A Proof of the Müntz–Szász Theorem

We give a proof of the Müntz–Szász theorem. The two halves use different ideas: Gram determinants for necessity, and the Riesz representation theorem together with the Blaschke condition for sufficiency.

1. Introduction

Stone-Weierstrass tells us that ordinary polynomials are dense in \(C([0,1])\). So every continuous function on \([0,1]\) can be approximated by linear combinations of \(1,x,x^2,\dots\). The Müntz–Szász theorem asks what happens if we keep only some of these powers.

The answer is sharp. Let

\[ 0=\lambda_0<\lambda_1<\lambda_2<\cdots. \]

We assume that \(\lambda_k\to\infty\). Then the span of \(x^{\lambda_0},x^{\lambda_1},x^{\lambda_2},\dots\) is dense in \(C([0,1])\) exactly when

\[ \sum_{k=1}^{\infty}\frac{1}{\lambda_k} \]

diverges. So \(1,x,x^2,x^3,\dots\) works because \(\sum 1/n\) diverges. But \(1,x,x^4,x^9,\dots\) does not work because \(\sum 1/n^2\) converges.

The theorem goes back to Müntz and Szász. For a short survey, see Moragues.

2. Setup and Standard Facts

We work with real-valued functions throughout. We fix a strictly increasing sequence

\[ 0=\lambda_0<\lambda_1<\lambda_2<\cdots, \qquad \lambda_k\to\infty, \]

and we write

\[ M_\Lambda=\operatorname{span}\lbrace x^{\lambda_k}:k\ge 0\rbrace\subset C([0,1]). \]

The space \(C([0,1])\) is equipped with the uniform norm

\[ \lVert f\rVert_\infty=\sup_{x\in[0,1]}\lvert f(x)\rvert. \]

We also use the real Hilbert space \(L^2([0,1])\) with inner product

\[ \langle f,g\rangle=\int_0^1 f(x)g(x)\,dx. \]

For real numbers \(a,b\ge 0\), we have

\[ \langle x^a,x^b\rangle=\int_0^1 x^{a+b}\,dx=\frac{1}{a+b+1}. \]

We now recall the facts that will be used later.

Let \(\mathbb D=\lbrace z\in\mathbb C:\lvert z\rvert<1\rbrace\). We write \(H^\infty(\mathbb D)\) for the space of bounded holomorphic functions on \(\mathbb D\).

If \(X\) is a proper closed subspace of a normed space \(Y\), then there is a nonzero continuous linear functional \(\varphi\colon Y\to\mathbb R\) such that \(\varphi(x)=0\) for every \(x\in X\).

Every continuous linear functional \(\varphi\colon C([0,1])\to\mathbb R\) has the form

\[ \varphi(f)=\int_0^1 f\,d\mu \]

for a unique finite signed Borel measure \(\mu\) on \([0,1]\).

Let \(A\subset C([0,1])\) be a subalgebra that contains the constant functions and separates points. Then \(A\) is dense in \(C([0,1])\).

This is the real Stone-Weierstrass theorem. See Rudin, Theorem 7.32.

3. The Theorem

Let \((\lambda_k)_{k\ge 0}\) be a strictly increasing sequence of nonnegative real numbers with \(\lambda_0=0\) and \(\lambda_k\to\infty\). Then the following are equivalent.

The space \(M_\Lambda\) is dense in \(C([0,1])\).

The series

\[ \sum_{k=1}^{\infty}\frac{1}{\lambda_k} \]

diverges.

The condition \(\lambda_0=0\) puts the constant function \(1\) into the span. Without \(1\), we could not be dense in all of \(C([0,1])\).

4. Necessity

In this section we assume that

\[ \sum_{k=1}^{\infty}\frac{1}{\lambda_k}<\infty. \]

We will show that \(M_\Lambda\) is not dense in \(C([0,1])\).

We start with a Hilbert space formula. The short version of the argument is this. We first replace the spanning family by an orthonormal basis of the same subspace. Then the determinant becomes easy to read.

Let \(H\) be a real Hilbert space. Let \(f_1,\dots,f_n\) be a basis of a finite-dimensional subspace \(V\subset H\). Let \(g\in H\). Then

\[ \operatorname{dist}(g,V)^2 = \frac{\Gamma(f_1,\dots,f_n,g)}{\Gamma(f_1,\dots,f_n)}, \]

where

\[ \operatorname{dist}(g,V)=\inf_{v\in V}\lVert g-v\rVert, \]

and \(\Gamma(h_1,\dots,h_r)\) is the Gram determinant

\[ \det(\langle h_i,h_j\rangle)_{1\le i,j\le r}. \]

Choose another basis \(u_1,\dots,u_n\) of \(V\). Then there is an invertible matrix \(A\) with

\[ (u_1,\dots,u_n)=(f_1,\dots,f_n)A. \]

The Gram matrix for \(u_1,\dots,u_n\) is \(A^\ast G A\), where \(G\) is the Gram matrix for \(f_1,\dots,f_n\). So its determinant is \(\lvert\det A\rvert^2\det G\). The same factor appears in the numerator. Hence the quotient in the lemma does not change when we replace \(f_1,\dots,f_n\) by another basis of \(V\).

Now choose an orthonormal basis \(e_1,\dots,e_n\) of \(V\). Write

\[ p=\sum_{i=1}^n c_i e_i \]

and

\[ g=p+h, \]

where \(h\perp V\). Since \(p\in V\), every \(v\in V\) satisfies

\[ g-v=(p-v)+h. \]

Here \(p-v\in V\), so \((p-v)\perp h\). Therefore

\[ \lVert g-v\rVert^2=\lVert p-v\rVert^2+\lVert h\rVert^2\ge \lVert h\rVert^2. \]

Equality holds when \(v=p\). So

\[ \operatorname{dist}(g,V)=\lVert h\rVert. \]

The denominator is now

\[ \Gamma(e_1,\dots,e_n)=1. \]

Let \(B\) be the Gram matrix of \(e_1,\dots,e_n,g\). Then

\[ B_{ij}=\delta_{ij} \qquad (1\le i,j\le n), \]

and

\[ B_{i,n+1}=c_i, \qquad B_{n+1,i}=c_i, \qquad B_{n+1,n+1}=\lVert g\rVert^2. \]

So the first \(n\) rows and columns look like the identity matrix. Now subtract \(c_i\) times row \(i\) from the last row for each \(i=1,\dots,n\). This changes the last row to

\[ (0,\dots,0,\lVert g\rVert^2-\sum_{i=1}^n c_i^2). \]

Since \(p=\sum c_i e_i\), we have \(\sum c_i^2=\lVert p\rVert^2\). Also \(g=p+h\) with \(p\perp h\), so

\[ \lVert g\rVert^2-\lVert p\rVert^2=\lVert h\rVert^2. \]

Now expand the determinant along the last row. The numerator is \(\lVert h\rVert^2\). Since \(\operatorname{dist}(g,V)=\lVert h\rVert\), this gives the formula.

We also need the next determinant identity. Since it will be used in the main computation, it is worth proving it here.

Let \(a_0,\dots,a_n\) and \(b_0,\dots,b_n\) satisfy \(a_i+b_j\neq 0\) for all \(i,j\). Then

\[ \det\left(\frac{1}{a_i+b_j}\right)_{0\le i,j\le n} = \frac{\prod_{0\le i<j\le n}(a_j-a_i)\prod_{0\le i<j\le n}(b_j-b_i)}{\prod_{i,j=0}^n(a_i+b_j)}. \]

Here \(\det\) means the determinant of the displayed matrix.

Let

\[ D_n(a,b)=\det\left(\frac{1}{a_i+b_j}\right)_{0\le i,j\le n}. \]

We argue by induction on \(n\). The case \(n=0\) is clear.

For \(n\ge 1\), subtract column \(0\) from columns \(1,\dots,n\). For \(j\ge 1\), the new entry in row \(i\), column \(j\) is

\[ \frac{1}{a_i+b_j}-\frac{1}{a_i+b_0} = \frac{b_0-b_j}{(a_i+b_j)(a_i+b_0)}. \]

So we can factor out \(b_0-b_j\) from column \(j\) and \(1/(a_i+b_0)\) from row \(i\). This gives

\[ D_n(a,b) = \frac{\prod_{j=1}^n (b_0-b_j)}{\prod_{i=0}^n (a_i+b_0)} \det N, \]

where the first column of \(N\) is all \(1\)’s and

\[ N_{ij}=\frac{1}{a_i+b_j} \qquad (j\ge 1). \]

Now subtract row \(0\) from rows \(1,\dots,n\). For \(i\ge 1\) and \(j\ge 1\),

\[ \frac{1}{a_i+b_j}-\frac{1}{a_0+b_j} = \frac{a_0-a_i}{(a_i+b_j)(a_0+b_j)}. \]

So we can factor out \(a_0-a_i\) from row \(i\) and \(1/(a_0+b_j)\) from column \(j\) for \(j\ge 1\). The first column now has a single \(1\) at the top and zeros below it, and the lower right \(n\times n\) block is

\[ \left(\frac{1}{a_i+b_j}\right)_{1\le i,j\le n}. \]

Hence

\[ D_n(a,b) = \frac{\prod_{i=1}^n(a_0-a_i)\prod_{j=1}^n(b_0-b_j)}{\prod_{i=0}^n(a_i+b_0)\prod_{j=1}^n(a_0+b_j)} D_{n-1}(a_1,\dots,a_n;b_1,\dots,b_n). \]

Now apply the induction hypothesis to \(D_{n-1}\). Each factor \(a_0-a_i\) and \(b_0-b_j\) contributes one minus sign, so the signs cancel in pairs. This gives exactly the stated formula.

We will also need a short fact about infinite products.

Let \(0<u_k<1\) and assume \(u_k\to 0\). Then

\[ \prod_{k=1}^{\infty}(1-u_k)=0 \quad\Longleftrightarrow\quad \sum_{k=1}^{\infty}u_k=\infty. \]

For large \(k\), we have \(u_k\le 1/2\). So

\[ -2u_k\le \log(1-u_k)\le -u_k. \]

This shows that \(\sum \log(1-u_k)=-\infty\) exactly when \(\sum u_k\) diverges.

Now we compute the distance from a monomial to the span of the allowed powers.

Fix \(m\in\lbrace 0,1,2,\dots\rbrace\). Let

\[ M_n=\operatorname{span}\lbrace x^{\lambda_0},\dots,x^{\lambda_n}\rbrace. \]

If \(m\notin\lbrace \lambda_0,\dots,\lambda_n\rbrace\), then

\[ \operatorname{dist}(x^m,M_n)^2 = \frac{\prod_{k=0}^n(m-\lambda_k)^2}{(2m+1)\prod_{k=0}^n(m+\lambda_k+1)^2}. \]

By ,

\[ \operatorname{dist}(x^m,M_n)^2 = \frac{\Gamma(x^{\lambda_0},\dots,x^{\lambda_n},x^m)}{\Gamma(x^{\lambda_0},\dots,x^{\lambda_n})}. \]

For the denominator, apply with

\[ a_i=\lambda_i+1 \qquad b_j=\lambda_j. \]

Here \(0\le i,j\le n\).

This gives

\[ \Gamma(x^{\lambda_0},\dots,x^{\lambda_n}) = \frac{\prod_{0\le i<j\le n}(\lambda_j-\lambda_i)^2}{\prod_{i,j=0}^n(\lambda_i+\lambda_j+1)}. \]

For the numerator, we add one more exponent, namely \(m\). So the same formula gives

\[ \Gamma(x^{\lambda_0},\dots,x^{\lambda_n},x^m) = \frac{\prod_{0\le i<j\le n}(\lambda_j-\lambda_i)^2\prod_{k=0}^n(m-\lambda_k)^2}{(2m+1)\prod_{i,j=0}^n(\lambda_i+\lambda_j+1)\prod_{k=0}^n(m+\lambda_k+1)^2}. \]

Now divide the second formula by the first one. The common factors cancel, and the stated distance formula follows.

We can now prove the first half of the theorem.

If

\[ \sum_{k=1}^{\infty}\frac{1}{\lambda_k}<\infty, \]

then \(M_\Lambda\) is not dense in \(C([0,1])\).

Since \(\sum_{n=1}^{\infty}1/n\) diverges, the set \(\lbrace \lambda_k:k\ge 1\rbrace\) cannot contain every positive integer. So we can choose \(m\in\mathbb N\) with \(m\notin\lbrace \lambda_k:k\ge 0\rbrace\).

For this \(m\), gives

\[ \operatorname{dist}(x^m,M_n)^2 = \frac{1}{2m+1} \prod_{k=0}^n \left(\frac{\lvert m-\lambda_k\rvert}{m+\lambda_k+1}\right)^2. \]

Choose \(K\) such that \(\lambda_k>m\) whenever \(k\ge K\). Then

\[ \frac{\lvert m-\lambda_k\rvert}{m+\lambda_k+1} = 1-\frac{2m+1}{m+\lambda_k+1}. \]

For \(k\ge K\), set

\[ u_k=\frac{2m+1}{m+\lambda_k+1}. \]

Then \(0<u_k<1\), \(u_k\to 0\), and

\[ \lambda_k u_k\longrightarrow 2m+1. \]

Since \(\sum 1/\lambda_k\) converges, \(\sum_{k=K}^{\infty}u_k\) also converges. By ,

\[ \prod_{k=K}^{\infty}(1-u_k)>0. \]

The factors with \(k<K\) form a fixed positive number because \(m\neq\lambda_k\). Thus the distance in the displayed formula stays bounded away from zero.

Therefore \(x^m\) is not in the \(L^2\)-closure of \(M_\Lambda\). In particular, it is not in the uniform closure either, because

\[ \lVert x^m-f\rVert_2\le \lVert x^m-f\rVert_\infty \]

on \([0,1]\). So \(M_\Lambda\) is not dense in \(C([0,1])\).

5. Sufficiency

Now assume that

\[ \sum_{k=1}^{\infty}\frac{1}{\lambda_k}=\infty. \]

We will show that \(M_\Lambda\) is dense in \(C([0,1])\).

The complex analysis input is the following fact.

Let \(g\in H^\infty(\mathbb D)\), and let \(a_k\in(0,1)\) be distinct zeros of \(g\). If

\[ \sum_{k=1}^{\infty}(1-a_k)=\infty, \]

then \(g\equiv 0\).

This is the theorem on zero sets of bounded holomorphic functions on the disk. See Rudin, Theorem 15.23.

Next we turn a measure on \([0,1]\) into a holomorphic function.

Let \(\mu\) be a finite signed Borel measure on \([0,1]\). Define

\[ F(z)=\int_0^1 x^z\,d\mu(x), \qquad \Re z>0. \]

Here \(x^z=e^{z\log x}\) for \(x>0\), and \(0^z=0\).

Then \(F\) is bounded and holomorphic on the right half-plane.

Write \(z=\sigma+it\) with \(\sigma>0\). Then \(\lvert x^z\rvert=x^\sigma\le 1\) on \([0,1]\), so the integral is bounded by the total variation of \(\mu\).

Now fix \(\delta>0\). The function \(x^\delta\lvert\log x\rvert\), extended by \(0\) at the origin, is bounded on \([0,1]\). Thus \(x^z\log x\) is bounded uniformly when \(\Re z\ge\delta\). We can differentiate under the integral on every compact subset of the right half-plane, so \(F\) is holomorphic there.

Now comes the main step.

Assume

\[ \sum_{k=1}^{\infty}\frac{1}{\lambda_k}=\infty. \]

Let \(\mu\) be a finite signed Borel measure on \([0,1]\) such that

\[ \int_0^1 x^{\lambda_k}\,d\mu(x)=0. \]

This holds for every \(k\ge 0\).

Then \(\mu=0\).

Define \(F\) as in . Then \(F\) is bounded and holomorphic on \(\Re z>0\). Also

\[ F(\lambda_k)=0. \]

This holds for every \(k\ge 1\).

We move to the unit disk because the Blaschke condition lives there. Let

\[ \phi(w)=\frac{1+w}{1-w}, \qquad w\in\mathbb D. \]

This map sends the unit disk onto the right half-plane. So

\[ G(w)=F(\phi(w)) \]

is a bounded holomorphic function on \(\mathbb D\).

The zeros \(\lambda_k\), for \(k\ge 1\), of \(F\) give zeros

\[ a_k=\phi^{-1}(\lambda_k)=\frac{\lambda_k-1}{\lambda_k+1} \]

of \(G\). Choose \(K\) such that \(a_k\in(0,1)\) whenever \(k\ge K\). For these indices,

\[ 1-a_k=\frac{2}{\lambda_k+1}. \]

Since \(\sum 1/\lambda_k\) diverges, the tail \(\sum_{k=K}^{\infty}(1-a_k)\) also diverges. Apply to the distinct zeros \((a_k)_{k\ge K}\). We get \(G\equiv 0\), so \(F\equiv 0\) on the right half-plane.

In particular,

\[ \int_0^1 x^n\,d\mu(x)=F(n)=0. \]

This holds for every integer \(n\ge 1\). The constant term is separate. Since \(\lambda_0=0\), the assumption also gives

\[ \int_0^1 1\,d\mu=0. \]

So \(\mu\) kills every polynomial. Let \(A\) be the algebra of ordinary polynomials. It contains the constants and it separates points because it contains the function \(x\). By , \(A\) is dense in \(C([0,1])\). Since integration against \(\mu\) is continuous in the uniform norm, it follows that

\[ \int_0^1 f\,d\mu=0. \]

This holds for every \(f\in C([0,1])\).

By uniqueness in the Riesz representation theorem, this forces \(\mu=0\).

Now we finish the theorem.

If

\[ \sum_{k=1}^{\infty}\frac{1}{\lambda_k}=\infty, \]

then \(M_\Lambda\) is dense in \(C([0,1])\).

Let \(X\) be the uniform closure of \(M_\Lambda\) in \(C([0,1])\). Suppose \(X\neq C([0,1])\). By , there is a nonzero continuous linear functional \(\varphi\) on \(C([0,1])\) that vanishes on \(X\). By , there is a finite signed Borel measure \(\mu\) such that

\[ \varphi(f)=\int_0^1 f\,d\mu \]

for every \(f\in C([0,1])\). Since \(\varphi\) vanishes on \(M_\Lambda\), we get

\[ \int_0^1 x^{\lambda_k}\,d\mu(x)=0. \]

This holds for every \(k\ge 0\).

By , this forces \(\mu=0\). So \(\varphi=0\), which is impossible. Therefore \(X=C([0,1])\).

Together with , this proves .

6. Large Sets and Stone-Weierstrass

The theorem has a clean form when the exponents are integers. Here \(\mathbb N={1,2,3,\dots}\).

A set \(S\subset\mathbb N\) is called large if

\[ \sum_{n\in S}\frac{1}{n}=\infty. \]

Let \(S\subset{0,1,2,\dots}\) with \(0\in S\). Then

\[ \operatorname{span}\lbrace x^n:n\in S\rbrace \]

is dense in \(C([0,1])\) if and only if \(S\setminus{0}\) is large.

If \(S\) is finite, its span is finite-dimensional and therefore closed. It is a proper subspace of \(C([0,1])\), so it is not dense. Also \(S\setminus{0}\) is not large.

Now suppose \(S\) is infinite. List its elements in increasing order as

\[ 0=\lambda_0<\lambda_1<\lambda_2<\cdots \]

and apply .

For algebras, much less is needed.

Let \(A\subset C([0,1])\) be a closed subalgebra. If \(1\in A\) and \(x^\lambda\in A\) for some \(\lambda>0\), then \(A=C([0,1])\).

The function \(x\mapsto x^\lambda\) is strictly increasing on \([0,1]\), so it separates points. Now apply .

Here are three quick examples.

If \(S={0,1,2,3,\dots}\), then \(\sum_{n\in S\setminus{0}}1/n\) diverges. So ordinary polynomials are dense in \(C([0,1])\).

If \(S\) is \(\lbrace 0\rbrace\) together with the prime numbers, then \(\sum_p 1/p\) still diverges. So the span of \(\lbrace 1,x^2,x^3,x^5,x^7,\dots\rbrace\) is dense in \(C([0,1])\).

If \(S=\lbrace 0\rbrace\cup\lbrace n^2:n\ge 1\rbrace\), then \(\sum 1/n^2\) converges. So \(\operatorname{span}\lbrace 1,x,x^4,x^9,\dots\rbrace\) is not dense in \(C([0,1])\).

7. References

Andreu Ferré Moragues, What is the Müntz–Szász theorem?, Ohio State University, 2018. https://math.osu.edu/sites/math.osu.edu/files/What%20is%202018%20Muntz%20Szasz%20Theorem.pdf.
Ch. H. Müntz, Über den Approximationssatz von Weierstraß, in Mathematische Abhandlungen Hermann Amandus Schwarz, Springer, Berlin, 1914, pp. 303-312. https://doi.org/10.1007/978-3-642-50735-9_22.
O. Szász, Über die Approximation stetiger Funktionen durch lineare Aggregate von Potenzen, Mathematische Annalen 77 (1916), 482-496. https://doi.org/10.1007/BF01456964.
Walter Rudin, Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, 1976, Chapter 7.
Walter Rudin, Real and Complex Analysis, 3rd ed., McGraw-Hill, 1987, Theorems 15.23 and 15.26(a), pp. 313-314.