Multiplication Operators

Suppose that $\mathbf{H}$ is a Hilbert space and that $\left\{ e_j\right\}$ is a family of vectors that constitute an orthonormal basis for $\mathbf{H}$. An operator $A$ is called a diagonal operator if $Ae_j$ is a scalar multiple of $e_j$, say $Ae_j = \alpha_j e_j$, for each $j$; the family $\left\{ \alpha_j\right\}$ may properly be called the diagonal of $A$. It is sometimes convenient to use the symbol $\operatorname {diag}\left\langle \alpha_0, \alpha_1, \alpha_2, \ldots \right\rangle$ to denote the diagonal operator with the indicated diagonal terms.

The set of all bounded sequences $\left\{ \alpha_n\right\}$ of complex numbers is an algebra (pointwise operations), with unit ($\alpha_n = 1$ for all $n$), with a conjugation ($\left\{ \alpha_n\right\} \to \left\{ \alpha_n^{*}\right\}$), and with a norm ($\left\| \left\{ \alpha_n\right\} \right\| = \sup_n \left|\alpha_n\right|$). A bounded sequence $\left\{ \alpha_n\right\}$ is called invertible if it has an inverse in this algebra, i.e., if there exists a bounded sequence $\left\{ \beta_n\right\}$ such that $\alpha_n \beta_n = 1$ for all $n$. A necessary and sufficient condition for this to happen is that $\left\{ \alpha_n\right\}$ be bounded away from $0$, i.e., that there exist a positive number $\delta$ such that $\left|\alpha_n\right| \geq \delta$ for all $n$.

Suppose that $X$ is a measure space with measure $\mu$. If $\varphi$ is a complex-valued bounded (i.e., essentially bounded) measurable function on $X$, then the multiplication operator (or just multiplication, for short) induced by $\varphi$ is the operator $A$ on $L^{2}\left(\mu \right)$ defined by $(Af)(x) = \varphi (x) f(x)$ for all $x$ in $X$. (Here, as elsewhere in measure theory, two functions are identified if they differ on a set of measure zero only. This applies to the bounded $\varphi$'s as well as to the square-integrable $f$'s.) If $X$ is the set of all positive integers and $\mu$ is the counting measure (the measure of every set is the number of elements in it), then multiplication operators reduce to diagonal operators.

The set of all bounded measurable functions (identified modulo sets of measure zero) is an algebra (pointwise operations), with unit ($\varphi (x) = 1$ for all $x$), with a conjugation ($\varphi \mapsto \varphi^{*}$), and with a norm ($\left\| \varphi \right\|_{\infty }$). A bounded measurable function is invertible if it has an inverse in this algebra, i.e., if there exists a bounded measurable function $\psi$ such that $\varphi (x)\psi (x) = 1$ for almost every $x$. A necessary and sufficient condition for this to happen is that $\varphi$ be bounded away from $0$ almost everywhere, i.e., that there exist a positive number $\delta$ such that $\left|\varphi (x)\right| \geq \delta$ for almost every $x$.

The essential range of a function $\varphi$ is, by definition, the set of all complex numbers $\lambda$ such that for each neighborhood $N$ of $\lambda$ the set $\varphi^{-1}(N)$ has positive measure.

If a function $\varphi$ multiplies $L^{2}$ into itself, then $\varphi$ is necessarily bounded (Solution 66), and therefore multiplication by $\varphi$ is necessarily an operator on $L^{2}$. This problem asks whether the analogues of these assertions are true for functional Hilbert spaces.

A function $\varphi$ on a set $X$ is a multiplier of a functional Hilbert space $\mathbf{H}$ over $X$ if $\varphi \cdot f \in \mathbf{H}$ for every $f$ in $\mathbf{H}$. Solution 68 shows that every multiplier is (necessarily) bounded.

To say that every element of $\mathbf{H}$ is a multiplier is the same as to say that $\mathbf{H}$ is closed under multiplication, i.e., that $\mathbf{H}$ is an algebra. The constant function $1$ is a multiplier of every $\mathbf{H}$; hence, to say that every multiplier of $\mathbf{H}$ belongs to $\mathbf{H}$ is the same as to say that $1 \in \mathbf{H}$. If $1 \in \mathbf{H}$, then, of course, the algebra $\mathbf{H}$ has a unit, but trivial examples show that the converse is not true. Thus, the construction of an infinite-dimensional functional Hilbert space that is an algebra with unit (under pointwise functional multiplication) is not quite, but almost, what the problem asks for.