Power-Function Membership Thresholds on Rn

Why Location Changes the Threshold

A power of the distance to a point can be integrable near that point and nonintegrable at infinity for different reasons. In \(\mathbb{R}^n\), the volume of a thin spherical shell of radius \(r\) is proportional to \(r^{n-1}\,dr\). Multiplying this volume factor by the \(p\)-th power of a function behaving like \(r^{-a}\) produces the decisive radial power \(r^{n-1-ap}\). Its integral has different convergence conditions at zero and at infinity.

Throughout, \(n\ge1\), and \(m\) denotes Lebesgue measure on \(\mathbb{R}^n\). For \(0<p<\infty\), a measurable real- or complex-valued function \(f\) belongs to \(L^p(\mathbb{R}^n)\) when \(\int_{\mathbb{R}^n}|f|^p\,dm<\infty\); functions that agree \(m\)-a.e. represent the same element. Membership in \(L^\infty(\mathbb{R}^n)\) means that \(|f|\) is bounded outside a Lebesgue-null set. The power parameter \(a\) below is any real number, and a value assigned at the single point \(x=0\) does not affect finite-\(p\) integrability.

Definition (Power behavior near a point or at infinity): We say that \(f\) has power behavior \(r^{-a}\) near the origin if there are constants \(c,C,r_0>0\) such that \(c|x|^{-a}\le |f(x)|\le C|x|^{-a}\) for \(m\)-a.e. \(x\) with \(0<|x|<r_0\). Power behavior \(r^{-a}\) at infinity means the same pair of inequalities holds for \(m\)-a.e. \(x\) with \(|x|>R\), for some \(c,C,R>0\). Two-sided comparison is important: an upper bound proves integrability, while a lower bound proves divergence.

The Radial Test and the Two Thresholds

For a nonnegative radial function \(g(x)=G(|x|)\), the polar-coordinate change of variables gives \(\int_{\mathbb{R}^n}g\,dm=\omega_{n-1}\int_0^\infty G(r)r^{n-1}\,dr\), whenever either side is defined, including the value \(+\infty\). Here \(\omega_{n-1}\) is the positive finite surface area of the unit sphere in \(\mathbb{R}^n\) (in dimension one it equals \(2\)). The factor \(r^{n-1}\) is the Jacobian of polar coordinates; the formula also applies on radial subregions such as balls, annuli, and their complements. Applying it to \(|x|^{-ap}\) reduces the question to the one-dimensional integral of \(r^{n-1-ap}\).

Theorem (Power thresholds): Let \(0<p<\infty\). If \(f\) has power behavior \(|x|^{-a}\) near the origin, then \(\int_{\{|x|<r_0\}}|f|^p\,dm<\infty\) if and only if \(ap<n\). If \(f\) has power behavior \(|x|^{-a}\) at infinity, then \(\int_{\{|x|>R\}}|f|^p\,dm<\infty\) if and only if \(ap>n\). At equality \(ap=n\), the corresponding integral diverges logarithmically.

Proof. By the two-sided comparison in the definition, the integral of \(|f|^p\) on the region under consideration is finite exactly when the integral of \(|x|^{-ap}\) there is finite. Indeed, raising the comparison to the positive power \(p\) bounds \(|f|^p\) above and below by positive constant multiples of \(|x|^{-ap}\). The polar-coordinate formula reduces the test near zero to \(\omega_{n-1}\int_0^{r_0}r^q\,dr\), and the tail test to \(\omega_{n-1}\int_R^\infty r^q\,dr\), where \(q=n-1-ap\).

If \(q\ne-1\), an antiderivative is \(r^{q+1}/(q+1)\). At zero, \(\int_\varepsilon^{r_0}r^q\,dr=(r_0^{q+1}-\varepsilon^{q+1})/(q+1)\); as \(\varepsilon\downarrow0\), this has a finite limit precisely when \(q+1>0\), or \(ap<n\). If \(q=-1\), the truncated integral is \(\log(r_0/\varepsilon)\), which tends to infinity. At infinity, \(\int_R^T r^q\,dr=(T^{q+1}-R^{q+1})/(q+1)\); as \(T\to\infty\), this has a finite limit precisely when \(q+1<0\), or \(ap>n\). When \(q=-1\), it is \(\log(T/R)\), which diverges. The positive finite factor \(\omega_{n-1}\) and the comparison constants do not alter finiteness. This proves both equivalences and the critical cases. \(\square\)

The inequalities point in opposite directions because the endpoints of integration differ. Near zero, a positive power \(r^q\) can be integrable; at infinity it is the negative powers that decay sufficiently fast. A singularity at any fixed point \(x_0\in\mathbb{R}^n\) has the same local test: translation \(y=x-x_0\) preserves Lebesgue measure and changes \(|x-x_0|^{-a}\) into \(|y|^{-a}\).

Worked Example: A Locally Integrable Singularity in Three Dimensions

Consider \(f(x)=|x|^{-1}\) on the punctured unit ball \(\{x\in\mathbb{R}^3:0<|x|<1\}\), with any finite value assigned at \(x=0\). Test membership there for \(p=2\) and compute the integral.

Here \(n=3\), \(a=1\), and \(ap=1\cdot2=2<3=n\), so the theorem predicts finite integral. In three dimensions \(\omega_2=4\pi\), and the radial formula gives

\int_{\{|x|<1\}}|f(x)|^2\,dm =4\pi\int_0^1 r^{-2}r^2\,dr =4\pi\int_0^1 1\,dr =4\pi.

Thus \(f\) belongs to \(L^2\) on this ball and its \(L^2\) norm there is \(\sqrt{4\pi}\). Check: integrating first over the punctured ball with inner radius \(\varepsilon>0\) gives \(4\pi\int_\varepsilon^1 1\,dr=4\pi(1-\varepsilon)\), which tends to \(4\pi\) as \(\varepsilon\downarrow0\), agreeing with the improper radial integral.

Worked Example: A Tail at the Critical Exponent

In \(\mathbb{R}^2\), let \(f(x)=|x|^{-1}\) for \(|x|>1\). Test whether \(f\) is integrable on this tail, and compute the integral up to radius \(T>1\).

Here \(n=2\), \(a=1\), and \(p=1\), so \(ap=1<2\). The tail condition requires \(ap>n\), and therefore predicts divergence. Since \(\omega_1=2\pi\), the truncated integral is

\int_{\{1<|x|<T\}}|f(x)|\,dm =2\pi\int_1^T r^{-1}r\,dr =2\pi\int_1^T 1\,dr =2\pi(T-1).

As \(T\to\infty\), this tends to \(+\infty\), so the tail is not in \(L^1\). Check: the radial integrand is exactly \(1\), so every additional unit of radial length adds \(2\pi\) to the truncated integral; the expression \(2\pi(T-1)\) has exactly that growth.

Worked Example: Equality Produces Logarithmic Divergence

In \(\mathbb{R}^2\), let \(f(x)=|x|^{-1}\) on the punctured unit ball and test \(L^2\) membership. Since \(a=1\), \(p=2\), and \(ap=2=n\), this is the critical case, not either strict inequality in the theorem.

For \(0<\varepsilon<1\), integrate on the annulus \(\varepsilon<|x|<1\):

\int_{\{\varepsilon<|x|<1\}}|f(x)|^2\,dm =2\pi\int_\varepsilon^1 r^{-2}r\,dr =2\pi\int_\varepsilon^1\frac{1}{r}\,dr =2\pi\log(1/\varepsilon).

This tends to \(+\infty\) as \(\varepsilon\downarrow0\), so \(f\) is not in \(L^2\) near the origin. Check: taking \(\varepsilon=e^{-k}\), where \(k\) is any positive integer, gives the truncated value \(2\pi k\), which grows without bound. The logarithm is the exact boundary behavior when \(ap=n\).

Combining Local and Tail Behavior

For a function defined on all of \(\mathbb{R}^n\), its \(L^p\) integral can be split into the integral on a ball and the integral outside that ball. Thus behavior at zero alone does not determine global membership if the function also has a tail; both regions must pass their own tests. On the intermediate annulus, a function bounded there has finite integral because that annulus has finite measure. More generally, the theorem applies separately wherever the stated two-sided power comparisons hold.

Worked Example: Different Powers Near Zero and at Infinity

Define a radial function on \(\mathbb{R}^3\) by \(f(x)=|x|^{-1}\) when \(0<|x|<1\), \(f(x)=|x|^{-2}\) when \(|x|\ge1\), and assign any finite value at zero. Determine whether \(f\in L^2(\mathbb{R}^3)\) and compute its squared norm.

Near zero, \(a=1\) gives \(ap=2<3\), so the local integral is finite. At infinity, \(a=2\) gives \(ap=4>3\), so the tail is finite. The function is exactly given by these powers on the respective regions, and \(\omega_2=4\pi\). Hence

\begin{aligned} \int_{\mathbb{R}^3}|f|^2\,dm &=4\pi\int_0^1 r^{-2}r^2\,dr +4\pi\int_1^\infty r^{-4}r^2\,dr\\ &=4\pi\int_0^1 1\,dr +4\pi\int_1^\infty r^{-2}\,dr\\ &=4\pi+4\pi=8\pi. \end{aligned}

Therefore \(f\in L^2(\mathbb{R}^3)\) and \(\|f\|_2=\sqrt{8\pi}\). Check: replacing the infinite endpoint by \(T>1\) gives \(4\pi+4\pi(1-1/T)=4\pi(2-1/T)\), which tends to \(8\pi\) as \(T\to\infty\), the same value as the improper integral.

Corollary (A pure power on all of \(\mathbb{R}^n\)): For \(0<p<\infty\), the function \(|x|^{-a}\) is not in \(L^p(\mathbb{R}^n)\) for any real \(a\). Indeed, local integrability requires \(ap<n\), while tail integrability requires \(ap>n\), and both cannot hold. This does not prevent a power restricted to a ball or to an exterior region from belonging to \(L^p\) on that region.

For \(p=\infty\), the question is essential boundedness rather than finiteness of the integral. Near zero, \(|x|^{-a}\) is essentially bounded exactly when \(a\le0\); on an exterior region it is essentially bounded exactly when \(a\ge0\). For example, if \(a>0\), its values on every sufficiently small punctured ball exceed any fixed bound on a set of positive measure. If \(a<0\), its values exceed any fixed bound sufficiently far out, again on sets of positive measure. When \(a=0\), the function is constant away from the origin. Consequently, a pure power on all of \(\mathbb{R}^n\) is in \(L^\infty\) exactly when \(a=0\), up to its value at the origin.

A Common Mistake: Reversing the Tail Inequality

A tempting but incorrect rule is to use \(ap<n\) for tail integrability as well as for a singularity at zero. The example in \(\mathbb{R}^2\) with \(a=1\) and \(p=1\) exposes the error: the mistaken test gives \(ap=1<2=n\) and would predict an integrable tail, but the actual truncated tail integral is \(2\pi(T-1)\), which diverges. The correct condition at infinity is \(ap>n\). At the origin it is \(ap<n\); at the shared boundary \(ap=n\), both radial tests diverge logarithmically.

When the singularity is centered at \(x_0\) rather than at the origin, use \(|x-x_0|\) and apply the local threshold without changing \(n\), \(a\), or \(p\). The location changes which region is being tested, not the dimension-dependent calculation. The threshold gives a precise membership test for two-sided power behavior; if a function has only an upper comparison, it yields a sufficient condition for integrability, but not a necessary one.

Check Your Understanding

For each question, identify whether the issue is local integrability, tail integrability, or essential boundedness, and check the relevant strict inequality.

In \(\mathbb{R}^4\), for which finite \(p>0\) is \(|x|^{-1}\) integrable on the unit ball?
In \(\mathbb{R}^3\), for which finite \(p>0\) is \(|x|^{-2}\) integrable on \(\{|x|>1\}\)?
What happens to the local \(L^p\) integral when \(ap=n\), and what is the radial integrand in that case?
For a function behaving like \(|x-x_0|^{-a}\) near a fixed point \(x_0\), does moving \(x_0\) change the threshold? Explain.
For which real \(a\) is \(|x|^{-a}\) essentially bounded on \(\{|x|>1\}\)?