Proof of Hoelder's Inequality, for Two and for Finitely Many Factors

Each result cited is universally quantified over the data appearing in its own statement, and is applied here to the data named in the statement above.

We record three facts used repeatedly. First, for $h\in\mathcal{L}^{1}$ one has $\lVert h\rVert_{1}=\int_{X}|h|\,d\mu$, because $(|h(x)|)^{1}=|h(x)|$ and $t^{1}=t$ for nonnegative $t$, by Properties of Real Powers of Nonnegative Real Numbers §agreement. Second, for nonnegative real numbers $a_{1},\dots,a_{n}$ and a positive real $c$,

by induction on $n$: for $n=1$ both sides are $a_{1}^{c}$ by claim 1 of Properties of Finite Products, and the induction step follows from that same recursion together with Properties of Real Powers of Nonnegative Real Numbers §product. Third, and by the same induction using claim 4 of Properties of the Absolute Value in an Ordered Field, the absolute value of a finite product of real numbers is the product of their absolute values.

Claim 1. The map $fg$ is measurable by claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and $|fg|=|f|\,|g|$ pointwise by claim 4 of Properties of the Absolute Value in an Ordered Field. Write $A=\lVert f\rVert_{p}$ and $B=\lVert g\rVert_{q}$, nonnegative real numbers.

Suppose first that $A=0$. By Elementary Properties of the p-Seminorm §vanishing we have $f=0$ almost everywhere, hence $|fg|=0$ almost everywhere, so $\int_{X}|fg|\,d\mu=0$ by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §vanishing. Thus $fg\in\mathcal{L}^{1}$ and $\lVert fg\rVert_{1}=0\le AB$, the product $AB$ being nonnegative. The case $B=0$ is identical.

Suppose now that $0<A$ and $0<B$. Let $F$ and $G$ be the maps $x\mapsto |f(x)|/A$ and $x\mapsto |g(x)|/B$; they are measurable and nonnegative. For every $x$, Properties of Real Powers of Nonnegative Real Numbers §product gives $(F(x))^{p}=(1/A)^{p}(|f(x)|)^{p}$, and $(1/A)^{p}A^{p}=(A^{-1}A)^{p}=1^{p}$ by the same claim, while $1^{p}=\exp(p\log 1)=\exp(0)=1$ by Real Power of a Nonnegative Real Number §power, by $\log 1=0$ in claim 2 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities and by claim 1 of Basic Properties of the Exponential Function; so $(1/A)^{p}=1/A^{p}$. Using the homogeneity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral and Elementary Properties of the p-Seminorm §power,

and in the same way $\int_{X}G^{q}\,d\mu=1$.

By Conjugate Exponents and Young's Inequality §young, applied at each $x$ to the nonnegative numbers $F(x)$ and $G(x)$,

All the maps appearing here are measurable and nonnegative, so the monotonicity, additivity and homogeneity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral give

Finally $FG$ is the map $x\mapsto\frac{1}{AB}|f(x)|\,|g(x)|=\frac{1}{AB}|(fg)(x)|$, so by homogeneity again $\int_{X}FG\,d\mu=\frac{1}{AB}\int_{X}|fg|\,d\mu$. Multiplying the previous display by the positive number $AB$ gives $\int_{X}|fg|\,d\mu\le AB$; in particular this integral is finite, so $fg\in\mathcal{L}^{1}$, and $\lVert fg\rVert_{1}=\int_{X}|fg|\,d\mu\le\lVert f\rVert_{p}\lVert g\rVert_{q}$.

Claim 2. We argue by induction on $n$, the assertion for a given $n$ being understood as quantified over all admissible exponents and functions.

For $n=1$ the hypothesis reads $1/p_{1}=1$, so $p_{1}=1$, and $F=f_{1}$ by claim 1 of Properties of Finite Products. Hence $F\in\mathcal{L}^{1}$ and $\lVert F\rVert_{1}=\lVert f_{1}\rVert_{p_{1}}$, which is $\prod_{k=1}^{1}\lVert f_{k}\rVert_{p_{k}}$ by the same claim.

Let $n\in\mathbb{N}$ and assume the assertion for $n$. Let exponents $p_{1},\dots,p_{n+1}$, each at least $1$, with $\sum_{k=1}^{n+1}1/p_{k}=1$, and functions $f_{k}\in\mathcal{L}^{p_{k}}$ be given. Each summand $1/p_{k}$ is positive, and there are at least two of them, so each satisfies $1/p_{k}<1$ and hence $1<p_{k}$. Define $r$ by

which is positive and at most $1$; thus $1\le r$, and $p_{n+1}$ and $r$ are conjugate exponents. Put $h=\prod_{k=1}^{n}f_{k}$, the pointwise finite product, which is measurable by claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions.

For each $k\in[n]$ set $s_{k}=p_{k}/r$. Then $1/p_{k}\le\sum_{i=1}^{n}1/p_{i}=1/r$, so $r\le p_{k}$ and $1\le s_{k}$; moreover $\sum_{k=1}^{n}1/s_{k}=r\sum_{k=1}^{n}1/p_{k}=1$. By Elementary Properties of the p-Seminorm §rescaling, applied with the exponents $r$ and $s_{k}$, whose product is $p_{k}\ge 1$, the map $|f_{k}|^{r}$ lies in $\mathcal{L}^{s_{k}}$ and

By the two elementary product identities recorded at the start, $|h(x)|^{r}=\bigl(\prod_{k=1}^{n}|f_{k}(x)|\bigr)^{r}=\prod_{k=1}^{n}\bigl(|f_{k}(x)|\bigr)^{r}$ for every $x$, so the pointwise product of the $n$ functions $|f_{k}|^{r}$ is $|h|^{r}$. The inductive hypothesis, applied to these $n$ functions and the exponents $s_{k}$, therefore gives $|h|^{r}\in\mathcal{L}^{1}$ and

In particular $\int_{X}|h|^{r}\,d\mu$ is finite, so $h\in\mathcal{L}^{r}$, and applying Properties of Real Powers of Nonnegative Real Numbers §monotone with the positive exponent $1/r$ and then the product identity and Properties of Real Powers of Nonnegative Real Numbers §exponents,

By the recursion in claim 1 of Properties of Finite Products, the pointwise product $F$ of $f_{1},\dots,f_{n+1}$ equals $h\,f_{n+1}$. Applying claim 1 above to $h\in\mathcal{L}^{r}$ and $f_{n+1}\in\mathcal{L}^{p_{n+1}}$, which carry conjugate exponents, gives $F\in\mathcal{L}^{1}$ and

the last step again by the recursion in claim 1 of Properties of Finite Products, and the middle step being legitimate because $\lVert f_{n+1}\rVert_{p_{n+1}}$ is nonnegative. This is the assertion for $n+1$, and the induction is complete.