%%  chapter-11.tex
%%  Approximation Theory: Chapter 11
%%  Neal L. Carothers
%%  Bowling Green State University
%%  Bowling Green, Ohio  43403
%%  carother@math.bgsu.edu
%%  http://www.bgsu.edu/~carother/


\input 680-setup.tex

\chaptertitle = {M\"untz Theorems}

\centerline{\hfil\tf The M\"untz Theorems\hfil}
\vskip-\baselineskip
\line{\sc Math 680 \hfil 7/28/94}

\noindent
For several weeks now we've taken advantage of the
fact that the monomials $1,x,x^2,\ldots$ have
{\sl dense linear span\/} in $C[\,0,1\,]$.  What,
if anything, is so special about these particular
powers?  How about if we consider polynomials of
the form $\sum_{k=0}^na_kx^{k^2}$; are they dense,
too?  More generally, what can be said about the
span of a sequence of monomials $(x^{\lambda_n})$,
where $\lambda_0<\lambda_1<\lambda_2<\cdots$?  
Of course, we'll have to assume that $\lambda_0\ge0$,
but it's not hard to see that we will actually
need $\lambda_0=0$, for otherwise each of the
polynomials $\sum_{k=0}^na_kx^{\lambda_k}$ vanishes
at $x=0$ (and so has distance
at least $1$ from the constant $1$ function, for
example).  If the $\lambda_n$'s are integers,
it's also clear that we'll have to have
$\lambda_n\to\infty$ as $n\to\infty$.  But
what else is needed?  The answer comes to us from 
M\"untz in 1914.  
(You sometimes see the name Otto Sz\'asz associated with
M\"untz's theorem, because Sz\'asz proved a similar theorem
at nearly the same time (1916).)

\proclaim Theorem.
Let\/ $0\le\lambda_0<\lambda_1<\lambda_2<\cdots$.  Then, 
the functions $(x^{\lambda_n})$
have dense linear span in $C[\,0,1\,]$ if and only if\/
$\lambda_0=0$ and\/ 
$\sum_{n=1}^\infty\lambda_n^{-1}=\infty$.

What M\"untz is trying to tell us here is that the
$\lambda_n$'s can't get big too quickly.  In particular,
the polynomials of the form $\sum_{k=0}^na_kx^{k^2}$
are evidently {\sl not\/} dense in $C[\,0,1\,]$.
On the other hand, the $\lambda_n$'s don't have to be
unbounded; indeed, M\"untz's theorem implies an earlier
result of Bernstein from 1912: If $0<\alpha_1<\alpha_2<\cdots<K$
(some constant), then $1,x^{\alpha_1},x^{\alpha_2},\ldots$
have dense linear span in $C[\,0,1\,]$.   

Before we give the proof of M\"untz's theorem, 
let's invent a bit of notation: We write
$$X_n \ = \ \left\{\sum_{k=0}^na_kx^{\lambda_k}:
	a_0,\ldots,a_n\in\R\right\}$$
and, given $f\in C[\,0,1\,]$, we write 
${\rm dist}(f,X_n)$ to denote the distance from $f$
to the space spanned by $1,x^{\lambda_1},\ldots,x^{\lambda_n}$.
Let's also write $X=\bigcup_{n=0}^\infty X_n$.  That is,
$X$ is the linear span of the entire sequence 
$(x^{\lambda_n})_{n=0}^\infty$.  The question here is
whether $X$ is dense, and we'll address the problem by
determining whether ${\rm dist}(f,X_n)\to0$, as $n\to\infty$,
for every $f\in C[\,0,1\,]$.

If we can show that each (fixed) power $x^m$ can be
uniformly approximated by a linear combination of $x^{\lambda_n}$'s,
then the Weierstrass theorem will tell us that $X$
is dense in $C[\,0,1\,]$.  
(How?)  Surprisingly, the numbers ${\rm dist}(x^m,X_n)$ can
be estimated.  Our proof won't give the best
estimate, but it will show how the condition 
$\sum_{n=1}^\infty\lambda_n^{-1}=\infty$ comes into the
picture.

\proclaim Lemma. 
Let $m>0$.  Then, 
$\displaystyle {\rm dist}(x^m,X_n)
	\ \le \ \prod_{k=1}^n\left|1-{m\over{\lambda_k}}\right|$.

\proof
We may certainly assume that $m\ne\lambda_n$ for any $n$.
Given this, we inductively define a sequence of functions by 
setting $P_0(x)=x^m$ and 
$$P_n(x) \ = \ 
	(\lambda_n-m)\,x^{\lambda_n}\int_x^1t^{-1-\lambda_n}\,P_{n-1}(t)\,dt$$
for $n\ge1$.  For example, 
$$P_1(x) \ = \ (\lambda_1-m)\,x^{\lambda_1}\int_x^1t^{-1-\lambda_1}\,t^m\,dt
	\ = \ -x^{\lambda_1}\,t^{m-\lambda_1}\bigr]_x^1 \ = \ x^m-x^{\lambda_1}.$$
By induction, each $P_n$ is of the form $x^m-\sum_{k=0}^na_kx^{\lambda_k}$ for
some scalars $(a_k)$:
$$\eqalign{
P_n(x) \ 
	&= \ (\lambda_n-m)\,x^{\lambda_n}\int_x^1t^{-1-\lambda_n}\,P_{n-1}(t)\,dt\cr
	&= \ (\lambda_n-m)\,x^{\lambda_n}\int_x^1t^{-1-\lambda_n}\,
		\left[\,t^m-\sum_{k=0}^{n-1}a_kt^{\lambda_k}\right]\,dt\cr
	&= \ x^m-x^{\lambda_n}+(\lambda_n-m)\,\sum_{k=0}^{n-1}\,
		{{a_k}\over{\lambda_n-\lambda_k}}\,(x^{\lambda_k}-x^{\lambda_n}).\cr
}$$
Finally, $\Vert P_0\Vert=1$ and 
$\Vert P_n\Vert\le|1-{m\over{\lambda_n}}|\,\Vert P_{n-1}\Vert$, because 
$$|\lambda_n-m|\,x^{\lambda_n}\int_x^1t^{-1-\lambda_n}\,dt
	\ = \ {|\lambda_n-m|\over{\lambda_n}}\,(1-x^{\lambda_n})
	\ \le \ \left|1-{m\over{\lambda_n}}\right|.$$
Thus,
$${\rm dist}(x^m,X_n) \ \le \ \Vert P_n\Vert \ \le \ 
	\prod_{k=1}^n\left|1-{m\over{\lambda_k}}\right|.\eqno\qed$$

The preceding result is due to v.\ Golitschek.
A slightly better estimate, also due to v.\ Golitschek
(1970), is \ 
${\rm dist}(x^m,X_n)\le\prod_{k=1}^n{{|m-\lambda_k|}\over{m+\lambda_k}}$.

Now a well-known fact about infinite products is that 
for positive $a_k$'s, the product 
$\prod_{k=1}^\infty\bigl|1-a_k\bigr|$ diverges (to $0$\/)
if and only if the series $\sum_{k=1}^\infty a_k$ diverges 
(to $\infty$\/) if and only if the product 
$\prod_{k=1}^\infty\bigl|1+a_k\bigr|$ diverges (to $\infty$\/).
In particular, $\prod_{k=1}^n\bigl|1-{m\over{\lambda_k}}\bigr|\to0$
if and only if $\sum_{k=1}^n{1\over{\lambda_k}}\to\infty$.
That is,  ${\rm dist}(x^m,X_n)\to0$ if and only
if $\sum_{k=1}^\infty{1\over{\lambda_k}}=\infty$.  This
proves the ``backward'' direction of M\"untz's theorem.

We'll prove the ``forward'' direction of M\"untz's 
theorem by proving a version of M\"untz's theorem
for the space $L_2[\,0,1\,]$.   
For our purposes, $L_2[\,0,1\,]$ denotes the space $C[\,0,1\,]$
endowed with the norm 
$$\Vert f\Vert_2 \ = \ \left(\int_0^1|f(x)|^2dx\right)^{1/2},$$
although our results are equally valid in the ``real'' 
space $L_2[\,0,1\,]$ (consisting of square-integrable,
Lebegue measurable functions).  In the latter case, 
we no longer need to assume that $\lambda_0=0$, but 
we do need to assume that each $\lambda_n>-1/2$ (in order that 
$x^{2\lambda_n}$ be integrable on $[\,0,1\,]$\/).

Remarkably, the distance from $f$ to the span of 
$x^{\lambda_0},x^{\lambda_1},\ldots,x^{\lambda_n}$ can be
computed {\sl exactly\/} in the $L_2$ norm.  For
this we'll need some more notation: Given linearly
independent vectors $f_1,\ldots,f_n$ in an inner product
space, we call  
$$G(f_1,\ldots,f_n) \ = \ \left|\matrix{
	\langle\,f_1,f_1\,\rangle &\cdots & \langle\,f_1,f_n\,\rangle\cr
	\vdots & \ddots & \vdots\cr
	\langle\,f_n,f_1\,\rangle &\cdots & \langle\,f_n,f_n\,\rangle\cr
	}\right| \ = \ 
	{\rm det}\bigl[\,\langle\,f_i,f_j\,\rangle\,\bigr]_{i,j}$$
the {\sl Gram determinant\/} of the $f_k$'s.

\proclaim Lemma. {\rm(Gram)} \ 
Let $F$ be a finite dimensional subspace of an inner product
space $V$, and let $g\in V\setminus F$.  Then, the distance
$d$ from $g$ to $F$ is given by 
$$d^{\,2} \ = \ {{G(g,f_1,\ldots,f_n)}\over{G(f_1,\ldots,f_n)}},$$
where $f_1,\ldots,f_n$ is any basis for $F$.

\proof
Let $f=\sum_{i=1}^na_if_i$ be the best approximation to $g$ 
out of $F$.  Then, since $g-f$ is orthogonal to $F$, we have,
in particular, $\langle\,f_j,g\,\rangle=\langle\,f_j,g\,\rangle$
for all $j$; that is, 
$$\sum_{i=1}^na_i\langle\,f_j,f_i\,\rangle \ = \ \langle\,f_j,g\,\rangle,
	\quad j=1,\ldots,n.\eqno(\ast)$$
Since this system of equations
always has a unique solution $a_1,\ldots,a_n$, we must have 
$G(f_1,\ldots,f_n)\ne0$ (and so the formula in our Lemma at least
makes sense).

Next, notice that 
$$d^{\,2} \ = \ \langle\,g-f,g-f\,\rangle \ = \ 
	\langle\,g-f,g\,\rangle \ = \ \langle\,g,g\,\rangle 
	\ - \ \langle\,g,f\,\rangle;$$
in other words,
$$d^{\,2} \ + \ \sum_{i=1}^na_i\langle\,g,f_i\,\rangle
	= \ \langle\,g,g\,\rangle.\eqno(\ast\ast)$$
Now consider $(\ast)$ and $(\ast\ast)$ as a
system of $n+1$ equations in the $n+1$ unknowns $a_1,\ldots,a_n$,
and $d^{\,2}$; in matrix form we have 
$$\left[\ \matrix{
	1 & \langle\,g,f_1\,\rangle &\cdots & \langle\,g,f_n\,\rangle\cr
	\noalign{\smallskip}
	0 & \langle\,f_1,f_1\,\rangle &\cdots & \langle\,f_1,f_n\,\rangle\cr
	\vdots & \ddots & \vdots & \vdots\cr
	0 & \langle\,f_n,f_1\,\rangle &\cdots & \langle\,f_n,f_n\,\rangle\cr
	}\ \right]\left[\,\matrix{
	 d^{\,2}\cr \noalign{\smallskip} a_1\cr \vdots\cr a_n\cr
	}\,\right] \ = \ \left[\,\matrix{
	\langle\,g,g\,\rangle\cr \noalign{\smallskip} 
	\langle\,f_1,g\,\rangle\cr \vdots\cr \langle\,f_n,g\,\rangle\cr
	}\,\right].$$
Solving for $d^{\,2}$ using Cramer's rule gives the desired result;
expanding along the first column shows that 
the matrix of coefficients has determinant $G(f_1,\ldots,f_n)$,
while the matrix obtained by replacing the ``$d$ column'' by the 
right-hand side has determinant $G(g,f_1,\ldots,f_n)$.~\qed

\noindent
{\bf Note}: By our last Lemma and induction, every Gram determinant is positive!

In what follows, we will still use
$X_n$ to denote the span of $x^{\lambda_0},\ldots,x^{\lambda_n}$,
but now we'll write ${\rm dist\,}_2(f,X_n)$ to denote the distance 
from $f$ to $X_n$ in the $L_2$ norm.  

\proclaim Theorem.
Let $m$, $\lambda_k>-1/2$ for $k=0,1,2,\ldots$.  Then,
$${\rm dist\,}_2(x^m,X_n) \ 
	= \ {1\over\sqrt{2m+1}}\,\prod_{k=0}^n{{|m-\lambda_k|}\over{m+\lambda_k+1}}.$$

\proof
The proof is based on a determinant formula due to Cauchy:
$$\prod_{i,j}(a_i+b_j)\,\left|\,\matrix{
	{1\over{a_1+b_1}} & \cdots & 
		{1\over{a_1+b_n}}\cr
	\vdots & \ddots & \vdots\cr
	\noalign{\medskip}
	{1\over{a_n+b_1}} & \cdots & 
		{1\over{a_n+b_n}}\cr
	}\,\right| \ = \ 
	\prod_{i>j}(a_i-a_j)(b_i-b_j).$$
If we consider each of the $a_i$'s and $b_j$'s as
``variables,'' then each side of the equation is
a {\sl polynomial\/} in $a_1,\ldots,a_n,b_1,\ldots,b_n$.
(Why?)  Now the right-hand side clearly vanishes if
$a_i=a_j$ or $b_i=b_j$ for some $i\ne j$, but the
left-hand side also vanishes in any of these cases.
Thus, the right-hand side divides the left-hand side.
But both polynomials have degree $n-1$ in each of the
$a_i$'s and $b_j$'s.  (Why?)  Thus, the left-hand side
is a constant multiple of the right-hand side.  To
show that the constant must be $1$, write the
left-hand side as 
$$\prod_{i\ne j}(a_i+b_j)\,\left|\,\matrix{
	1 & {{a_1+b_1}\over{a_1+b_2}} & \cdots & 
		{{a_1+b_1}\over{a_1+b_n}}\cr
	\noalign{\medskip}
	{{a_2+b_2}\over{a_2+b_1}} & 1 & \cdots & 
		{{a_2+b_2}\over{a_2+b_n}}\cr
	\vdots && \ddots & \vdots\cr
	\noalign{\medskip}
	{{a_n+b_n}\over{a_n+b_1}} & \cdots & 
		{{a_n+b_n}\over{a_n+b_{n-1}}} & 1\cr
	}\,\right|$$
and now take the limit as $b_1\to-a_1$, $b_2\to-a_2$, etc.  The 
expression above tends to $\prod_{i\ne j}(a_i-a_j)$, as does the
right-hand side of Cauchy's formula.

Now, $\langle\,x^p,x^q\,\rangle = \int_0^1x^{p+q}\,dx={1\over{p+q+1}}$
for $p$, $q>-1/2$, so
$$G(x^{\lambda_0},\ldots,x^{\lambda_n}) \ = \ 
	{\rm det}\left(\left[{1\over{\lambda_i+\lambda_j+1}}\right]_{i,j}\right) \ = \ 
	{{\prod_{i>j}(\lambda_i-\lambda_j)^2}
		\over{\prod_{i,j}(\lambda_i+\lambda_j+1)}},$$
with a similar formula holding for $G(x^m,x^{\lambda_0},\ldots,x^{\lambda_n})$.
Substituting these expressions into our distance formula and taking
square roots finishes the proof.~\qed

Now we can determine exactly when $X$ 
is dense in $L_2[\,0,1\,]$.  For easier comparison to the $C[\,0,1\,]$
case, we suppose that the $\lambda_n$'s are nonnegative.

\proclaim Theorem.
Let\/ $0\le\lambda_0<\lambda_1<\lambda_2<\cdots$.  Then, 
the functions $(x^{\lambda_n})$
have dense linear span in $L_2[\,0,1\,]$ if and only if\/
$\sum_{n=1}^\infty\lambda_n^{-1}=\infty$.

\proof
If $\sum_{n=1}^\infty{1\over{\lambda_n}}<\infty$, then each of
the products $\prod_{k=1}^n\bigl|1-{m\over{\lambda_k}}\bigr|$
and $\prod_{k=1}^n\bigl|1+{{(m+1)}\over{\lambda_k}}\bigr|$
converges to some nonzero limit for any $m$ not equal to any 
$\lambda_k$.  Thus, ${\rm dist\,}_2(x^m,X_n)\not\to0$,
as $n\to\infty$, for any $m\ne\lambda_k$, $k=0,1,2,\ldots$.   
In particular, the functions $(x^{\lambda_n})$ cannot  
have dense linear span in $L_2[\,0,1\,]$.

Conversely, if $\sum_{n=1}^\infty{1\over{\lambda_n}}=\infty$, 
then $\prod_{k=1}^n\bigl|1-{m\over{\lambda_k}}\bigr|$ diverges
to $0$ while $\prod_{k=1}^n\bigl|1+{{(m+1)}\over{\lambda_k}}\bigr|$
diverges to $+\infty$.  Thus, ${\rm dist\,}_2(x^m,X_n)\to0$,
as $n\to\infty$, for every $m>-1/2$.  Since the polynomials
are dense in $L_2[\,0,1\,]$, this finishes the proof.~\qed

Finally, we can finish the proof of M\"untz's theorem
in the case of $C[\,0,1\,]$.  Suppose that the functions 
$(x^{\lambda_n})$ have dense linear span in $C[\,0,1\,]$.
Then, since 
$\Vert f\Vert_2\le\Vert f\Vert$,
it follows that the functions $(x^{\lambda_n})$ must also 
have dense linear span in $L_2[\,0,1\,]$.  (Why?)  Hence, 
$\sum_{n=1}^\infty{1\over{\lambda_n}}=\infty$.~\qed

Just for good measure, here's a second proof of the
``backward'' direction for $C[\,0,1\,]$ based on the
$L_2[\,0,1\,]$ version.  Suppose that 
$\sum_{n=1}^\infty{1\over{\lambda_n}}=\infty$, and let
$m\ge1$.  Then,
$$\openup2\jot\eqalignno{
\left|\,x^m-\sum_{k=0}^na_kx^{\lambda_k}\,\right| 
	\ & = \ \left|\,{1\over m}\int_0^xt^{m-1}\,dt
		-\sum_{k=0}^n{{a_k}\over{\lambda_k}}\int_0^xt^{\lambda_k-1}\,dt\,\right|\cr
	&\le \ \int_0^1\left|\,{1\over m}\,t^{m-1}
		-\sum_{k=0}^n{{a_k}\over{\lambda_k}}\,t^{\lambda_k-1}\,\right|\,dt\cr
	&\le \ \left(\int_0^1\left|\,{1\over m}\,t^{m-1}
		-\sum_{k=0}^n{{a_k}\over{\lambda_k}}\,t^{\lambda_k-1}\,
			\right|^2dt\right)^{1/2}.\cr
}$$
Now the functions
$(x^{\lambda_k-1})$ have dense linear span in $L_2[\,0,1\,]$
because $\sum_{\lambda_n>1}{1\over{\lambda_n-1}}=\infty$.
Thus, we can find $a_k$'s so that the right-hand side of this
inequality is less than some $\eps$.  Since this estimate
is independent of $x$, we've shown that 
$$\Max_{\strut 0\le x\le1}
	\left|\,x^m-\sum_{k=0}^na_kx^{\lambda_k}\,\right|<\eps.
	\eqno\qed$$

\proclaim Application.
Let $0=\lambda_0<\lambda_1<\lambda_2<\cdots$ 
with $\sum_{n=1}^\infty \lambda_n^{-1}=\infty$, 
and let $f$ be a continuous function on $[\,0,\infty)$ 
for which $c=\lim\limits_{t\to\infty}f(t)$ exists.
Then, $f$ can be
uniformly approximated by finite linear combinations of 
the exponentials $(e^{-\lambda_nt})_{n=0}^\infty$.

\proof
The function $g(x)=f(-\log x)$, for
$0<x\le1$, and $g(0)=c$, is continuous on $[\,0,1\,]$.  
In other words, $g(e^{-t})=f(t)$ for each $0\le t<\infty$.  
Thus, given 
$\eps>0$, we can find $n$ and $a_0,\ldots,a_n$ such that
$$\max_{\strut 0\le x\le1}\left|g(x)-\sum_{k=0}^na_kx^{\lambda_k}\right| \ 
	= \ \max_{\strut 0\le t<\infty}\left|f(t)-\sum_{k=0}^na_ke^{-\lambda_kt}\right|
	\ < \ \eps.\eqno\qed$$



\bye


%%  end of chapter-11.tex



