关于希尔伯特空间无界稠定算子位置算子定义域的疑问

Great question about this foundational unbounded operator example from Reed & Simon! Let's break down how to confirm that $D(T)$ is the largest possible domain where the position operator maps into $L^2(\mathbb{R})$, plus the underlying theory.

What "Maximal Domain" Means Here #

First, let's clarify the definition: A domain $D'$ for $T$ is maximal if any other domain $D''$ where $T: D'' \to L^2(\mathbb{R})$ (i.e., $T\varphi \in L^2(\mathbb{R})$ for all $\varphi \in D''$) must satisfy $D'' \subseteq D'$. So we need to show two core things:

1. $T$ maps every element of $D(T)$ into $L^2(\mathbb{R})$.
2. No larger domain than $D(T)$ can have this property.

Step 1: Confirm $D(T)$ is a Valid Domain #

By definition, $D(T) = { \varphi \in L^2(\mathbb{R}) : \int_\mathbb{R} x^{2|\varphi(x)|}2 {\rm d}x < \infty }$. For any $\varphi \in D(T)$, compute the $L^2$-norm of $T\varphi$:
$$
|T\varphi|{L^2}2 = \int\mathbb{R} |T\varphi(x)|^2 dx = \int_\mathbb{R} |x\varphi(x)|^2 dx = \int_\mathbb{R} x^{2|\varphi(x)|}2 dx
$$
Since this integral is finite by the definition of $D(T)$, $T\varphi$ is clearly an element of $L^2(\mathbb{R})$. So $T$ is well-defined on $D(T)$.

Step 2: Prove $D(T)$ is Maximal #

Suppose there exists a domain $D' \supseteq D(T)$ such that $T: D' \to L^2(\mathbb{R})$. Take any $\varphi \in D'$—by assumption, $T\varphi \in L^2(\mathbb{R})$, which means:
$$
\int_\mathbb{R} x^{2|\varphi(x)|}2 dx = |T\varphi|_{L^2}2 < \infty
$$
But this is exactly the condition that defines membership in $D(T)$! So $\varphi$ must be in $D(T)$. Since this holds for every $\varphi \in D'$, we have $D' \subseteq D(T)$. Combining with our initial assumption that $D' \supseteq D(T)$, we get $D' = D(T)$. That's the maximality proof wrapped up.

Underlying Theoretical Basis #

This result is a special case of a general fact about multiplication operators on $L^2$ spaces:

• For any measurable function $f: \mathbb{R} \to \mathbb{C}$, the multiplication operator $M_f: \varphi \mapsto f\varphi$ has a maximal domain given by:
  $$
  D(M_f) = { \varphi \in L^2(\mathbb{R}) : f\varphi \in L^2(\mathbb{R}) }
  $$
  The position operator is just the multiplication operator where $f(x) = x$. The maximality follows directly from the definition of $L^2$: an element is in $L^2$ iff its square is integrable. For $M_f\varphi$ to be in $L^2$, $|f\varphi|^2 = |f|^2|\varphi|2$ must be integrable—this is a direct translation of the domain condition.

Also, note that $D(T)$ is dense in $L^2(\mathbb{R})$ (which is why we call it a densely defined operator): spaces like compactly supported smooth functions $C_c^\infty(\mathbb{R})$ are contained in $D(T)$, and $C_c^\infty(\mathbb{R})$ is dense in $L^2(\mathbb{R})$ (a standard result in measure theory and functional analysis).

内容的提问来源于stack exchange，提问作者Rodrigo Dias