From an \(L^q\) Function to a Functional on \(L^p\)
Hölder’s inequality does more than bound an integral: it turns each function \(g\) in a conjugate space into a continuous linear functional on \(L^p\). The bound is sharp. In fact, for every \(1\le p<\infty\), the norm of this functional is exactly the \(L^q\) norm of \(g\), with \(q\) the conjugate exponent. The endpoint \(p=1,\ q=\infty\) requires care: sigma-finiteness ensures that the \(L^1\) functions can test the essential supremum of \(g\).
Let \((X,\mathcal{M},\mu)\) be a measure space. Functions may be real- or complex-valued, and \(L^p(\mu)\) consists of measurable functions with finite \(p\)-norm, identified when they agree \(\mu\)-a.e. For \(1<p<\infty\), set \(q=p/(p-1)\), so \(1/p+1/q=1\). At \(p=1\), set \(q=\infty\). The \(L^\infty\) norm is the essential supremum: \(\|g\|_\infty\) is the least \(C\ge0\) such that \(|g|\le C\) \(\mu\)-a.e.
For complex-valued functions we use \(fg\), without a conjugate in the definition. The norm calculation below chooses \(f\) with the conjugate phase of \(g\), so that \(fg\) is nonnegative. Using the convention \(\int f\overline{g}\,d\mu\) instead gives the same norm.
The Bound and Its Sharpness
We use Hölder’s inequality: if \(1<p<\infty\), \(q=p/(p-1)\), \(u\in L^p(\mu)\), and \(v\in L^q(\mu)\), then \(\int |uv|\,d\mu\le\|u\|_p\|v\|_q\). At the endpoint, if \(u\in L^1(\mu)\) and \(v\in L^\infty(\mu)\), then \(\int |uv|\,d\mu\le\|u\|_1\|v\|_\infty\). For completeness, in the interior case normalize both norms to one and apply Young’s inequality \(ab\le a^p/p+b^q/q\) pointwise to \(|u|\) and \(|v|\), then integrate. If either norm is zero, the corresponding function vanishes a.e. and the inequality follows directly. The endpoint estimate follows from \(|v|\le\|v\|_\infty\) a.e. and monotonicity of the integral.
Proof. First, Hölder’s inequality gives \(\int |fg|\,d\mu\le\|f\|_p\|g\|_q\) in both cases. Thus the integral defining \(T_g(f)\) is absolutely convergent, and \(|T_g(f)|\le\|f\|_p\|g\|_q\). In particular, changing either representative on a null set does not change the integral. Linearity of the integral gives linearity of \(T_g\), and the bound proves \(\|T_g\|\le\|g\|_q\).
Now suppose \(1<p<\infty\), and write \(A=\|g\|_q\). If \(A=0\), then \(g=0\) a.e., so \(T_g=0\) and the claimed equality holds. If \(A>0\), define \(h=0\) where \(g=0\), and \(h=\overline{g}|g|^{q-2}\) where \(g\ne0\). Then \(hg=|g|^q\). Since \((q-1)p=q\), we have \(\|h\|_p^p=\int|g|^q\,d\mu=A^q\), hence \(\|h\|_p=A^{q/p}=A^{q-1}\). Set \(f=h/A^{q-1}\). This gives \(\|f\|_p=1\) and $$ |T_g(f)|=\frac{\int |g|^q\,d\mu}{A^{q-1}}=\frac{A^q}{A^{q-1}}=A. $$ Thus \(\|T_g\|\ge A\), proving equality in the interior case.
For \(p=1\), let \(M=\|g\|_\infty\). If \(M=0\), then \(g=0\) a.e. and both norms are zero. Suppose \(M>0\), and fix any real \(a\) with \(0\le a<M\). By the definition of essential supremum, \(E=\{|g|>a\}\) has positive measure. Sigma-finiteness gives a countable cover \(X=\bigcup_{n\ge1}X_n\) with \(\mu(X_n)<\infty\). Since \(E=\bigcup_n(E\cap X_n)\), at least one set \(F=E\cap X_n\) has positive measure; it also has finite measure. Define \(f=\overline{g}|g|^{-1}\chi_F/\mu(F)\) where \(g\ne0\), and \(f=0\) elsewhere. Then \(\|f\|_1=1\) and \(T_g(f)=\mu(F)^{-1}\int_F|g|\,d\mu>a\). Therefore \(\|T_g\|\ge a\) for every \(a<M\), which implies \(\|T_g\|\ge M\). Combined with the upper bound, \(\|T_g\|=M\). \(\square\)
The proof also identifies norming functions. When \(1<p<\infty\) and \(g\ne0\), the constructed \(f\) has norm one and attains the functional norm. At \(p=1\), the proof uses sets on which \(|g|\) is close to its essential supremum. A norm-attaining \(L^1\) function need not exist.
Worked Norm Calculations
Worked Example: A Pairing on a Two-Point Space
Give \(X=\{1,2\}\) counting measure, take \(p=q=2\), and let \(g(1)=3,\ g(2)=4\). For \(f(1)=a,\ f(2)=b\), the pairing is \(T_g(f)=3a+4b\), while \(\|g\|_2=(3^2+4^2)^{1/2}=5\). Hölder’s inequality gives \(|T_g(f)|\le5(a^2+b^2)^{1/2}\), so \(\|T_g\|\le5\).
Choose \(f(1)=3/5\) and \(f(2)=4/5\). Then \(\|f\|_2=((3/5)^2+(4/5)^2)^{1/2}=(9/25+16/25)^{1/2}=1\), and \(T_g(f)=3(3/5)+4(4/5)=9/5+16/5=5\). Thus \(\|T_g\|=5=\|g\|_2\). Check: the norm-one test function gives exactly the upper bound, so the computed operator norm is sharp.
Worked Example: The \(L^1\)-to-\(L^\infty\) Endpoint
On \((0,1)\) with Lebesgue measure, let \(g(x)=x\). Since \(x\le1\) everywhere and \(x>1-\varepsilon\) on \((1-\varepsilon,1)\) for every \(0<\varepsilon<1\), \(\|g\|_\infty=1\). The endpoint theorem gives \(\|T_g\|=1\).
To see the lower bound directly, set \(f_\varepsilon=\varepsilon^{-1}\chi_{(1-\varepsilon,1)}\). The interval has measure \(\varepsilon\), so \(\|f_\varepsilon\|_1=\varepsilon^{-1}\varepsilon=1\). Its pairing is $$ T_g(f_\varepsilon)=\frac1\varepsilon\int_{1-\varepsilon}^1x\,dx =\frac1\varepsilon\left[\frac{x^2}{2}\right]_{1-\varepsilon}^1 =\frac{1-(1-\varepsilon)^2}{2\varepsilon} =1-\frac{\varepsilon}{2}. $$ For \(\varepsilon=1/4\), this is \(1-1/8=7/8\); as \(\varepsilon\) decreases to zero, it tends to \(1\). Check: the upper bound is \(1\), and these norm-one test functions give values arbitrarily close to \(1\), so the supremum is \(1\).
No norm-one \(f\in L^1(0,1)\) attains this value. Indeed, for any such \(f\), \(|T_g(f)|\le\int_0^1x|f(x)|\,dx<\int_0^1|f(x)|\,dx=1\). The strict inequality holds because \((1-x)|f(x)|>0\) on a set of positive measure whenever \(f\) is nonzero in \(L^1\). Thus the norm is a supremum, not necessarily a maximum.
Worked Example: An Interior Norming Function
On \([0,\infty)\) with Lebesgue measure, take \(p=3,\ q=3/2\), and \(g(x)=e^{-x}\). Then $$ \|g\|_{3/2}=\left(\int_0^\infty e^{-3x/2}\,dx\right)^{2/3} =\left(\frac{2}{3}\right)^{2/3}. $$ The norming function from the proof is proportional to \(g^{q-1}=e^{-x/2}\). Set \(f(x)=(3/2)^{1/3}e^{-x/2}\). Its norm is $$ \|f\|_3^3=\frac32\int_0^\infty e^{-3x/2}\,dx =\frac32\cdot\frac23=1, $$ so \(\|f\|_3=1\). The pairing is $$ T_g(f)=\left(\frac32\right)^{1/3}\int_0^\infty e^{-3x/2}\,dx =\left(\frac32\right)^{1/3}\frac23 =\left(\frac23\right)^{2/3}. $$ Check: this equals the previously computed \(\|g\|_{3/2}\), and the test function has \(L^3\) norm one. Therefore the functional norm is exactly \((2/3)^{2/3}\).
Why the Endpoint Hypothesis Matters
The sigma-finiteness condition in the endpoint statement cannot simply be dropped. Consider a nonempty set \(X\), its full sigma-algebra \(\mathcal{M}\), and the measure \(\mu\) given by \(\mu(\varnothing)=0\) and \(\mu(E)=\infty\) for every nonempty \(E\). This is a measure: a countable disjoint union has measure zero if every set in it is empty, and otherwise both the union’s measure and the sum of the measures are infinite. The space is not sigma-finite, since no countable union of finite-measure sets can cover \(X\).
For this measure, every nonzero measurable function has infinite \(L^1\) integral: if it is nonzero at a point, some set \(\{|f|>1/n\}\) is nonempty and has infinite measure. Thus \(L^1(\mu)=\{0\}\). But \(g=1\) has \(\|g\|_\infty=1\), while \(T_g\) acts only on the zero space and has operator norm zero. This is why the endpoint norm identity above includes sigma-finiteness. It also illustrates a broader point: the theorem identifies the norm of the integration functional \(T_g\); it does not, by itself, say that every bounded functional on \(L^p\) arises from such a pairing.
Common mistake. The estimate \(|\int fg\,d\mu|\le\|f\|_p\|g\|_q\) proves only \(\|T_g\|\le\|g\|_q\), not equality. For example, the upper bound \(\|T_g\|\le1\) in the endpoint example would not establish that the norm is \(1\). The correction is to produce norm-one test functions: the functions \(f_\varepsilon\) give \(T_g(f_\varepsilon)=1-\varepsilon/2\), approaching \(1\). One must also distinguish the supremum from attainment: every one of those values is strictly below \(1\).
Check Your Understanding
Use the definition of operator norm and the hypotheses of the theorem to answer the following.
- For \(p=4\), what is the conjugate exponent \(q\), and which \(L^q\) norm bounds the pairing?
- For \(1<p<\infty\), why does the function \(h=\overline{g}|g|^{q-2}\), set to zero where \(g=0\), belong to \(L^p\)?
- On \((0,1)\), take \(g(x)=1-x\). What is \(\|g\|_\infty\), and how might you build norm-one \(L^1\) tests with pairings approaching that norm?
- Which step in the endpoint lower-bound proof uses sigma-finiteness?
- Does the exact norm identity imply that every bounded functional on \(L^p\) has the form \(T_g\)? Explain what the proved result does and does not establish.