A generalized Hölder inequality

[gugo82]

A basic exercise in [tex]$L^p$[/tex] spaces theory.
Hope someone will answer to improve English.

***

Let us recall some basic facts.

Let [tex]$(X,\mathfrak{M} ,\mu)$[/tex] be a non trivial measure space; w.l.o.g. suppose that [tex]$\mathfrak{M}$[/tex] is [tex]$\mu$[/tex]-complete (i.e. the class [tex]$\mathcal{N} (\mu ) :=\{ E\in \mathfrak{M} :\ \mu (E)=0\}$[/tex] of [tex]$\mu$[/tex]-negligible sets is an ideal in [tex]$P(X)$[/tex]*).

A function [tex]$f:X\to \widehat{\mathbb{C}}$[/tex]** is measurable iff [tex]$\{ x\in X:\ f(x)\in U\} \in \mathfrak{M}$[/tex] for all open sets [tex]$U\in \widehat{\mathbb{C}}$[/tex].
A measurable function [tex]$f:X\to \widehat{\mathbb{C}}$[/tex] is said to be [tex]$p$[/tex]-summable (or [tex]$p$[/tex]th power integrable) iff [tex]\int_X |f|^p\ \text{d} \mu <+\infty[/tex]: in this case we put:

[tex]$\lVert f\rVert_p :=\left\{ \int_X |f|^p\ \text{d} \mu\right\}^{1/p}$[/tex]

(here [tex]$p>0$[/tex]).
A measurable function [tex]$f:X \to \widehat{\mathbb{C}}$[/tex] is said essentially bounded iff there exists [tex]$\alpha \geq 0$[/tex] s.t.

(m) [tex]$\{ |f|>\alpha\} :=\{ x\in X:\ |f(x)|>\alpha \} \in \mathcal{N} (\mu)$[/tex]:

in this case the class [tex]$\mathcal{M}_f:=\{\alpha \geq 0:\ \text{(m) holds}\}$[/tex] admits minimum, which is denoted by [tex]$\lVert f\rVert_\infty$[/tex].

The class of [tex]$p$[/tex]-summable [resp. essentially bounded] functions is usually denoted by [tex]$L^p (\mu )$[/tex] [resp. [tex]$L^\infty (\mu )$[/tex]]: it isn't empty, because it contains the null function.
If [tex]$p\in [1,+\infty]$[/tex], when we identify [tex]$L^p$[/tex] functions which are equal a.e. (w.r.t. [tex]$\mu$[/tex]), we turn [tex]$L^p (\mu )$[/tex] into a Banach space with norm [tex]$\lVert \cdot \rVert_p$[/tex].

The following Hölder inequality holds:

(H) [tex]$\lVert f\ g\rVert_1 \leq \lVert f\rVert_p \ \lVert g \rVert_{p^\prime}$[/tex]

where [tex]$f\in L^p(\mu )$[/tex], [tex]$g\in L^{p^\prime} (\mu )$[/tex] and:

[tex]$p^\prime :=\begin{cases} +\infty &\text{, if $p=1$} \\ \frac{p}{p-1} &\text{, if $1
is the Hölder conjugate exponent of [tex]$p$[/tex] (note that relation [tex]$\tfrac{1}{p} +\tfrac{1}{p^\prime} =1$[/tex] holds, at least formally).

__________
* In a set theoretic contest, a family of subsets is said to be an ideal if it has some kind of "absorption property"; in the present case a family [tex]$\mathfrak{F} \subseteq P(X)$[/tex] is said an ideal iff for each [tex]$E\in P(X)$[/tex] s.t. exists [tex]$F_E\in \mathfrak{F}$[/tex] with [tex]$E\subseteq F_E$[/tex], then [tex]$E\in \mathfrak{F}$[/tex].

** [tex]$\widehat{\mathbb{C}}$[/tex] is the standard (Alexandroff) one-point compactification of the complex field [tex]$\mathbb{C}$[/tex]: roughly speaking [tex]$\widehat{\mathbb{C}} :=\mathbb{C} \cup \{ \infty\}$[/tex], where [tex]$\infty$[/tex] is the point at infinity.

***

Exercise:

Let [tex]$(X,\mathfrak{M} ,\mu )$[/tex] be a nontrivial measure space, [tex]$N \in \mathbb{N}$[/tex] a number [tex]$\geq 2$[/tex] and let [tex]$p_1,\ldots ,p_N \in [1,+\infty]$[/tex] s.t. [tex]$\sum_{n=1}^N \frac{1}{p_n} =1$[/tex]***.

Prove that the generalized Hölder inequality:

(gH) [tex]$\left\lVert \prod_{n=1}^N f_n \right\rVert_1 \leq \prod_{n=1}^N \lVert f_n\rVert_{p_n}$[/tex]

holds true for all [tex]$f_1\in L^{p_1} (\mu ), \ldots ,f_n\in L^{p_n} (\mu )$[/tex].

__________
*** If one or more [tex]$p_n$[/tex]s are equal to [tex]$+\infty$[/tex] this relation has to be understood as [tex]\sum_{n=1,\ldots N,\ p_n\neq +\infty} \frac{1}{p_n} =1[/tex].
On the other hand, if one [tex]$p_\nu$[/tex] is equal to [tex]$1$[/tex], then relation [tex]\sum_{n=1}^N \frac{1}{p_n} =1[/tex] forces the other [tex]$p_n$[/tex]s to be equal to [tex]$+\infty$[/tex].

[gugo82]

"gugo82":Exercise:

Let [tex]$(X,\mathfrak{M} ,\mu )$[/tex] be a nontrivial measure space, [tex]$N \in \mathbb{N}$[/tex] a number [tex]$\geq 2$[/tex] and let [tex]$p_1,\ldots ,p_N \in [1,+\infty]$[/tex] s.t. [tex]$\sum_{n=1}^N \frac{1}{p_n} =1$[/tex]***.

Prove that the generalized Hölder inequality:

(gH) [tex]$\left\lVert \prod_{n=1}^N f_n \right\rVert_1 \leq \prod_{n=1}^N \lVert f_n\rVert_{p_n}$[/tex]

holds true for all [tex]$f_1\in L^{p_1} (\mu ), \ldots ,f_n\in L^{p_n} (\mu )$[/tex].

__________
*** If one or more [tex]$p_n$[/tex]s are equal to [tex]$+\infty$[/tex] this relation has to be understood as [tex]\sum_{n=1,\ldots N,\ p_n\neq +\infty} \frac{1}{p_n} =1[/tex].
On the other hand, if one [tex]$p_\nu$[/tex] is equal to [tex]$1$[/tex], then relation [tex]\sum_{n=1}^N \frac{1}{p_n} =1[/tex] forces the other [tex]$p_n$[/tex]s to be equal to [tex]$+\infty$[/tex].

Throughout we assume [tex]$p_n \in ]1,+\infty[$[/tex] for each [tex]$n$[/tex].

We're going to make induction on [tex]$N$[/tex].
For [tex]$N=2$[/tex], (gH) reduces to the standard Hölder inequality, so we have the basis.

Now we assume (gH) to be true for [tex]$N\geq 3$[/tex] and prove it holds also for [tex]$N+1$[/tex].
Put [tex]$\frac{1}{q} =\sum_{n=1}^N \tfrac{1}{p_n}$[/tex] so that the number [tex]$q$[/tex] is the Hölder conjugate of [tex]$p_{N+1}$[/tex].
Note that:

(*) [tex]$\left\lVert \prod_{n=1}^{N+1} f_n\right\rVert_1 =\int_X \Big\{\prod_{n=1}^N |f_n|\Big\} \cdot |f_{N+1}|\ \text{d} \mu$[/tex]

so it seems a good strategy using the standard Hölder inequality to find an upper bound for the r.h.s.; in order to do this we have to show that [tex]$\prod_{n=1}^N f_n \in L^q$[/tex]. Ok, let's do it!

For [tex]$n=1,\ldots , N$[/tex] we have:

[tex]$f_n \in L^{p_n} \quad \Rightarrow \quad |f_n|^q \in L^\frac{p_n}{q}$[/tex]

and [tex]$\sum_{n=1}^N\frac{1}{\frac{p_n}{q}} =q\ \sum_{n=1}^N \frac{1}{p_n} =1$[/tex], so we can apply the induction hypothesis to write:

[tex]$\left\lVert \prod_{n=1}^N f_n \right\rVert_q^q =\int_X \prod_{n=1}^N |f_n|^q\ \text{d} \mu$[/tex]
[tex]$\stackrel{\text{{\bf gH}}}{\leq} \prod_{n=1}^N \lVert |f_n|^q\rVert_\frac{p_n}{q}$[/tex]
[tex]$=\prod_{n=1}^N \left\{ \int_X (|f_n|^q)^\frac{p_n}{q}\ \text{d} \mu\right\}^\frac{q}{p_n}$[/tex]
[tex]$=\prod_{n=1}^N \left\{ \int_X |f_n|^{p_n} \ \text{d} \mu \right\}^\frac{q}{p_n}$[/tex]
[tex]$=\prod_{n=1}^N \lVert f_n\rVert_{p_n}^q <+\infty$[/tex]

hence [tex]$\prod_{n=1}^N f_n \in L^q$[/tex] and:

(**) [tex]$\left\lVert \prod_{n=1}^N f_n \right\rVert_q \leq \prod_{n=1}^N \lVert f_n\rVert_{p_n}$[/tex].

Hence the standard Hölder inequality applies to (*) and (**) implies:

[tex]$\left\lVert \prod_{n=1}^{N+1} f_n\right\rVert_1 \stackrel{\text{{\bf H}}}{\leq} \left\lVert \prod_{n=1}^N f_n\right\rVert_q\ \lVert f_{N+1}\rVert_{p_{N+1}} \leq \prod_{n=1}^{N+1} \lVert f_n\rVert_{p_n}$[/tex],

which is the inductive step.

Finally, Mathematical Induction Principle does its magic and the theorem follows.


[size=67]Post written while listening to this tune. If there are mistakes, blame them on Slowhand! [/size]

[gugo82]

Ok fu^2!
Your solution covers all the "easy" cases.

For the general case (when [tex]$p_n\in ]1,+\infty[$[/tex] for [tex]$n=1,\ldots ,N$[/tex]), may I suggest to put [tex]\frac{1}{q} :=\sum_{n=2}^N \frac{1}{p_n}[/tex] and to investigate the relationship between [tex]$p_1$[/tex] and [tex]$q$[/tex]?

[fu^2]

if $N=2$ the problem is trivial (i.e. we find the classic Holder inequality)

We go on using the induction method, therefore we suppose true the sentence for $N-1$.

If $f=\prod_{j=1}^{N}f_j$ and exist $j$ s.t. $p_j=+oo$, the sentence follows.
In fact, we can suppose that $j=1$, and we have $\sum_{j=2}^{N}=1$ so $\int|g|=\int|f_2...f_{N}|<=||f_2||_{p_2}...||f_{N}||_{p_N}$ by induction hipothesis.
Then, using the classic Holder inequality since $g=f_2...f_{N}\in L^1(\mu)$ and $f=f_1g$, we obtain that $||f||_1<=||f_1||_{oo}...||f_{N}||_{p_N}$.

So we can suppose that $p_j\ne +oo$ and $p_j\ne 1$ for each $j=1,...,N$.

but now i don't understand how i can solve this case (this is the interesting case ).

Maybe this night i'll came back to this post...