Intermediate Lp Membership from Intersections

Two Endpoint Norms Control the Exponents Between Them

On a measure space of infinite measure, membership in $L^p$ alone does not generally imply membership in $L^q$ for $p<q$, nor does membership in $L^q$ alone generally imply membership in $L^p$. Yet having both kinds of control is enough to fill the gap: a function that belongs to $L^p$ and $L^q$ belongs to every $L^r$ with $p<r<q$. The proof works without a finite-measure or sigma-finiteness assumption. Its central step is to use the $p$-power where the function is at most $1$, and the $q$-power where it is greater than $1$.

Throughout, $(X,\mathcal{M},\mu)$ is an arbitrary measure space; it need not be finite, sigma-finite, or complete. Functions may be real-valued or complex-valued and are measurable. For $1\le s<\infty$, $L^s(\mu)$ is the space of measurable functions $f$ for which $\int_X |f|^s\,d\mu<\infty$, with functions equal $\mu$-a.e. identified. Its norm is $\|f\|_s=(\int_X |f|^s\,d\mu)^{1/s}$. Thus membership and norm statements concern equivalence classes, and pointwise estimates used in proofs can be arranged to hold outside a measurable null set.

Theorem (Intermediate $L^r$ membership): Let $(X,\mathcal{M},\mu)$ be any measure space, let $1\le p<r<q<\infty$, and let $f$ be a measurable real-valued or complex-valued function. If $f\in L^p(\mu)\cap L^q(\mu)$, then $f\in L^r(\mu)$, and $$ \|f\|_r\le \bigl(\|f\|_p^p+\|f\|_q^q\bigr)^{1/r}. $$ The membership and norm inequality are statements about $f$ up to $\mu$-a.e. equality. No assumption that $\mu(X)$ is finite is needed.

Proof. Since $\int_X|f|^p\,d\mu<\infty$, the set $N=\{x:|f(x)|=\infty\}$ has measure zero: for every positive integer $k$, $k\mu(N)\le\int_X|f|^p\,d\mu$, so $\mu(N)=0$. We may redefine $f$ to be zero on the measurable set $N$. The resulting representative is finite everywhere and has the same $L^p$ and $L^q$ norms.

Partition $X$ into the measurable sets $A=\{x:|f(x)|\le1\}$ and $B=\{x:|f(x)|>1\}$. On $A$, because $r>p$, raising a number in $[0,1]$ to the larger power gives a value no larger: $|f|^r\le |f|^p$. On $B$, because $r<q$, raising a number greater than $1$ to the smaller power gives a value no larger: $|f|^r\le |f|^q$. At $|f|=1$, the point is in $A$ and the first inequality is equality. These inequalities also hold when $f=0$. Therefore, by monotonicity and additivity of the nonnegative Lebesgue integral,

\int_X |f|^r\,d\mu =\int_A |f|^r\,d\mu+\int_B |f|^r\,d\mu \le \int_A |f|^p\,d\mu+\int_B |f|^q\,d\mu \le \|f\|_p^p+\|f\|_q^q<\infty.

Thus $f\in L^r(\mu)$. Taking the $r$-th root of the displayed inequality gives the claimed norm bound. This reasoning also covers the zero function and cases where either endpoint norm is zero: the integrals and inequalities remain valid, and a zero endpoint norm means $f=0$ $\mu$-a.e. $\square$

Reading the Split and the Bound

The two parts of the estimate have different jobs. On $A$, the lower exponent $p$ controls the $r$-power because $|f|$ is no larger than $1$. On $B$, the higher exponent $q$ controls it because $|f|$ is larger than $1$. The cutoff is useful precisely because the direction of the power comparison changes at $1$. No estimate of $\mu(A)$ or $\mu(B)$ is needed, which is why the argument remains valid even if either set has infinite measure.

The bound is explicit but need not be sharp. It uses the entire $L^p$ integral to bound the low part and the entire $L^q$ integral to bound the high part, even though each of those integrals may also include contributions from the other part of the domain. Its advantage is that it is direct and needs no additional information about the measure space.

Worked Example: A Function on a Two-Point Measure Space

Let $X=\{a,b\}$, let $\mathcal{M}$ be its power set, and give the points measures $\mu(\{a\})=4$ and $\mu(\{b\})=1$. Define $f(a)=1/2$ and $f(b)=2$. Take $p=1$, $r=2$, and $q=3$, so $1<2<3$.

At $a$, $|f(a)|=1/2\le1$, so it is in the low part; at $b$, $|f(b)|=2>1$, so it is in the high part. The endpoint integrals are

\|f\|_1=\int_X |f|\,d\mu =\left(\frac12\right)4+2\cdot1 =2+2=4,

and

\|f\|_3^3=\int_X |f|^3\,d\mu =\left(\frac12\right)^3 4+2^3\cdot1 =\frac18\cdot4+8 =\frac12+8=\frac{17}{2}.

The intermediate integral, computed directly, is

\|f\|_2^2=\int_X |f|^2\,d\mu =\left(\frac12\right)^2 4+2^2\cdot1 =\frac14\cdot4+4 =1+4=5.

The theorem gives $\|f\|_2^2\le \|f\|_1+\|f\|_3^3=4+17/2=25/2$, hence $\|f\|_2\le\sqrt{25/2}=5/\sqrt{2}$. The directly computed norm is $\|f\|_2=\sqrt{5}$, and indeed $5\le25/2$. Check: the low-part contribution to the squared integral is $(1/2)^2\cdot4=1$, not $(1/2)\cdot4$; the high-part contribution is $2^2\cdot1=4$. Their sum is $5$, as computed above.

Worked Example: Exponential Decay on an Infinite-Measure Space

On $X=[0,\infty)$ with the Lebesgue sigma-algebra and Lebesgue measure $m$, let $f(x)=e^{-x}$. This measure space has infinite total measure, but $f$ belongs to $L^1(m)$ and $L^3(m)$. Again choose $p=1$, $r=2$, and $q=3$.

Since $0<e^{-x}\le1$ for every $x\ge0$, the whole domain lies in $A=\{|f|\le1\}$, including $x=0$, where $|f(0)|=1$. Direct integration gives

\|f\|_1=\int_0^\infty e^{-x}\,dx =\lim_{T\to\infty}(1-e^{-T})=1,

and

\|f\|_3^3=\int_0^\infty e^{-3x}\,dx =\lim_{T\to\infty}\frac{1-e^{-3T}}{3} =\frac13.

The intermediate integral is also directly computable:

\|f\|_2^2=\int_0^\infty e^{-2x}\,dx =\lim_{T\to\infty}\frac{1-e^{-2T}}{2} =\frac12.

The bound asserts $\|f\|_2^2\le1+1/3=4/3$, which is verified by $1/2\le4/3$. In norm form, the actual value $1/\sqrt{2}$ is at most the bound $(4/3)^{1/2}=2/\sqrt{3}$. Check: the low-part estimate alone gives $\int |f|^2\,dm\le\int |f|\,dm$, or $1/2\le1$; it is consistent with the full bound, which also includes the nonnegative $L^3$ contribution.

Worked Example: Why Membership at Only One Exponent Is Not Enough

On $(0,1)$ with Lebesgue measure, define $g(x)=x^{-1/2}$. This function belongs to $L^1(m)$, but not to $L^2(m)$. The calculation for the first exponent is

\int_0^1 |g(x)|\,dx =\lim_{\varepsilon\downarrow0}\int_\varepsilon^1 x^{-1/2}\,dx =\lim_{\varepsilon\downarrow0}2(1-\sqrt{\varepsilon}) =2.

For the second exponent,

\int_0^1 |g(x)|^2\,dx =\lim_{\varepsilon\downarrow0}\int_\varepsilon^1 x^{-1}\,dx =\lim_{\varepsilon\downarrow0}(-\ln\varepsilon) =+\infty.

Thus $g\in L^1(m)$ does not imply $g\in L^2(m)$, even though the underlying interval has finite measure. There is no contradiction with the theorem: the required second endpoint membership is missing. Check: the first integral is finite, while the truncated second integral grows without bound as $\varepsilon$ decreases to zero.

Common Errors and the Role of the Intersection

A frequent mistake is to use the low-part inequality in the wrong direction. For $0\le t\le1$ and $p<r$, the correct comparison is $t^r\le t^p$, not $t^p\le t^r$. For example, with $t=1/2$, $p=1$, and $r=2$, the powers are $t^r=1/4$ and $t^p=1/2$, so $1/4\le1/2$. On the high part the comparison reverses: for $t=2$ and $r=2<q=3$, $t^r=4\le8=t^q$. These checks show why the domain must be split at $1$ rather than using one power inequality everywhere.

Another mistake is to assume that a finite-measure embedding applies on every measure space. The argument here does not compare $\|f\|_p$ and $\|f\|_q$ by using $\mu(X)$; it uses both integrability assumptions on different regions. The exponential example shows that this distinction matters: its domain has infinite measure, yet both endpoint integrals and the intermediate integral are finite.

The intersection result is useful whenever a problem gives two integrability estimates but asks for an exponent between them. It supplies membership and a quantitative bound without requiring a formula for $f$, a finite underlying measure, or any convergence theorem. The cutoff argument also makes the mechanism transparent: small values are controlled by the lower exponent, while large values are controlled by the higher exponent.

Check Your Understanding

In each question, keep track of the measure space, the exponent ordering, and which side of the threshold $1$ is being used.

State the hypotheses and conclusion of the intermediate $L^r$ membership theorem, including its norm bound.
For $0\le t\le1$ and $p<r$, which is larger, $t^p$ or $t^r$? Verify your answer at $t=1/2$, $p=1$, $r=2$.
On the two-point space in the first example, calculate separately the contributions of $a$ and $b$ to $\int_X|f|^2\,d\mu$, then compare their sum with the theorem's upper bound.
Why does $g(x)=x^{-1/2}$ on $(0,1)$ not contradict the theorem when it belongs to $L^1(m)$ but not $L^2(m)$?
Does the theorem require $\mu(X)<\infty$ or sigma-finiteness? Identify where the proof avoids using either assumption.

Intermediate Lp Membership from Intersections

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