# on completeness of R_mn, the set of all rational functions of type (m,n)

It is known from finite dimensionality of $P_r$, the space of all polynomials of degree less than or equal to $r$, that $P_r$ is complete with respect to uniform norm.

Considering $R_{m,n}[a,b]=\{p/q: p \in P_m[a,b],q\in P_n[a,b], q|_{[a,b]}\neq 0\}$ (which is not a linear space) endowed with the uniform metric, is it a complete metric space? if not, exhibit a Cauchy sequence that is not convergent.

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This question does not seem too trivial to me. – Joseph Van Name Jun 13 '13 at 12:08
I gave it some thought, but I did not yet come up with an easy way to do it. Rational functions are very special things: meromorphic on the Riemann sphere. With at most a fixed number of poles (counted with multiplicity). – Gerald Edgar Jun 14 '13 at 14:30
Maybe not trivial, but the imperative formulation still suggests it's an exercise and thus off topic here. – Noam D. Elkies Jun 14 '13 at 21:42
I have my doubts it's homework. The imperative formulation might be a result of less than perfect command of English -- we've talked about this sort if thing before at MO. – Todd Trimble Jun 15 '13 at 14:28

At the risk that I'm doing somebody's homework...

Yes, $R_{m,n}[a,b]$ is complete with respect to the uniform metric $\left\| \cdot \right\|$.

Let $f \in {\cal C}([a,b])$ be a uniform limit of rational functions $f_j = p_j \phantom./\phantom. q_j \in R_{m,n}$ with each $p_j$ (respectively $q_j$) a polynomial of degree at most $m$ (resp. $n$), and $q_j(x) \neq 0$ for all $x \in [a,b]$. We shall show that $f \in R_{m,n}[a,b]$.

We would like to write $f = p/q$ with $p = \lim_j p_j$ and $q = \lim_j q_j$, but this doesn't work in general because $f_j$ does not determine $p_j,q_j$ uniquely: we can multiply $p_j,q_j$ by the same scalar, and if $f_j$ is actually a quotient of polynomials of degrees strictly less than $m$ and $n$ then we could even introduce a nonconstant common factor into $p_j$ and $q_j$.

We deal with the first ambiguity by scaling each $p_j$ and $q_j$ so that $\left\| q_j \right\| = 1$. Then the $q_j$ are all on the unit sphere in the finite-dimensional space $P_n[a,b]$, which is compact. This lets us finesse the second ambiguity by passing to a subsequence that converges in ${\cal C}([a,b])$.

Let $q$, then, be the limit of such a subsequence $\lbrace q_k \rbrace$, and let $p = f q$. Then $p$ is the uniform limit of polynomials $p_k = f_k q_k$, because $f$ and $q$ are uniform limits of $f_k$ and $q_k$ respectively. But each $p_k$ is in the finite-dimensional vector space $P_m$, which is complete in ${\cal C}([a,b])$; so the uniform limit $p$ is also a polynomial, of degree at most $m$.

Let $f_\infty \in R_{m,n}$ be the rational function $p/q$, reduced to lowest terms. We claim that $f = f_\infty$ on $[a,b]$. This is not quite immediate from what we've done so far, because $q$ might have a zero $x_0 \in [a,b]$ even though none of the $q_k$ does. [For example, let $m=n=2$ and $f_k = 1 + 1/(k^2 (x-x_0)^2 + k)$.] Still, since $f_k = p_k/q_k$ with $p_k \rightarrow p$ and $q_k \rightarrow q$, it does follow that $f_\infty(x) = f(x)$ for all $x \in [a,b]$ where $q(x) \neq 0$. Since $f$ is continuous on $[a,b]$, it follows that $f_\infty(x) = f(x)$ even at each of the finitely many $x$ where $q(x) = 0$. QED

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Not my area of expertise, but appears to be well-known in the literature that this is a complete metric space. From here for instance, we have that any continuous function $f$ on $[-1, 1]$ has a best uniform approximation by rational functions $r = p/q$ with $\deg(p) \leq m$, $\deg(q) \leq n$, and $q$ non-vanishing on $[-1, 1]$, in the sense that for some such (unique) $r^*$ we have

$$\|f - r^*\| \leq \|f - r\|$$

in sup norm for any other such $r$. Hence the infimum of $\|f - r\|$ cannot be zero, unless of course $f = r^*$ already belongs to the space.

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