in \mathopen [ 0,n\mathclose ] \right \}$ for $f,g \in C$, then set $d\left ( f,g \right )= \sum_{n=1}^{\infty} {2^{-n}}d_n\left ( f,g \right )$. One proves: $d$ is a metric on $C$ that induces the topology of compact convergence. However, why is $d$ complete? Supposedly, this is a trivial matter - it never gets explained - but I don't see why. If $\left ( f_k\right )$ is a Cauchy sequence in $C$ for the metric $d$, apparently, then one should prove that the sequence converges pointwise (that's easy enough) but what next?

Romanda de Gore
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  • The convergence to the pointwise limit is in fact uniform on all intervals. Therefore, the limit is continuous and convergence happen with respect to $d$. – Jochen Jun 18 '18 at 11:25
  • It's best not to use $n$ as both the index of metrics and the index of functions. These are different indices. –  Jun 18 '18 at 11:40

1 Answers1

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First note that $$ \lim_{k\to\infty} d(f_k, f) = 0 \iff \forall n \ \lim_{k\to\infty} d_n(f_k, f) = 0 \tag1 $$ If $(f_k)$ is a Cauchy sequence with respect to $d$, then by (1) their restrictions to $[0, n]$ form a Cauchy sequence with respect to $d_n$. Since $C[0, n]$ is complete (reference), we have $g_n\in C[0, n]$ such that $f_k\to g_n$ uniformly on $[0, n]$.

Next, observe that the functions $g_n$ are such that the restriction of $g_n$ to $[0, m]$ with $m<n$ is exactly $g_m$. (This follows from the uniqueness of pointwise limits.) Thus, we can define $$ g(x) = g_n(x) \quad \text{where $n\in \mathbb{N}$ is any such that $n\ge x$} $$ which is a continuous function because each $g_n$ is continuous. Finally, $d(f_k, g)\to 0$ by (1).