A counterexample for the Riesz integral
An easy counterexample from my talk in Cortona (the topic was Riesz rearrangement inequality; I focused on a proof with GMT's tools).
Enjoy.
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Let [tex]$f,g,h$[/tex] be measurable non-negative functions defined in [tex]$\mathbb{R}$[/tex] such that:
(F) for every [tex]$t>0$[/tex] the sets [tex]$\{ f>t\},\ \{ g>t\},\ \{ h>t\} $[/tex] have finite measure.
From Young's and Holder inequalities (and Fubini's theorem) it follows that the Riesz integral of [tex]$f,g,h$[/tex], i.e. the double integral:
[tex]$\mathfrak{I} (f,g,h)= \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(x)\ g(x-y)\ h(y)\ \text{d} x\ \text{d} y$[/tex]
is finite if [tex]$f\in L^\infty$[/tex] and [tex]$g,h\in L^1$[/tex].*
If one uses a more general version of Young's inequality, then it is possible to prove that the Riesz integral is finite even if [tex]$f\in L^p,\ g\in L^q,\ h\in L^r$[/tex] with the three exponents chosen in such a way that [tex]$\tfrac{1}{p}+\tfrac{1}{q}+\tfrac{1}{r}=2$[/tex].
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Exercise:
1. Find three functions [tex]$f,g,h$[/tex] for which (F) holds and s.t. the Riesz integral [tex]$\mathfrak{I} (f,g,h)$[/tex] is not finite.
2. Generalize the previous result to dimension [tex]$N$[/tex] (i.e. find [tex]$f,g,h:\mathbb{R}^N \to [0,+\infty[$[/tex] measurable functions s.t. (F) holds and [tex]$\mathfrak{I} (f,g,h)$[/tex] is not finite).
______________
* Obviously, if we were interested only in functions belonging to [tex]$L^p$[/tex] with [tex]$1\leq p<+\infty$[/tex], then we would have dropped condition (F) (because of the Chebyshev's inequality; can you say why?).
Enjoy.
***
Let [tex]$f,g,h$[/tex] be measurable non-negative functions defined in [tex]$\mathbb{R}$[/tex] such that:
(F) for every [tex]$t>0$[/tex] the sets [tex]$\{ f>t\},\ \{ g>t\},\ \{ h>t\} $[/tex] have finite measure.
From Young's and Holder inequalities (and Fubini's theorem) it follows that the Riesz integral of [tex]$f,g,h$[/tex], i.e. the double integral:
[tex]$\mathfrak{I} (f,g,h)= \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(x)\ g(x-y)\ h(y)\ \text{d} x\ \text{d} y$[/tex]
is finite if [tex]$f\in L^\infty$[/tex] and [tex]$g,h\in L^1$[/tex].*
If one uses a more general version of Young's inequality, then it is possible to prove that the Riesz integral is finite even if [tex]$f\in L^p,\ g\in L^q,\ h\in L^r$[/tex] with the three exponents chosen in such a way that [tex]$\tfrac{1}{p}+\tfrac{1}{q}+\tfrac{1}{r}=2$[/tex].
***
Exercise:
1. Find three functions [tex]$f,g,h$[/tex] for which (F) holds and s.t. the Riesz integral [tex]$\mathfrak{I} (f,g,h)$[/tex] is not finite.
2. Generalize the previous result to dimension [tex]$N$[/tex] (i.e. find [tex]$f,g,h:\mathbb{R}^N \to [0,+\infty[$[/tex] measurable functions s.t. (F) holds and [tex]$\mathfrak{I} (f,g,h)$[/tex] is not finite).
______________
* Obviously, if we were interested only in functions belonging to [tex]$L^p$[/tex] with [tex]$1\leq p<+\infty$[/tex], then we would have dropped condition (F) (because of the Chebyshev's inequality; can you say why?).
Risposte
"gugo82":
1. Find three functions [tex]$f,g,h$[/tex] for which (F) holds and s.t. the Riesz integral [tex]$\mathfrak{I} (f,g,h)$[/tex] is not finite.
We said that:
[tex]$f\in L^\infty \text{ and } g,h\in L^1 \quad \Rightarrow \quad \mathfrak{I}(f,g,h) <+\infty$[/tex];
moreover, by Tonelli's theorem, the Riesz integral [tex]$\mathfrak{I} (f,g,h)$[/tex] is finite iff we can compute the value of [tex]$\mathfrak{I} (f,g,h)$[/tex] as iterated integrals (recall [tex]$f,g,h \geq 0$[/tex]): therefore we have the following identity:
[tex]$\mathfrak{I} (f,g,h) =\int_{-\infty}^{+\infty} f(x)\ g*h(x)\ \text{d}x$[/tex],
where [tex]$g*h$[/tex] denotes the convolution of [tex]$g$[/tex] and [tex]$h$[/tex] (which has to be finite a.e., otherwise the iterated integral w.r.t. [tex]$y$[/tex] would not exist).
Therefore in order to find a counterexample it seems a good idea to find two functions [tex]$g,h$[/tex] not in [tex]$L^1$[/tex] s.t. (F) holds but the convolution [tex]$g*h$[/tex] is not finite in a large set.
Now set:
[tex]$g(x):=\max \{ \tfrac{1}{|x|} -1 ,0\} =: h(x)$[/tex]
and note that [tex]$g=h$[/tex] is compactly supported in [tex]$[-1,1]$[/tex] and has the property (F); moreover [tex]$g=h\notin L^1([-1 ,1])$[/tex] 'cause it critically blows up in [tex]$0$[/tex].
The convolution [tex]$g*h$[/tex] is formally given by the integral [tex]$g*h(x):=\int_{-\infty}^{+\infty} g(x-y)\ h(y)\ \text{d} y$[/tex], but for [tex]$x$[/tex] sufficiently small we get:
[tex]$g*h(x)=\begin{cases} \int_{-1}^{1+x} \frac{1-|y|}{|y|}\ \frac{1-|y-x|}{|y-x|}\ \text{d} y &\text{, if $x<0$} \\ \int_{-1}^{1+x} (\tfrac{1-|y|}{|y|})^2\ \text{d} y &\text{, if $x=0$} \\ \int_{-1+x}^1 \frac{1-|y|}{|y|}\ \frac{1-|y-x|}{|y-x|}\ \text{d} y &\text{, if $0< x$}\end{cases}$[/tex]
[tex]$=\begin{cases} +\infty &\text{, if $x\neq 0$} \\ \text{a number} &\text{, if $x=0$} \end{cases}$[/tex].
Finally choose [tex]$\varepsilon >0$[/tex] much smaller than [tex]$1$[/tex] and set [tex]$f:=\chi_{[-\varepsilon ,\varepsilon]}$[/tex]; obviously [tex]$f$[/tex] has property (F) and the integral:
[tex]$\int_{-\infty}^{+\infty} f(x)\ g*h(x)\ \text{d} x =\int_{-\varepsilon}^{\varepsilon} g*h(x)\ \text{d} x$[/tex]
is not finite, as we wanted.
"gugo82":
2. Generalize the previous result to dimension [tex]$N$[/tex] (i.e. find [tex]$f,g,h:\mathbb{R}^N \to [0,+\infty[$[/tex] measurable functions s.t. (F) holds and [tex]$\mathfrak{I} (f,g,h)$[/tex] is not finite).
It suffices to choose [tex]$f,g,h$[/tex] to be radial, [tex]$f$[/tex] being the characteristic function of a small ball around [tex]$o$[/tex] and [tex]$g=h$[/tex] being as in 1 (with [tex]$|x|$[/tex] replaced by [tex]$|x|^N$[/tex]).
"gugo82":
* Obviously, if we were interested only in functions belonging to [tex]$L^p$[/tex] with [tex]$1\leq p<+\infty$[/tex], then we would have dropped condition (F) (because of the Chebyshev's inequality; can you say why?).
Chebyshev's inequality states that if [tex]$f\in L^p$[/tex] then:
[tex]$|\{ |f|>t\}| \leq \frac{1}{t^p}\ \lVert f\rVert_p^p$[/tex]
therefore if [tex]$f$[/tex] is non-negative and of class [tex]$L^p$[/tex] then it has property (F) "for free".
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