Introduction

We consider the integral operator $T$ on $L^2[0, \infty)$ defined by:

$$(Tf)(x) = \int_0^\infty J_0(|x - y|)f(y)dy$$

where $J_0$ is the Bessel function of the first kind of order zero. We aim to prove that $T$ is compact using the concept of Bochner V-boundedness.

Preliminaries

Lemma

For $x \neq 0$, $|J_0(x)| \leq \min(1, \sqrt{2/(\pi|x|)})$.

Proof

This follows from the asymptotic behavior of $J_0(x)$ and its maximum value of 1 at $x = 0$.

Lemma \label{lem:integral}

For any $\epsilon > 0$, the integral

$$\int_0^\infty (J_0(x) / (\epsilon + x))^2 dx$$

converges and is equal to:

$$\frac{1}{\pi^{3/2} \epsilon^2} G_{2, 4}^{2, 1} \left(\epsilon^2 \left| \begin{array}{c} 1, 1 \ \frac{3}{2}, 1, \frac{1}{2}, \frac{1}{2} \end{array} \right. \right)$$

where $G$ is the Meijer G-function. Moreover, $\epsilon = 0$ is the abscissa of convergence for this integral.

Bochner V-boundedness

Definition

An integral operator $T$ with kernel $K(x, y)$ is Bochner V-bounded if there exists a positive function $V(x)$ such that:

$$\int_0^\infty \sup_{y \geq 0} |K(x, y) / V(y)|^2 V(x)^2 dx < \infty$$

Theorem

If $T$ is Bochner V-bounded on $L^2[0, \infty)$, then $T$ is compact.

Proof

Let ${e_n}$ be an orthonormal basis for $L^2[0, \infty)$. Define the finite rank operators:

$$T_N f = \sum_{n=1}^N \langle Tf, e_n \rangle e_n$$

We will show that $T_N \to T$ in operator norm. Let $f \in L^2[0, \infty)$ with $|f| \leq 1$. Then:

$$|(T-T_N)f|^2 = \sum_{n>N} |\langle Tf, e_n \rangle|^2 = \sum_{n>N} \left|\int_0^\infty \int_0^\infty K(x, y)f(y)e_n(x) dy dx\right|^2$$

$$\leq \sum_{n>N} \left(\int_0^\infty \int_0^\infty |K(x, y)/V(y)| |V(y)f(y)| |e_n(x)| dy dx\right)^2$$

$$\leq \sum_{n>N} \left(\int_0^\infty \sup_{y \geq 0} |K(x, y) / V(y)| |V f| |e_n(x)| dx\right)^2$$

$$\leq |V f|^2 \sum_{n>N} \int_0^\infty \sup_{y \geq 0} |K(x, y) / V(y)|^2 |e_n(x)|^2 dx$$

$$= |V f|^2 \int_0^\infty \sup_{y \geq 0} |K(x, y) / V(y)|^2 \sum_{n>N} |e_n(x)|^2 dx$$

By Parseval's identity, for any fixed $x$, $\sum_{n=1}^\infty |e_n(x)|^2 = 1$ almost everywhere. Therefore, $\sum_{n>N} |e_n(x)|^2$ represents the tail of this series and converges to zero pointwise as $N \to \infty$ for almost every $x$. This sum is also bounded by 1 for all $N$ and $x$.

By the dominated convergence theorem and the Bochner V-boundedness condition, $|(T-T_N)f|^2 \to 0$ as $N \to \infty$, uniformly for $|f| \leq 1$. Thus, $T$ is the limit of finite rank operators and is therefore compact.

Proof of Compactness

We will show that $T$ is Bochner V-bounded with $V(x) = \epsilon + x$ for any $\epsilon > 0$.

Theorem

The operator $T$ defined by:

$$(Tf)(x) = \int_0^\infty J_0(|x-y|)f(y)dy$$

is compact on $L^2[0, \infty)$.

Proof

We need to show:

$$\int_0^\infty \sup_{y \geq 0} |J_0(|x-y|)/(\epsilon+y)|^2 (\epsilon+x)^2 dx < \infty$$

Using the result from Lemma \ref{lem:integral}, we have for any $\epsilon > 0$:

$$\int_0^\infty |J_0(x)/(\epsilon+x)|^2 dx = \frac{1}{\pi^{3/2} \epsilon^2} G_{2, 4}^{2, 1} \left(\epsilon^2 \left| \begin{array}{c} 1, 1 \ \frac{3}{2}, 1, \frac{1}{2}, \frac{1}{2} \end{array} \right. \right) < \infty$$

Now, observe that:

$$\sup_{y \geq 0} |J_0(|x-y|)/(\epsilon+y)|^2 \leq |J_0(x)/(\epsilon+x)|^2$$

Therefore,

$$\begin{align*} \int_0^\infty \sup_{y\geq0} |J_0(|x-y|)/(\epsilon+y)|^2 (\epsilon+x)^2 dx &\leq \int_0^\infty |J_0(x)/(\epsilon+x)|^2 (\epsilon+x)^2 dx < \infty \end{align*}$$

It's crucial to note that $\int_0^\infty |J_0(x)|^2 dx = \infty$, which means it diverges. However, the integral $\int_0^\infty |J_0(x)/(\epsilon+x)|^2 dx$ converges for any $\epsilon > 0$, including in the limit as $\epsilon \to 0$. This convergence, with the $(\epsilon+x)$ term in the denominator, is the key to the effectiveness of our V-boundedness approach.

This proves that $T$ is Bochner V-bounded with $V(x) = \epsilon + x$ for any $\epsilon > 0$, and therefore compact.

Remark

The choice of $V(x) = \epsilon + x$ for any $\epsilon > 0$ is crucial to prove compactness. The fact that the integral $\int_0^\infty |J_0(x)/(\epsilon+x)|^2 dx$ converges even as $\epsilon \to 0$ shows that this is the optimal family of functions for establishing the Bochner V-boundedness of $T$. This convergence in the limit is key to understanding why this approach works where simpler methods fail.