Best weighted rational approximants exist

Let $\gamma :={\min }_{x\in I}w(x)>0$ and $M:=\mathrm{\parallel }wf{\mathrm{\parallel }}_{\mathrm{\infty },I}$. Positivity of $\gamma$ follows because the continuous positive function $w$ attains its minimum on compact $I$ by [F3]. In the coefficient space for pairs $(p,q)$ of degrees at most $(m,n)$ and an extra scalar $t$, define $K$ by $\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{f},2}=1$, $q(x)\ge 0$ on $I$, $0\le t\le M$, and $\mathrm{\mid }w(x)(f(x)q(x)-p(x))\mathrm{\mid }\le tq(x)$ for every $x\in I$. The triple $(0,1,M)$ lies in $K$, so $K$ is nonempty.

The coefficient norm equation, the interval for $t$, and each pointwise inequality defining $K$ are closed conditions on the finite coefficient vector, so $K$ is closed. Its denominator coefficients have norm $1$, and for every $(p,q,t)\in K$, $\mathrm{\mid }p(x)\mathrm{\mid }\le (\mathrm{\mid }f(x)\mathrm{\mid }+M\mathrm{/}\gamma )q(x)$ on $I$. The coefficient norm bound gives $\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty },I}\le {\sum }_{j=0}^n\max (\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid }{)}^j$, hence $\mathrm{\parallel }p{\mathrm{\parallel }}_{\mathrm{\infty },I}$ is uniformly bounded. For $m=0$, evaluation at one node bounds the sole coefficient of $p$. For $m\ge 1$, evaluate at the $m+1$ distinct nodes ${\tau }_j=a+j(b-a)\mathrm{/}m$; each value is bounded by $\mathrm{\parallel }p{\mathrm{\parallel }}_{\mathrm{\infty },I}$, so [F7] bounds every coefficient. Thus $K$ is bounded.

By [F3], $K$ is compact, so its coordinate function $(p,q,t)\mapsto t$ has a minimum $t_{\ast }$ at some $(p_{\ast },q_{\ast },t_{\ast })\in K$. For every admissible rational $r=p\mathrm{/}q$ with error at most $M$, [F1] supplies a normalized positive reduced representing pair, unique by [F2]; with $t=\mathrm{\parallel }w(f-r){\mathrm{\parallel }}_{\mathrm{\infty },I}$ this triple belongs to $K$. Candidates with larger error exceed $M\ge t_{\ast }$. Therefore $t_{\ast }\le E_{m,n}(f,w;I)$.

For the minimizing pair, $q_{\ast }$ is a nonzero nonnegative polynomial because its coefficient norm is $1$. Its zero set in $I$ is finite by [F4]. The defining inequality of $K$ and $w\ge \gamma$ give $\mathrm{\mid }{p}_{\ast }(x)\mathrm{\mid }\le (\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },I}+M\mathrm{/}\gamma )q_{\ast }(x)$ for every $x\in I$. At any zero $c$ of $q_{\ast }$, repeated use of [F5] writes $q_{\ast }(x)=(x-c{)}^{s}u(x)$ with $u(c)\mathrm{\ne }0$ and, if $p_{\ast }\mathrm{\ne }0$, writes $p_{\ast }(x)=(x-c{)}^{r}v(x)$ with $v(c)\mathrm{\ne }0$. If $p_{\ast }\mathrm{\ne }0$ and $r<s$, then [F8] and [F4] give $q_{\ast }(x)>0$ on a punctured one-sided neighborhood of $c$ inside $I$. By continuity in [F6] and $u(c)v(c)\mathrm{\ne }0$, $\mathrm{\mid }v(x)\mathrm{\mid }\mathrm{/}\mathrm{\mid }u(x)\mathrm{\mid }$ is bounded below by a positive constant on a sufficiently small such neighborhood, so $\mathrm{\mid }{p}_{\ast }(x)\mathrm{\mid }\mathrm{/}{q}_{\ast }(x)=\mathrm{\mid }x-c{\mathrm{\mid }}^{r-s}\mathrm{\mid }v(x)\mathrm{\mid }\mathrm{/}\mathrm{\mid }u(x)\mathrm{\mid }$ is unbounded as $x\to c$ from within $I$, contradicting the displayed bound. Hence $p_{\ast }$ is divisible by the factor of $q_{\ast }$ at every interval zero; this is automatic if $p_{\ast }=0$. Thus every interval zero of $q_{\ast }$ cancels from the quotient.

For each interior zero $c$ of $q_{\ast }$, its multiplicity is even: an odd multiplicity would make $q_{\ast }$ change sign across $c$, contradicting $q_{\ast }\ge 0$ on $I$. Form $H$ from the factors $(x-c{)}^s$ at interior zeros (where $s$ is even), $(x-a{)}^{s_a}$ at $a$ if it is a zero, and $(b-x{)}^{s_b}$ at $b$ if it is a zero. Then $H\ge 0$ on $I$ and $H>0$ off the finite zero set of $q_{\ast }$. Step 3.2 shows that $H$ divides both $p_{\ast }$ and $q_{\ast }$. Write $p_{\ast }=H\bar{p}$ and $q_{\ast }=H\bar{q}$. The exact multiplicities included in $H$ make $\bar{q}$ nonzero at every zero of $q_{\ast }$; elsewhere, $q_{\ast }>0$ and $H>0$, so $\bar{q}>0$. Since $\bar{q}$ is continuous by [F6], positivity off the finite zero set makes it nonnegative at each removed zero; its nonvanishing there makes it positive on all of $I$. At every point outside those zeros, divide the defining inequality $\mathrm{\mid }w(f{q}_{\ast }-p_{\ast })\mathrm{\mid }\le t_{\ast }{q}_{\ast }$ by $H$ to obtain $\mathrm{\mid }w(f\bar{q}-\bar{p})\mathrm{\mid }\le t_{\ast }\bar{q}$. Thus $\mathrm{\mid }w(f-\bar{p}\mathrm{/}\bar{q})\mathrm{\mid }\le t_{\ast }$ there. The quotient $\bar{p}\mathrm{/}\bar{q}$ and $w(f-\cdot )$ are continuous on $I$ by [F6], so the same bound holds at the cancelled zeros by continuity. Reduce $(\bar{p},\bar{q})$ using [F2]. The reduced denominator still has no zero on $I$ because it divides $\bar{q}$; it is continuous by [F6] and has constant sign by [F9] (otherwise the intermediate value theorem would give a zero). A common sign change makes it positive, and a further positive scaling gives coefficient norm $1$. The resulting pair is admissible and represents the same quotient, hence has error at most $t_{\ast }$. Therefore $E_{m,n}(f,w;I)\le t_{\ast }$. Together with step 3.1, equality holds and the infimum is attained.