A generalization of Hardy’s operator and an asymptotic Müntz-Szász Theorem

In this note we shall give a novel proof that Hardy’s Operator $H$, defined on ${\rm L}^{2}([0,1])$ by the formula,

is bounded. This new proof relies only on algebra together with the observation that the monomial functions are eigenvectors for $H$. Specifically, for each $k\geq 0$,

Always intrigued by results in analysis whose proofs rely mainly on algebra, the new proof of Hardy’s Inequality prompts the authors to propose the following definition.

We shall call the number $m$ in (1.3) the order of $T$. It can be any complex number with non-negative real part, though in all our examples it will be a natural number. In addition to $H$, a monomial operator of order $0$, other examples of monomial operators are the multiplication operator $M_{x}$ that sends a function $f(x)$ to the function $xf(x)$, and the Volterra operator $V=M_{x}H$, the operator given by

Both $M_{x}$ and $V$ are of order $1$.

For $0\leq s\leq 1$ we shall use $L^{2}[s,1]$ to denote the closed subspace of $L^{2}[0,1]$ of functions that are $0$ on $[0,s]$.

In this note we shall prove the following result.

Why might such a theorem be true? If $T$ is a monomial operator, and $f$ is a polynomial that vanishes at $0$ to some high order $M$, then $Tf$ also vanishes to order at least $M$. So if one thinks of vanishing on $[0,s]$ as an extreme case of vanishing to high order, one might believe that monomial operators preserve this property.

Our proof of Theorem 1.5 relies on a new type of Müntz-Szász Theorem, wherein the monomial sequence is allowed to drift. This may be more interesting than the theorem itself!

For a continuous function $f$ on $[0,1]$ consider the continuous function $Hf$ on $(0,1]$ defined by the formula

Noting that as $x\to 0$, $\frac{1}{x}\to\infty$ and $\int_{0}^{x}f(t)dt\to 0$, the following question arises:

Invoking the Mean Value Theorem for Integrals yields that for each $x\in[0,1]$, there exists $c\in[0,x]$ such that $Hf(x)=f(c)$. Thus,

so that in particular, $Hf$ is bounded near 0. More delicately, if we apply the MVT to the function $f(x)-f(0)$, we obtain the estimate

which implies that $Hf(x)\to f(0)$ as $x\to 0$. Therefore, if we agree to extend the definition of $Hf$ at the point $x=0$ by setting $Hf(0)=f(0)$, then our observations imply the following proposition.

Hardy [5] was the first to study the local behavior of $H$ at $0$ for functions equipped with norms other than the supremum norm. His result when specialized to ${\rm L}^{2}([0,1])$, the Hilbert space of square integrable Lesbesgue measurable functions on $[0,1]$, is as follows.

A linear operator $T$ on a normed vector space ${\mathcal{V}}$ is called bounded¹¹1It is a straightforward exercise to show that a linear operator is bounded if and only if it is continuous. if there is some constant $C$ so that

The infimum of all $C$ for which (2.8) holds is called the norm of $T$, and written $\|T\|$. Using this terminology, (2.7) says $\|H\|\leq 2$.

Our proof of Proposition 2.6 in Section 4 relies on a new “sum of squares identity” involving the operator $H$, proved using (1.1).

Let $h_{1},\ldots,h_{n}$ be $n$ vectors in a hilbert space $\mathcal{H}$. We may associate to these vectors their Gram matrix, i.e., the $n\times n$ matrix $G[h_{1},\ldots,h_{n}]$ defined by

An often used application is the following elegant formula for the distance to the span of $h_{1},\ldots,h_{n}$.

The key step in our proof is the following lemma.

Let $S$ be a subset of the non-negative integers. When is the linear span of the monomials $\{x^{n}:n\in S\}$ dense? The Müntz-Szász Theorem, proved by Müntz [6] and Szász [7], answers this question in both $L^{2}[0,1]$ and $C[0,1]$, the continuous functions on $[0,1]$. The answer is basically the same in both cases, but the constant function $1$ plays a special role in $C[0,1]$, since it cannot be approximated by any polynomial that vanishes at $0$ (which all the other monomials do).

What happens if the approximants come from a set of linear combinations of monomials that is losing as well as gaining members? Fix an increasing sequence $\{N_{n}\}$ of natural numbers and for each $n$ define

For each $n$ let $\rho_{n}$ denote the fraction of the non-negative integers less than or equal to $n+N_{n}$ that do not lie in $S_{n}$, i.e.,

Finally, with this setup, let

We wish to characterize $\mathcal{M}_{\infty}$ (Theorem 5.12 below). We shall follow Müntz’s original proof of Theorem 5.1 [6]. His argument involved an ingenious calculation using Theorem 3.1 and the following venerable formula of Cauchy [3].

We need two more auxiliary results.

We call Proposition 5.5 the Baby Brodskii-Donoghue Theorem because Brodskii and Donoghue independently proved the far deeper fact that the only closed invariant subspaces of $V$ are $L^{2}[s,1]$ [2, 4]. The operator $M_{x}$ has other invariant subspaces. Indeed the ideas in the preceding proof can be adapted to show that the invariant subspaces of $M_{x}$ are the spaces $\{f\in L^{2}[0,1]:f(x)=0{\rm\ a.e.\ on\ }F\}$, where $F$ is any measurable subset of $[0,1]$.

The $C[0,1]$ Müntz-Szász Theorem can be deduced from the $L^{2}$ version. What about the asymptotic version? Let $\mathcal{M}_{n}$ be as in Section 5, and let ${\mathcal{M}}_{\infty}^{\rm unif}$ be

where in this section all distances are with respect to the supremum norm³³3This means $\|f\|_{C[0,1]}=\sup_{0\leq x\leq 1}|f(x)|$..

One way to prove the Weierstraß approximation theorem is to use the Bernstein polynomials. For each $n$, these are the $n+1$ polynomials defined by

Bernstein proved in 1912 [1] that for every continuous function $f\in C[0,1]$, the polynomials

converge uniformly on $[0,1]$ to $f$.

As the lowest order term in $b_{k,n}$ is $x^{k}$, if $f$ vanished on $[0,\rho_{n}]$ and one used the Bernstein formula (6.2) to approximate it, the corresponding polynomial $p_{n+N_{n}+1}$ would lie in the span of $\{x^{n+1},\dots,x^{n+N_{n}+1}\}$ which is in $x\mathcal{M}_{n}\subseteq\mathcal{M}_{n+1}$. So one immediately gets that ${\mathcal{M}}_{\infty}^{\rm unif}$ contains all the continuous functions that vanish on $[0,\rho]$.

This construction seems natural, and could lead one to suspect that ${\mathcal{M}}_{\infty}^{\rm unif}$ should be the functions that vanish on $[0,\rho]$. However, Theorem 6.3 shows this is incorrect.