Contents

# Contents

## Idea

Hölder’s inequality is a basic inequality in analysis, which can be interpreted as saying that the $\mathbf{C}$-graded *-algebra of L^p-spaces

$L^{1/p}(X,\mu)$

(the unfortunate reciprocal in the grading is explained in the article Lebesgue space) is a $\mathbf{C}$-graded normed *-algebra. That is, the canonical bilinear multiplication map

$L^{1/p}(X,\mu)\otimes L^{1/q}(X,\mu)\to L^{1/(p+q)}(X,\mu)$

is a contractive map, i.e., if $f\in L^{1/p}(X,\mu)$ and $g\in L^{1/q}(X,\mu)$, then

$f g \in L^{1/(p+q)}(X,\mu)$

and

$\|f g \|\le \|f\|\cdot\|g\|,$

where the norms are p-norms for the corresponding values of $p$.

As a consequence, it implies that the canonical pairing

$L^{1/p}(X,\mu)\otimes L^{1/(1-p)}(X,\mu)\to L^1(X,\mu)\to \mathbf{C}$

exhibits $L^{1/(1-p)}(X,\mu)$ as the Banach space dual of the Banach space $L^{1/p}(X,\mu)$.

## Statements

Let $(X, \mu)$ be a measure space, and for $p \in \mathbf{C}$ a complex number with a nonnegative real part let $L^{1/p}$ denote $L^{1/p}(X, \mu)$, the Banach space of complex-valued functions on $X$ with finite 1/p-norm modulo equality almost everywhere.

Suppose $p,q,r\in \mathbf{C}$ have nonnegative real parts and $p+q=r$. Then Hölder’s inequality states that for any $f \in L^{1/p}$, $g \in L^{1/q}$ we have

$f g \in L^{1/r}(X,\mu)$

and

$\|f g\|_{1/r}\le\|f\|_{1/p}\cdot \|g\|_{1/q}.$

## Duality

In particular, if $r=1$ we have

$\int_X \left| f g \right| \leq {\|f\|_{1/p}} {\|g\|_{1/q}}$

(in particular, $f g$ is an $L^1$ function).

The “nPOV” meaning is this: in this situation there is a canonical pairing $\langle -, - \rangle$ between $L^{1/p}$ and $L^{1/q}$,

$L^{1/p} \times L^{1/q} \to \mathbb{C}: (f, g) \mapsto \langle f, g \rangle \coloneqq \int_X f \cdot g,$

which gives a bounded linear map

$L^{1/p} \otimes L^{1/q} \to \mathbb{C}$

between Banach spaces. The point of Hölder’s inequality is that this pairing is a short map, i.e., a map of norm bounded above by $1$. In other words, this is morphism in the symmetric monoidal closed category Ban consisting of Banach spaces and short linear maps between them. Accordingly, the map

$L^{1/p} \otimes L^{1/q} \to \mathbb{C}$

induces (by currying) a map from $L^{1/p}$ to the Banach dual of $L^{1/(1-p)}$:

$L^{1/p} \to (L^{1/q})^\ast \coloneqq [L^{1/q}, \mathbb{C}]$

(again a short map of course), and reciprocally a map $L^{1/q} \to (L^{1/p})^\ast$.

It is a short step to prove that in fact the norm of the pairing $L^{1/p} \otimes L^{1/q} \to \mathbb{C}$ is exactly $1$, and even better that the maps $L^{1/p} \to (L^{1/q})^\ast$ and $L^{1/q} \to (L^{1/p})^\ast$ are in fact isometric embeddings.

If the real parts of $p$ and $q$ are nonzero, then with a little more work (with the help of the Radon-Nikodym theorem; see for example here), one sees these maps are surjective and thus isomorphisms in $Ban$.

###### Remark

It is true also that $(L^1)^\ast \cong L^\infty$, but it is not true that $(L^\infty)^\ast$ is isomorphic to $L^1$. Or, it is at least not true in ZFC, although it may be true in dream mathematics.

## The noncommutative case

Remarkably, L^p-spaces can be defined for arbitrary von Neumann algebras (Haagerup, 1979) and the Hölder inequality continues to hold in this generality (Kosaki, 1984).

The $L^{1/p}$-spaces are now bimodules over the underlying von Neumann algebra.

If the real part of $p$ is zero and the von Neumann algebra is commutative, than the space $L^{1/p}$ is (noncanonically) isomorphic to $L^\infty$.

In the noncommutative case this is no longer true, which is the starting point of the Tomita–Takesaki theory.

## Proof of Hölder’s inequality in the commutative case

The proof is remarkably simple. First, if $p, q \gt 0$ and $\frac1{p} + \frac1{q} = 1$, then we have Young's inequality, viz. for $a, b \gt 0$

$a b \;\leq\; \frac{a^p}{p} + \frac{b^q}{q} \,,$

with equality precisely when $a^p = b^q$. This is quickly derived from the (strict) convexity of the exponential function, that $0 \leq t \leq 1$ implies

$e^{t x + (1-t)y} \leq t e^x + (1-t)e^y \,,$

where equality holds iff $e^x = e^y$. All one has to do is put $t = \frac{1}{p}$ and arrange $x, y$ so that $e^x = a^p$ and $e^y = b^q$.

Then, to prove ${|\langle f, g \rangle|} \leq {\|f\|_p} {\|g\|_q}$, we may assume $f, g$ nonzero (so their norms are positive) and normalize them to unit vectors $u = f/{\|f\|_p}, v = g/{\|g\|_q}$, so that now the object is to prove

$\int_X {|u|} \cdot {|v|} \leq 1.$

But since we are dealing with unit vectors, we have $\int_X {|u|^p} = 1$ and $\int_X {|v|^q} = 1$, and now what we want follows straightaway from Young’s inequality applied to integrands:

$\int_X {|u|} \cdot {|v|} \leq \int_X \frac{|u|^p}{p} + \frac{|v|^q}{q} = \frac1{p} + \frac1{q} = 1$

and so the proof of Hölder’s inequality is complete.

To prove that the norm of the pairing $\langle -, - \rangle$ is exactly $1$ (is not less than $1$), it’s enough to take any $u \in L^p$ of norm $1$, so $f = {|u|}$ is a nonnegative function of norm $1$, and then put $g = f^{p-1} = f^{p/q}$. We then have $f^p = g^q$ (almost) everywhere, where we then have $f g = \frac{f^p}{p} + \frac{g^q}{q}$, and now

$\int_X f g = \int_X \frac{f^p}{p} + \frac{g^q}{q} = \frac1{p} + \frac1{q} = 1.$

Actually these calculations do a little better: they show that upon currying, the map

$\begin{array}{ccc} L^p &\longrightarrow& [L^q, \mathbb{C}] \\ f &\mapsto& \lambda g. \langle f, g \rangle \end{array}$

preserves the norm, so that $L^p$ isometrically embeds into $(L^q)^\ast$.

## Properties

### Relation to Minkowski’s inequality

Recall that Minkowski's inequality is just the triangle inequality in the context of L-p space. There is a well-known trick, covered in just about every functional analysis text, that allows one to deduce Minkowski’s inequality as a corollary of Hölder’s inequality. You can look it up for instance in the English Wikipedia, here.

What seemingly most such presentations lack is motivation for the trick, so let us try to say something about this.

First, Minkowski's inequality can be restated as asserting the convexity of the unit ball $B = \{f \in L^p: {\|f\|_p} \leq 1\}$ of $L^p$. If we place ourselves for a moment in the context of $L^p$ real-valued functions, then it suffices to show that $B$ is the intersection of a collection of affine half-spaces, say $H_\lambda = \{f \in L^p: \lambda(f) \leq 1\}$ where $\lambda: L^p \to \mathbb{R}$ is a (continuous) linear functional. But with hindsight into the meaning of Hölder’s inequality, seen as paving the way to characterizing linear functionals on $L^p$ as those of the form $\lambda(g) = (f \mapsto \langle f, g \rangle)$ for some $g \in L^q$, it’s only natural to see whether we can find a sufficiently large collection $B'$ of such $g$ such that $B = \bigcap_{g \in B'} H_{\lambda(g)}$, and in fact the intuition is that the unit ball $B'$ in $L^q$ ought to work.

Thus the idea is clear, and it’s just a matter of technique from here. We let the relation ${|\langle f, g \rangle|} \leq 1$ on $L^p \times L^q$ set up a Galois connection between subsets of $L^p$ and subsets of $L^q$. The connection takes the unit ball $B'$ in $L^q$ to

$(B')^\perp \coloneqq \big\{ f \in L^p \colon (\forall_{g \in B'}) \; {|\langle f, g \rangle|} \leq 1 \big\} \,,$

which is clearly convex, being an intersection of convex sets $\{f: {|\langle f, g \rangle|} \leq 1\}$, one for each $g \in B'$. Hölder’s inequality itself just asserts the containment $B \subseteq (B')^\perp$. If we show the other inclusion $(B')^\perp \subseteq B$, then $B = (B')^\perp$ is convex. So we want to show that if ${|\langle f, g \rangle|} \leq 1$ whenever ${\|g\|_q} \leq 1$, then ${\|f\|_p} \leq 1$. But we already did that calculation when we proved $L^p \hookrightarrow (L^q)^\ast$ is an isometry. Explicitly: take $h = {|f|^p}/f$ (with $h = 0$ where $f = 0$). Then ${|h|} = {|f|^{p-1}} = {|f|^{p/q}}$, so ${|h|^q} = {|f|^p}$ whence ${\|h\|_q^q} = {\|f\|_p^p}$. Put $g = \frac{h}{{\|h\|_q}}$; since ${\|g\|_q} \leq 1$, it follows by the hypothesis on $f$ that $1 \geq {|\langle f, g \rangle|}$. But this gives

$1 \geq \frac1{{\|h\|_q}} \int_X f h = \frac1{{\|f\|_p^{p/q}}} \int_X {|f|^p} = {\|f\|_p^{p - p/q}} = {\|f\|_p}$

as was to be shown.

The standard derivation of Minkowski’s inequality from Hölder’s inequality is nothing more than a very tidied-up rendering of this argument, but without the additional conceptual explanation given here.

### Relation to Log-convex functions

Let $D$ be a convex space, e.g. an affine space. We say that a function $f: D \to (0, \infty)$ is log-convex if its logarithm $\log(f)$ is a convex function.

Hölder’s inequality is closely related to the notion of log-convexity. On the one hand, we saw that the inequality follows from the convexity of the exponential function, which is the most basic log-convex function of all. On another hand, we have the following result which uses Hölder’s inequality.

###### Theorem

The collection of log-convex positive functions on a convex domain $D$ is closed under pointwise multiplication, pointwise addition, and pointwise sups.

###### Proof

The statement for multiplication is clear since $\log(f \cdot g) = \log(f) + \log(g)$ and any sum of convex functions is convex.

Similarly, $\log: (0,\infty) \to \mathbb{R}$ is an isomorphism of partially ordered sets and so $\log (\sup\{f_i\}) = \sup\{\log(f_i)\}$. It thus suffices to show that if $f_i$ is a collection of convex functions on $D$, then so is $\sup\{f_i\}$. For $x, y \in D$ and $a, b \geq 0$ such that $a + b = 1$, we must show

$\sup\{f_i\}(a x + b y) \leq a \sup\{f_i\}(x) + b\sup\{f_i\}(y);$

letting $c$ denote the right side, this holds iff $f_i(a x + b y) \leq c$ for all $i$ (by definition of $\sup$). But

$\array{ f_i(a x + b y) & \leq & a f_i(x) + b f_i(y) & since\; f_i\; is\; convex \\ & \leq & a\sup\{f_i\}(x) + b\sup\{f_i\}(y) & }.$

Finally, for the sum $f + g$, in order to show $\log(f + g)$ is convex, it suffices to show that

(1)$(f + g)\left(\frac1{p}x + \frac1{q}y\right) \leq (f+g)(x)^{\frac1{p}} (f+g)(y)^{\frac1{q}}$

for $p, q \gt 1$ such that $\frac1{p} + \frac1{q} = 1$. But setting

$s = f(x)^{\frac1{p}}, \qquad t = g(x)^{\frac1{p}}, \qquad u = f(y)^{\frac1{q}}, \qquad v = g(y)^{\frac1{q}},$

the right side of (1) is $(s^p + t^p)^{\frac1{p}} \cdot (u^q + v^q)^{\frac1{q}}$. By Hölder’s inequality, this is greater than or equal to

$\array{ s u + t v & = & f(x)^{\frac1{p}} f(y)^{\frac1{q}} + g(x)^{\frac1{p}} g(y)^{\frac1{q}} \\ & \geq & f\left(\frac1{p} x + \frac1{q} y\right) + g\left(\frac1{p} x + \frac1{q} y\right) }$

where the last inequality is by log-convexity of $f$ and $g$.

This last theorem enters into the account of Artin (1931) of the basic theory of the Gamma function:

###### Corollary
$\Gamma(x) = \int_0^\infty t^x e^{-t} \frac{d t}{t}$

is log-convex over the domain $x \gt 0$.

###### Proof

The function $x \mapsto t^{x-1}$ is log-linear, hence log-convex. Hence the integral defining $\Gamma(x)$ over $x \gt 0$ is a sup over suitable Riemann sums that are positive-linear combinations of the form

$\sum_{i=1}^n t_i^{x-1} e^{-t_i} \Delta t_i$

and these, and together with their sup, are log-convex by the Theorem.

The main fact underlying Artin’s 1931 discussion is the Bohr-Haagerup theorem (Artin (1931), Thm. 2.1):

###### Theorem

The Gamma function is characterized as the unique function $\Gamma: \{x \in \mathbb{R}|\; -x \notin \mathbb{N}\} \to \mathbb{R}$ satisfying the following conditions:

• $\Gamma(1) = 1$,

• $\Gamma(x+1) = x\Gamma(x)$,

• $\Gamma$ is log-convex over $(0, \infty)$.

## References

• Emil Artin, Einführung in die Theorie der Gammafunktion, Hamburger Mathematische Einzelschriften

l. Heft, Verlag B. G. Teubner, Leipzig (1931)

English translation by Michael Butler: The Gamma Function, Holt, Rinehart and Winston (1931) [pdf]

Last revised on December 27, 2022 at 21:48:59. See the history of this page for a list of all contributions to it.