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 (ii). Second, for nonnegative real numbers $a_{1},\dots,a_{n}$ and a positive real $c$,

$$\Bigl(\prod_{k=1}^{n}a_{k}\Bigr)^{c}=\prod_{k=1}^{n}a_{k}^{\,c},$$

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 (iii). 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 of Measurable Real- and [0,∞]-Valued Maps, 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 (v) 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 (iv). 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 (iii) 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 Powers (i), 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 the Integrability Criterion and Elementary Properties of the p-Seminorm (i),

$$\int_{X}F^{p}\,d\mu=\frac{1}{A^{p}}\int_{X}|f|^{p}\,d\mu=\frac{1}{A^{p}}\,A^{p}=1 ,$$

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

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

$$F(x)G(x)\le\frac{(F(x))^{p}}{p}+\frac{(G(x))^{q}}{q}\qquad(x\in X).$$

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, and the Integrability Criterion give

$$\int_{X}FG\,d\mu\le\frac{1}{p}\int_{X}F^{p}\,d\mu+\frac{1}{q}\int_{X}G^{q}\,d\mu=\frac{1}{p}+\frac{1}{q}=1 .$$

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

$$\frac{1}{r}=\sum_{k=1}^{n}\frac{1}{p_{k}}=1-\frac{1}{p_{n+1}} ,$$

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 of Measurable Real- and [0,∞]-Valued Maps.

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 (vi), 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

$$\bigl\lVert\,|f_{k}|^{r}\,\bigr\rVert_{s_{k}}=\bigl(\lVert f_{k}\rVert_{p_{k}}\bigr)^{r}.$$

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

$$\int_{X}|h|^{r}\,d\mu\le\prod_{k=1}^{n}\bigl\lVert\,|f_{k}|^{r}\,\bigr\rVert_{s_{k}}=\prod_{k=1}^{n}\bigl(\lVert f_{k}\rVert_{p_{k}}\bigr)^{r}.$$

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

$$\lVert h\rVert_{r}=\Bigl(\int_{X}|h|^{r}\,d\mu\Bigr)^{1/r}\le\prod_{k=1}^{n}\Bigl(\bigl(\lVert f_{k}\rVert_{p_{k}}\bigr)^{r}\Bigr)^{1/r}=\prod_{k=1}^{n}\lVert f_{k}\rVert_{p_{k}} .$$

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

$$\lVert F\rVert_{1}\le\lVert h\rVert_{r}\,\lVert f_{n+1}\rVert_{p_{n+1}}\le\Bigl(\prod_{k=1}^{n}\lVert f_{k}\rVert_{p_{k}}\Bigr)\lVert f_{n+1}\rVert_{p_{n+1}}=\prod_{k=1}^{n+1}\lVert f_{k}\rVert_{p_{k}},$$

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.