Strict Lp Inclusions from Power Singularities

Power Singularities and Integrability Near Zero

A function can be finite at every point of $(0,1)$ and still fail to be integrable because it grows without bound near the missing endpoint $0$. The powers $x^{-a}$ give a precise test of how much growth $L^p$ can accommodate. The resulting threshold also supplies a direct way to see that the finite-measure inclusion from $L^q$ into $L^p$, when $p<q$, can be strict.

Throughout, the measure space is $((0,1),\mathcal{L}_{(0,1)},m)$, where $\mathcal{L}_{(0,1)}$ is the Lebesgue sigma-algebra restricted to $(0,1)$, and $m$ is Lebesgue measure. This is a complete finite measure space with $m((0,1))=1$. For $a\in\mathbb{R}$, let $f_a(x)=x^{-a}$ for $0<x<1$. Each $f_a$ is a finite, real-valued measurable function at every point of this interval.

Definition: For $1\le p<\infty$, $L^p(0,1)$ consists of the almost-everywhere equivalence classes of measurable real-valued functions $f$ for which $\int_0^1 |f|^p\,dm<\infty$; the norm is $\|f\|_p=(\int_0^1|f|^p\,dm)^{1/p}$. The space $L^\infty(0,1)$ consists of classes that are essentially bounded: there is a finite $C$ such that $|f|\le C$ almost everywhere. Its norm is the essential supremum $\|f\|_\infty$. Membership and norm statements concern equivalence classes, so changing values on a null set has no effect.

The Exact Membership Threshold

Theorem: For $a\in\mathbb{R}$ and $1\le p<\infty$, $f_a\in L^p(0,1)$ if and only if $ap<1$. When this condition holds, $$ \|f_a\|_p=\left(\frac{1}{1-ap}\right)^{1/p}. $$ For $p=\infty$, $f_a\in L^\infty(0,1)$ if and only if $a\le 0$; in that case $\|f_a\|_\infty=1$.

Proof. Fix $1\le p<\infty$. Since $f_a$ is positive, $|f_a(x)|^p=x^{-ap}$. The nonnegative Lebesgue integral can be computed by first integrating over $(\delta,1)$ and then letting $\delta\downarrow0$. If $ap\ne1$, direct integration gives $$ \int_\delta^1 x^{-ap}\,dx=\frac{1-\delta^{1-ap}}{1-ap}. $$ If $ap<1$, then $1-ap>0$ and $\delta^{1-ap}\to0$, so the integral over $(0,1)$ is $1/(1-ap)<\infty$. If $ap>1$, then $1-ap<0$, so $\delta^{1-ap}\to\infty$ and the displayed integrals tend to $+\infty$. At the critical value $ap=1$, direct integration instead gives $$ \int_\delta^1 \frac{1}{x}\,dx=-\ln\delta\longrightarrow+\infty. $$ Thus membership holds exactly when $ap<1$, and taking the $p$-th root of the finite integral gives the stated norm. For $p=\infty$, if $a\le0$, then $f_a(x)=x^{-a}\le1$ on $(0,1)$. If $a=0$, it is identically $1$. If $a<0$, write $b=-a>0$, so $f_a(x)=x^b$. For every $c<1$ with $c\ge0$, the set $\{x:x^b>c\}$ has positive measure: for $0<c<1$ it is $(c^{1/b},1)$, of measure $1-c^{1/b}>0$, and for $c=0$ it is all of $(0,1)$. Hence the essential supremum is $1$. If $a>0$, then for every $K\ge1$, $\{x:f_a(x)>K\}=(0,K^{-1/a})$, which has positive measure $K^{-1/a}$. No finite bound holds almost everywhere, so $f_a\notin L^\infty(0,1)$. This proves both cases. $\square$

The condition is strict: the critical exponent is not included. For finite $p$, the threshold is $a=1/p$; at that value the $p$-th power is $1/x$, whose integral diverges logarithmically. For $a<1/p$, the exact norm formula also records how the norm grows as $a$ approaches the threshold from below.

Worked Example: A Function in L2 but Not L4

Consider $f(x)=x^{-1/4}$ on $(0,1)$. For $p=2$, the product of the exponent and $p$ is $(1/4)2=1/2<1$. The theorem gives $$ \int_0^1 |f(x)|^2\,dm(x)=\int_0^1 x^{-1/2}\,dx=\frac{1}{1-1/2}=2, \qquad \|f\|_2=\sqrt{2}. $$ For $p=4$, the product is $(1/4)4=1$, so the critical case applies. More explicitly, for $0<\delta<1$, $$ \int_\delta^1 |f(x)|^4\,dx=\int_\delta^1\frac{1}{x}\,dx=-\ln\delta\longrightarrow+\infty. $$ Thus $f\in L^2(0,1)$ but $f\notin L^4(0,1)$. Check: the square has exponent $-1/2$, and its antiderivative is $2\sqrt{x}$, giving $2-0=2$; the fourth power has exponent $-1$, producing the divergent logarithm.

Worked Example: Integrable but Not Essentially Bounded

Let $g(x)=x^{-1/2}$. Its $L^1$ integral is $$ \int_0^1 g(x)\,dx=\frac{1}{1-1/2}=2, $$ so $g\in L^1(0,1)$ and $\|g\|_1=2$. For every $K\ge1$, the strict tail set is $$ \{x\in(0,1):|g(x)|>K\}=(0,K^{-2}), $$ which has measure $K^{-2}>0$. Consequently $g$ is not essentially bounded and does not belong to $L^\infty(0,1)$. Check: integrating only from $\delta$ to $1$ gives $2(1-\sqrt{\delta})$, which tends to $2$ as $\delta\downarrow0$; meanwhile, the positive-measure tail at every finite $K$ rules out an almost-everywhere bound.

Worked Example: A Negative Exponent and the L-infinity Norm

Take $h(x)=x^{1/2}=x^{-(-1/2)}$, so the power parameter is $a=-1/2$. For $p=3$, $ap=(-1/2)3=-3/2<1$. Directly, $$ \int_0^1 |h(x)|^3\,dx=\int_0^1 x^{3/2}\,dx=\frac{1}{1+3/2}=\frac{2}{5}, \qquad \|h\|_3=\left(\frac25\right)^{1/3}. $$ Also $0<h(x)<1$ everywhere, and for every $0\le K<1$, the set $\{x:h(x)>K\}$ has measure $1-K^2>0$ (for $K=0$, the set is all of $(0,1)$, also of measure $1$). Therefore $\|h\|_\infty=1$. Check: the antiderivative of $x^{3/2}$ is $(2/5)x^{5/2}$, yielding $2/5$ at $1$ and $0$ at $0$; the tail measures $1-K^2$ are positive for every $K<1$, verifying the essential supremum.

Finite-Measure Embeddings Can Be Strict

On a finite measure space, the finite-measure embedding says that if $1\le p<q\le\infty$, then $L^q$ is contained in $L^p$. Here, containment means that every $L^q$ equivalence class is also an $L^p$ equivalence class. In the present space, the containment follows from Hölder's inequality when $q<\infty$: for $f\in L^q(0,1)$, apply Hölder to $|f|^p$ and $1$, with exponents $q/p$ and $q/(q-p)$. Since $m((0,1))=1$, this gives $$ \int_0^1 |f|^p\,dm\le \left(\int_0^1|f|^q\,dm\right)^{p/q}. $$ For $q=\infty$, the defining bound $|f|\le\|f\|_\infty$ almost everywhere gives $\int_0^1|f|^p\,dm\le\|f\|_\infty^p$. These inclusions do not assert equality of the spaces.

Proposition: On $((0,1),\mathcal{L}_{(0,1)},m)$, $L^q(0,1)$ is a proper subset of $L^p(0,1)$ whenever $1\le p<q\le\infty$.

Proof. The finite-measure embedding just stated gives $L^q(0,1)\subseteq L^p(0,1)$. To prove that the inclusion is proper, we construct an element of $L^p(0,1)$ that is not in $L^q(0,1)$, treating the two possibilities for $q$. If $q<\infty$, take $a=1/q$ and $f(x)=x^{-1/q}$. Then $ap=p/q<1$, because $p<q$, so the membership theorem gives $f\in L^p(0,1)$. But $aq=1$, the critical case, so $f\notin L^q(0,1)$. If $q=\infty$, then $p<\infty$; take $a=1/(2p)>0$ and $f(x)=x^{-1/(2p)}$. Here $ap=1/2<1$, so $f\in L^p(0,1)$, while every positive $a$ gives $f_a\notin L^\infty(0,1)$. In either case there is a class in $L^p(0,1)$ that is not in $L^q(0,1)$, proving proper inclusion. $\square$

Worked Example: Choosing a Witness for L1 but Not L-infinity

For $p=1$ and $q=\infty$, the construction in the proof chooses $a=1/(2p)=1/2$, giving $f(x)=x^{-1/2}$. The calculations are $$ \int_0^1 |f(x)|\,dx=\frac{1}{1-1/2}=2<\infty, \qquad \{x:f(x)>K\}=(0,K^{-2})\quad(K\ge1). $$ Each tail has measure $K^{-2}>0$. Thus this particular function witnesses $L^\infty(0,1)\subsetneq L^1(0,1)$. Check: the finite-$p$ condition is $ap=(1/2)1=1/2<1$, while the essential-boundedness condition fails because $a=1/2>0$. The integral value and the tail-set test agree with the two membership conclusions.

The Critical Case and a Common Mistake

A frequent error is to apply the formula $1/(1-ap)$ at $ap=1$, or to treat the condition as $ap\le1$. At the boundary, that formula has a zero denominator and does not describe a finite integral. The correct calculation uses the logarithmic antiderivative. For example, with $p=2$ and $a=1/2$, the truncated integral is $$ \int_\delta^1 |x^{-1/2}|^2\,dx=\int_\delta^1\frac{1}{x}\,dx=-\ln\delta, $$ which tends to $+\infty$, not to a finite value. The condition $ap<1$ must therefore be strict.

The distinction between finite-$p$ integrability and essential boundedness is also important. A power with $0<a<1/p$ is in $L^p(0,1)$, yet it is not in $L^\infty(0,1)$. In contrast, when $a\le0$, the function is bounded by $1$, and is also in every finite $L^p(0,1)$. The inclusion $L^q\subseteq L^p$ for $p<q$ is consequently a one-way statement in general, not an equality of spaces.

Check Your Understanding

Use the strict threshold $ap<1$, and distinguish finite-$p$ integrability from essential boundedness.

For which real $a$ does $x^{-a}$ belong to $L^3(0,1)$? When it does, state its $L^3$ norm.
Determine whether $x^{-1/3}$ belongs to $L^3(0,1)$, and identify the integral that decides the critical case.
Give an explicit power in $L^2(0,1)$ that is not in $L^\infty(0,1)$, and verify both claims.
For finite $1\le p<q$, choose a power that belongs to $L^p(0,1)$ but not $L^q(0,1)$.
Explain why $x^{1/2}$ has essential supremum $1$ even though it is strictly less than $1$ at every point of $(0,1)$.