Classifying Lp Membership for Spikes, Tails, and Singularities

Membership Depends on Where the Mass Lies

An \(L^p\) test is a test of the size of \(|f|^p\) over the entire domain, not just of whether \(f\) is bounded or tends to zero. A function can be unbounded near one point and still have a finite \(p\)-th power integral; it can also tend to zero at infinity too slowly to be in \(L^p\). For spikes, the relevant quantity is both their height and the measure of the sets supporting them.

Unless stated otherwise, the domains below carry Lebesgue measure \(m\), and functions are measurable and real-valued except that an infinite value at a single point may be assigned arbitrarily without changing the \(L^p\) class. For \(1\le p<\infty\), membership means \(\int |f|^p\,dm<\infty\). For \(p=\infty\), membership means that \(f\) is essentially bounded: there is a finite \(C\) such that \(|f|\le C\) almost everywhere. Equalities and membership are unchanged by modifying a function on a null set. The tests below concern membership; they do not assert that a particular representative is bounded everywhere.

Definition: In a measure space \((X,\mathcal{M},\mu)\), for \(1\le p<\infty\), \(f\in L^p(\mu)\) exactly when \(f\) is measurable and \(\int_X |f|^p\,d\mu<\infty\). Also, \(f\in L^\infty(\mu)\) exactly when some \(C<\infty\) satisfies \(|f|\le C\) \(\mu\)-almost everywhere. On a finite-measure set \(E\), a constant function \(c\) satisfies \(\int_E|c|^p\,d\mu=|c|^p\mu(E)\); on an infinite-measure set this integral is infinite if \(c\ne0\).

Spikes: Height Must Be Paired With Support Size

Suppose \(E_1,E_2,\ldots\) are pairwise disjoint measurable sets in a measure space \((X,\mathcal{M},\mu)\), and let \(a_k\) be finite real numbers. Define \(f=\sum_{k=1}^\infty a_k\chi_{E_k}\), setting \(f=0\) off their union. At each point at most one summand is nonzero. The exact membership test is a weighted series, not a test of the heights alone.

Proposition: For \(1\le p<\infty\), with \(0\cdot\infty\) interpreted as zero, the disjoint-spike function above satisfies \(\int_X|f|^p\,d\mu=\sum_{k=1}^\infty |a_k|^p\mu(E_k)\). Thus \(f\in L^p(\mu)\) if and only if this series is finite. It belongs to \(L^\infty(\mu)\) if and only if the values \(|a_k|\) on sets \(E_k\) of positive measure have a finite supremum.

Proof. For each positive integer \(N\), disjointness gives \(\left|\sum_{k=1}^N a_k\chi_{E_k}\right|^p=\sum_{k=1}^N|a_k|^p\chi_{E_k}\) at every point. Integrating this finite sum gives \(\sum_{k=1}^N|a_k|^p\mu(E_k)\), by additivity of the integral for nonnegative measurable functions. As \(N\) increases, the right-hand side increases to the stated series, while the functions on the left increase pointwise to \(|f|^p\). The Monotone Convergence Theorem applies: all these functions are measurable and nonnegative, and the sequence is increasing. It gives the asserted equality, including when the value is infinite. For the \(L^\infty\) assertion, if the positive-measure heights are bounded by \(C\), then \(|f|\le C\) outside the union of any zero-measure \(E_k\), a countable union of null sets. Conversely, if those heights are unbounded, for every finite \(C\) some \(E_k\) of positive measure has \(|a_k|>C\); therefore \(|f|\le C\) cannot hold almost everywhere. \(\square\)

Worked Example: A Sequence of Narrowing Spikes

In \((0,1)\) with Lebesgue measure, define \(E_k=(2^{-k},\,2^{-k}+2^{-k-2})\) and \(f=\sum_{k=1}^\infty 2^{k/4}\chi_{E_k}\). The intervals are disjoint: the right endpoint of \(E_{k+1}\) is \((5/8)2^{-k}\), which is less than the left endpoint \(2^{-k}\) of \(E_k\). Each interval has measure \(2^{-k-2}\). For \(1\le p<\infty\), the spike criterion gives

\int_0^1 |f|^p\,dm =\sum_{k=1}^\infty 2^{kp/4}2^{-k-2} =\frac14\sum_{k=1}^\infty \left(2^{p/4-1}\right)^k.

This geometric series converges exactly when its ratio \(2^{p/4-1}\) is less than \(1\), which is equivalent to \(p<4\). At \(p=4\), each term is \(1/4\), so the sum diverges; for \(p>4\), the terms do not tend to zero. At \(p=2\), the ratio is \(2^{-1/2}=1/\sqrt2\), so the integral is \(\frac14\frac{1/\sqrt2}{1-1/\sqrt2}=(\sqrt2+1)/4\). Thus this example is in \(L^p(0,1)\) exactly for \(1\le p<4\). The heights \(2^{k/4}\) are unbounded on intervals of positive measure, so it is not in \(L^\infty\). Check: at \(p=2\), the computed integral is positive and finite; its geometric-series ratio \(1/\sqrt2\) is indeed less than \(1\).

For a single indicator, the same calculation reduces to \(\int|\chi_E|^p\,d\mu=\mu(E)\) for every finite \(p\): its height is \(1\), regardless of \(p\). Thus \(\chi_E\in L^p(\mu)\) exactly when \(\mu(E)<\infty\), for every finite \(p\), and \(\chi_E\in L^\infty(\mu)\) for every measurable \(E\). This separates two issues: infinite support measure rules out finite-\(p\) membership for a nonzero constant-height indicator, but does not rule out essential boundedness.

Power Singularities and Polynomial Tails

On a finite interval, a power singularity is tested near the point where it blows up. On an infinite interval, a decaying power is tested in the tail. The thresholds differ because the relevant endpoint of integration differs.

Theorem: In \(((0,1),\mathcal{L},m)\), \(x^{-\alpha}\) belongs to \(L^p\), for \(1\le p<\infty\) and \(\alpha\ge0\), exactly when \(\alpha p<1\). In \(([0,\infty),\mathcal{L},m)\), \((1+x)^{-\beta}\) belongs to \(L^p\), for \(1\le p<\infty\) and \(\beta\ge0\), exactly when \(\beta p>1\).

Proof. For the singularity, the integral to test is \(\int_0^1x^{-\alpha p}\,dx\). If \(\alpha p<1\), integration on \((\varepsilon,1)\) gives \((1-\varepsilon^{1-\alpha p})/(1-\alpha p)\), which tends to \(1/(1-\alpha p)\) as \(\varepsilon\downarrow0\). If \(\alpha p=1\), that integral is \(-\ln\varepsilon\), which tends to infinity. If \(\alpha p>1\), it is \((\varepsilon^{1-\alpha p}-1)/(\alpha p-1)\), which tends to infinity because \(1-\alpha p<0\). These limits determine the nonnegative Lebesgue integral, so the first condition is necessary and sufficient. For the tail, substitute \(u=1+x\); the integral becomes \(\int_1^\infty u^{-\beta p}\,du\). If \(\beta p>1\), integration up to \(R\) gives \((1-R^{1-\beta p})/(\beta p-1)\), tending to \(1/(\beta p-1)\). If \(\beta p=1\), it gives \(\ln R\), tending to infinity. If \(\beta p<1\), it gives \((R^{1-\beta p}-1)/(1-\beta p)\), also tending to infinity. Hence the second condition is necessary and sufficient. \(\square\)

Worked Example: A Singularity at Zero and a Power Tail

On \((\mathbb{R},\mathcal{L},m)\), set \(f(0)=0\), let \(f(x)=|x|^{-1/4}\) when \(0<|x|<1\), and let \(f(x)=|x|^{-2}\) when \(|x|\ge1\). For \(1\le p<\infty\), the singular part is integrable to the \(p\)-th power exactly when \(p/4<1\), or \(p<4\). The tail is integrable exactly when \(2p>1\), which holds for every \(p\ge1\). More explicitly, when \(p<4\), symmetry and the power integrals give

\int_{\mathbb{R}}|f|^p\,dm =2\int_0^1x^{-p/4}\,dx+2\int_1^\infty x^{-2p}\,dx =\frac{2}{1-p/4}+\frac{2}{2p-1}.

At \(p=2\), the first contribution is \(2/(1-1/2)=4\), and the second is \(2/(4-1)=2/3\). The total is \(4+2/3=14/3\), so \(f\in L^2(\mathbb{R})\). At \(p=4\), the singular integral diverges logarithmically, so \(f\notin L^4\), even though its tail passes the test. It is not essentially bounded: for every \(C>0\), the set \(0<|x|<C^{-4}\) (when \(C^{-4}\le1\)) has positive measure and \(f(x)>C\). Check: at \(p=2\), both required inequalities hold: \(1/2<1\) near zero and \(4>1\) at infinity; the two computed contributions sum to \(14/3\).

Exponential Tails and Finite Domains

Exponential decay on an infinite interval passes every finite-\(p\) test: raising the function to the \(p\)-th power only changes the decay rate. This is different from polynomial decay, whose exponent must cross a threshold. On a finite interval, a continuous exponential such as \(e^{cx}\) is bounded and has finite \(p\)-th power integral for each finite \(p\).

Worked Example: Exponential Versus Polynomial Decay

On \(([0,\infty),\mathcal{L},m)\), take \(g(x)=e^{-2x}\). For \(1\le p<\infty\), direct integration gives

\int_0^\infty |g(x)|^p\,dm =\int_0^\infty e^{-2px}\,dx =\frac{1}{2p}, \qquad \|g\|_p=\left(\frac{1}{2p}\right)^{1/p}.

The integral is finite for every such \(p\). For \(p=\infty\), \(0<g(x)\le1\) and values arbitrarily close to \(1\) occur on sets of positive measure near zero, so \(\|g\|_\infty=1\). As a numerical check, at \(p=2\) the integral is \(1/4\) and the norm is \(1/2\); squaring that norm gives the integral again. By contrast, \(h(x)=(1+x)^{-1}\) has \(p\)-th power integral \(\int_1^\infty u^{-p}\,du\), finite exactly when \(p>1\). At \(p=1\) it diverges as \(\ln R\) when the upper limit \(R\) tends to infinity, whereas \(g\) has integral \(1/2\). The exponential test uses the actual rate \(2p>0\), not a power-tail threshold.

Finite measure alone does not make every measurable function integrable: the singularity \(x^{-\alpha}\) on \((0,1)\) can fail the test. Conversely, an unbounded domain does not make every function nonintegrable: the exponential example is in every finite \(L^p\). Always identify the location and type of the possible obstruction before applying an integral test.

A Common Endpoint Error

A frequent mistake is to apply the finite-interval power rule to an infinite tail. For example, one might write \(\int_0^\infty(1+x)^{-p}\,dx=1/(1-p)\) by treating the integral as if its upper endpoint were \(1\). This expression does not evaluate the infinite-interval integral. The correct truncation is

\int_0^R(1+x)^{-p}\,dx = \begin{cases} \dfrac{(1+R)^{1-p}-1}{1-p},&p\ne1,\\[6pt] \ln(1+R),&p=1. \end{cases}

For \(p>1\), the first expression tends to \(1/(p-1)\); for \(p<1\), it tends to infinity; and for \(p=1\), the logarithm tends to infinity. The correction is to take the limit at the actual endpoint of the domain, including the critical case, rather than merely substitute an exponent into a formula from a different interval.

Check Your Understanding

For each question, identify the measure and domain, then test the integral of the \(p\)-th power; at \(p=\infty\), test essential boundedness instead.

For disjoint measurable \(E_k\) with \(\mu(E_k)=3^{-k}\) and heights \(a_k=2^k\), for which finite \(p\ge1\) is \(\sum_{k\ge1}a_k\chi_{E_k}\) in \(L^p(\mu)\)?
For which finite \(p\ge1\) does \(x^{-1/3}\) belong to \(L^p((0,1),m)\)? What happens at the critical exponent?
For which finite \(p\ge1\) does \((1+x)^{-1/2}\) belong to \(L^p(([0,\infty),m))\)?
Compute \(\int_0^\infty e^{-3px}\,dx\) for fixed \(p\ge1\), and state whether \(e^{-3x}\) is essentially bounded.
Explain why \(\chi_{[0,\infty)}\) is in \(L^\infty([0,\infty),m)\) but not in \(L^p([0,\infty),m)\) for finite \(p\).