Finite-Measure Embeddings from Lq into Lp

Why Finite Measure Controls the Smaller Norm

On a space of finite measure, a function whose $q$-th power is integrable also has an integrable $p$-th power whenever $p<q$. The size of the space matters: the estimate must account for how much measure is available to contribute to the lower-power integral. The result is a bounded inclusion, and the measure-dependent factor in its norm bound is the best possible one.

Let $(X,\mathcal{M},\mu)$ be a measure space with finite measure $\mu(X)<\infty$; it need not be complete. “Finite” means that the measure of all of $X$ is finite, a stronger condition than being sigma-finite. Functions in this tutorial are measurable real- or complex-valued functions, identified when they agree $\mu$-a.e. For $1\le r<\infty$, $L^r(\mu)$ consists of these equivalence classes with $\int_X |f|^r\,d\mu<\infty$, and $\|f\|_r=(\int_X|f|^r\,d\mu)^{1/r}$. The space $L^\infty(\mu)$ consists of classes with finite essential supremum norm $\|f\|_\infty$.

Definition: For $1\le p<q\le\infty$, the inclusion from $L^q(\mu)$ into $L^p(\mu)$ is the identity map on equivalence classes, once it has been shown that every class in $L^q(\mu)$ also belongs to $L^p(\mu)$. Its sharp norm bound is the least $C\ge0$ such that $\|f\|_p\le C\|f\|_q$ for every $f\in L^q(\mu)$. We use the convention $1/\infty=0$.

The Estimate and Its Proof

For finite $q$, the key ingredient is Hölder's inequality. We record the version needed here and justify it, so the estimate does not depend on an unstated inequality.

Lemma (Hölder's inequality): Let $(X,\mathcal{M},\mu)$ be any measure space, and let $1<r<\infty$ and $s=r/(r-1)$. If $u\in L^r(\mu)$ and $v\in L^s(\mu)$ are measurable real- or complex-valued functions, then $\int_X |uv|\,d\mu\le\|u\|_r\|v\|_s$.

Proof. For nonnegative numbers $a,b$, Young's inequality $ab\le a^r/r+b^s/s$ follows by maximizing $ab-a^r/r$ over $a\ge0$: when $b>0$, the maximum occurs at $a=b^{1/(r-1)}$ and is $b^s/s$; when $b=0$, the inequality holds directly. If either $\|u\|_r$ or $\|v\|_s$ is zero, the corresponding function is zero a.e., so the asserted integral is zero. Otherwise apply Young's inequality pointwise to $|u|/\|u\|_r$ and $|v|/\|v\|_s$. Integrating, each normalized power has integral one, and $1/r+1/s=1$, so the integral of the normalized product is at most one. Multiplying by $\|u\|_r\|v\|_s$ proves the claim. $\square$

Theorem (Finite-measure embedding): Let $(X,\mathcal{M},\mu)$ be a measure space with $\mu(X)<\infty$. If $1\le p<q\le\infty$, then every real- or complex-valued $f\in L^q(\mu)$ belongs to $L^p(\mu)$, and $$ \|f\|_p\le \mu(X)^{\,1/p-1/q}\|f\|_q, $$ where $1/q=0$ when $q=\infty$. If $0<\mu(X)<\infty$, the constant $\mu(X)^{1/p-1/q}$ is sharp. If $\mu(X)=0$, both spaces contain only the zero equivalence class and the least bound is $0$, which agrees with the displayed formula.

Proof. First suppose $q<\infty$. Set $r=q/p$, so $r>1$, and let $s=q/(q-p)$, so $1/r+1/s=1$. For $f\in L^q(\mu)$, apply Hölder's inequality to $u=|f|^p$ and $v=\chi_X$. These functions meet the hypotheses: $\int_X|u|^r\,d\mu=\int_X|f|^q\,d\mu<\infty$, and $\int_X|v|^s\,d\mu=\mu(X)<\infty$. Thus $$ \int_X|f|^p\,d\mu \le \left(\int_X|f|^q\,d\mu\right)^{p/q}\mu(X)^{1-p/q}. $$ Taking the $p$-th root gives the stated estimate and proves $f\in L^p(\mu)$.

Now suppose $q=\infty$. By the definition of essential supremum, $|f|\le\|f\|_\infty$ a.e. To see that the bound holds at the essential supremum itself, for each positive integer $n$ the inequality $|f|\le\|f\|_\infty+1/n$ holds outside a null set; the union of those countably many null sets is null, and letting $n$ increase gives $|f|\le\|f\|_\infty$ off that union. Therefore $$ \int_X|f|^p\,d\mu\le\|f\|_\infty^p\mu(X), $$ which gives the theorem's estimate. If $\mu(X)=0$, every measurable function is zero a.e., so the only equivalence class in either space is zero. Finally, if $\mu(X)>0$, the constant function $f=1$ has $\|1\|_q=\mu(X)^{1/q}$ when $q<\infty$, and $\|1\|_\infty=1$ when $q=\infty$; in both cases $\|1\|_p/\|1\|_q=\mu(X)^{1/p-1/q}$. No smaller constant can therefore work. $\square$

In the finite-$q$ proof, the exponent $1-p/q$ arises from the $L^s$ norm of $\chi_X$: that norm is $\mu(X)^{1/s}$, and $1/(ps)=(1/p-1/q)$. At $q=\infty$, the same final exponent is $1/p$, because the essential bound is multiplied by the measure of the whole space before taking a $p$-th root.

Worked Applications

Worked Example: A Nonconstant Function on a Finite Interval

Take $X=[0,4]$ with Lebesgue measure $m$, and choose $p=1$, $q=2$, and $f=\chi_{[0,1]}$. The space has measure $m(X)=4$, so the theorem gives $\|f\|_1\le4^{1-1/2}\|f\|_2=2\|f\|_2$.

The function equals one on a set of measure one and zero elsewhere. Consequently, $\int_X|f|\,dm=1$, so $\|f\|_1=1$; also $\int_X|f|^2\,dm=1$, so $\|f\|_2=1$. Substitution gives $1\le2$, as required. Check: the ratio $\|f\|_1/\|f\|_2=1$ is below the bound $2$. For the constant function $1$ on this same space, $\|1\|_1=4$ and $\|1\|_2=\sqrt4=2$, so the ratio is $4/2=2$. This verifies sharpness for these parameters.

Worked Example: A Weighted Two-Point Space

Let $X=\{a,b\}$, let $\mathcal{M}$ be all subsets of $X$, and define $\mu(\{a\})=1$, $\mu(\{b\})=3$. Then $\mu(X)=1+3=4$. For $p=2$, $q=4$, the theorem's factor is $4^{1/2-1/4}=4^{1/4}=\sqrt2$. Consider $f(a)=1$, $f(b)=0$. Every function on this finite measurable space is measurable.

Directly, $\|f\|_2=(1\cdot|1|^2+3\cdot|0|^2)^{1/2}=1$, and $\|f\|_4=(1\cdot|1|^4+3\cdot|0|^4)^{1/4}=1$. Thus the bound reads $1\le\sqrt2$. To check that the factor cannot be reduced, take the constant function $g(a)=g(b)=2$. Then $\|g\|_2=(4\cdot2^2)^{1/2}=\sqrt{16}=4$, while $\|g\|_4=(4\cdot2^4)^{1/4}=64^{1/4}=2\sqrt2$. Their ratio is $4/(2\sqrt2)=\sqrt2$, exactly the stated factor. Check: substituting this ratio into $\|g\|_2\le\sqrt2\|g\|_4$ gives $4\le\sqrt2(2\sqrt2)=4$.

Worked Example: The Endpoint q = Infinity

Let $X=[0,3]$ with Lebesgue measure, take $p=2$, and define $f=2$ on $[0,1]$ and $f=1$ on $(1,3]$. The endpoint estimate has factor $m(X)^{1/2}=3^{1/2}=\sqrt3$. The essential supremum of $f$ is $2$, since $f\le2$ everywhere and $f=2$ on a set of measure one.

The squared $L^2$ norm is $$ \|f\|_2^2=\int_0^1 2^2\,dm+\int_{(1,3]}1^2\,dm =4\cdot1+1\cdot2=6, $$ so $\|f\|_2=\sqrt6$. The right side of the estimate is $\sqrt3\cdot2=2\sqrt3$. Both sides are nonnegative, and their squares are $6$ and $12$, respectively; hence $\sqrt6\le2\sqrt3$. Check: the constant function $1$ on $[0,3]$ has $L^2$ norm $\sqrt3$ and $L^\infty$ norm $1$, so its ratio is $\sqrt3$. This is the claimed sharp factor at the endpoint.

Common Pitfall: Reversing the Measure Exponent

A frequent error is to write the factor as $\mu(X)^{1/q-1/p}$, reversing the exponents. This cannot be correct in general. On $X=[0,4]$ with Lebesgue measure, set $p=1$, $q=2$, and take $f=1$. The incorrect factor would be $4^{1/2-1}=4^{-1/2}=1/2$; since $\|f\|_2=2$, that would assert $\|f\|_1=4\le(1/2)2=1$, which is false. The correct factor is $4^{1-1/2}=2$, and the estimate gives $4\le2\cdot2=4$. The constant function both exposes the reversed-exponent error and confirms the sharp bound.

The role of the finite-measure hypothesis is also visible in the proof: it is precisely what makes $\chi_X$ belong to the conjugate space used in Hölder's inequality, and what makes the integral in the $L^\infty$ argument finite. The conclusion is an inclusion of equivalence classes with a norm bound; it does not assert pointwise boundedness everywhere, nor does it say that every $L^p$ function belongs to $L^q$.

Check Your Understanding

In each question, specify the measure space and verify which exponent and measure factor the theorem supplies.

On a measurable space of measure $9$, what is the sharp bound factor from $L^4$ into $L^2$?
For $q=\infty$, derive the $L^p$ bound directly by integrating the a.e. inequality $|f|^p\le\|f\|_\infty^p$.
On $[0,2]$ with Lebesgue measure, compute the $L^1$ and $L^2$ norms of the constant function $3$, and check sharpness for the inclusion $L^2$ into $L^1$.
What are the $L^p$ equivalence classes when $\mu(X)=0$, and what is the least embedding constant?
For finite $q$, identify the conjugate exponents used when applying Hölder to $|f|^p$ and $\chi_X$.

Finite-Measure Embeddings from Lq into Lp

What You'll Learn