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This is the second part of the Exercise 2 in Chapter 8 of Measure and Integral by Zygmund and Wheeden:

Show also that for real valued $f\notin L^p(E)$, there exist a function $g\in L^q(E)$, $\frac{1}{p}+\frac{1}{q}=1$, s.t. $fg\notin L^1(E)$. (Construct $g$ of the form $\sum a_kg_k$ for appropriate $a_k$ and $g_k$ satisfying $\int_E fg_k\to \infty$.)

In the following the integrals are taken over $E$.

Trying to follow the hint, if we assume $1\lt p\lt\infty$ note that $a_k=k^{-(q+1)}$, $b_k=k$ satisfies $$\begin{cases} \sum a_kb_k\lt\infty\\ \sum a_kb_k^q=\infty\end{cases}$$ In view of $$\begin{align*} &\Vert g\Vert_q=\Vert \sum a_kg_k\Vert_q\leq \sum a_k\Vert g_k\Vert_q\\ &\Vert fg \Vert_1=\int fg =\sum a_k\int fg_k \end{align*}$$ we just need $g_k$ such that $\Vert g_k\Vert_q=k$ and $\int fg_k=k^q$. Suppose that there exist a set $A_k\subseteq E$ such that $\int_{A_k} \vert f\vert^p=k^q$, then the sequence given by $$g_k=\vert f\vert^{p-1}\chi_{A_k}=\vert f\vert^{p/q}\chi_{A_k}$$ will do the job. I need to justify the existence of the $A_k$s. So, I'm looking for something like the following.

Let $E\subseteq \mathbb{R}^d$. Let $f$ a function with $$\int_E \vert f\vert=\infty.$$ My question is:
Given $t\in[0,\infty[$, does there exist a set $A_t\subseteq E$ s.t. $$\int_{A_t}\vert f\vert=t?$$

Let $C_t$ the cube around the origin with volume $t\geq 0$. I was trying to prove that the function $F:[0,\infty[\to\mathbb{R}\cup\{\infty\}$ given by $$F(t)=\int_{C_t\cap E}\vert f\vert$$ is continuous. My idea seems to have no future, because for example $f$ could be $\infty$ in any subset of positive measure.

If this is not possible, how can I approach this problem?

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Is the first integral supposed to be over the set $E$? –  Arturo Magidin Dec 10 '11 at 22:47
    
Yes @ArturoMagidin, I'll correct it, but feel free in take the integrals over all the space if necessary to deduce something satisfactory. –  leo Dec 10 '11 at 22:51
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You need some sort of conditions on $f$, and possibly on $E$; if you are allowing functions to take the value $\infty$, then simply defining $f$ as the constant function $\infty$ will make it impossible to find $A_t$ for any $t\gt 0$. –  Arturo Magidin Dec 10 '11 at 22:53
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If you assume that $f:E \to {\mathbb R}$ is measureable and the measure of $\{x: f(x)=\infty\}\cap E$ is zero, then it is true. –  Jeff Dec 10 '11 at 23:35
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@Jeff: can you please elaborate it in an answer? –  leo Dec 11 '11 at 0:13

1 Answer 1

up vote 5 down vote accepted

We first show that for every $(a_n)_{n}$ not in $\ell^p$, there exists $(b_n)_{n}$ in $\ell^q$ such that $(a_nb_n)_{n}$ is not in $\ell^1$. To this end, assume without loss of generality that every $a_n$ is positive and introduce the sequences $$ A_n=1+\sum\limits_{k\leqslant n}a_k^p,\qquad b_n=\frac{a_n^{p-1}}{A_n}. $$ To show that $(b_n)_{n}$ is in $\ell^q$, note that $$ (q-1)b_n^q=(q-1)\frac{a_n^{p}}{A_n^q}=(q-1)\frac{A_n-A_{n-1}}{A_n^q}\leqslant\frac1{A_{n-1}^{q-1}}-\frac1{A_{n}^{q-1}}. $$ The last inequality stems from the fact that $1+(q-1)u^q\geqslant qu^{q-1}$ for every $u$ in $[0,1]$, used for the value $u=A_{n-1}/A_n$.

Summing this, one gets $$ (q-1)\sum\limits_{n}b_n^q\leqslant1. $$ To show that $(a_nb_n)_{n}$ is not in $\ell^1$, note that $a_kb_k=a_k^p/A_k=(A_k-A_{k-1})/A_k$ for every $k$ hence, for every $m\geqslant n$, $$ \sum\limits_{k=n+1}^ma_kb_k\geqslant\sum\limits_{k=n+1}^m(A_k-A_{k-1})/A_m=(A_m-A_{n})/A_m. $$ By hypothesis $(A_k)_k$ is unbounded hence, for every $n$, there exists $m\geqslant n+1$ such that $A_m\geqslant2A_n$. For such $n$ and $m$, $\sum\limits_{k=n+1}^ma_kb_k\geqslant\frac12$. This proves that $(a_nb_n)_{n}$ is not in $\ell^1$.


Turning to the case of functions, assume without loss of generality that $f$ is nonnegative and consider the case when $E=\mathbb R_+$ with the Lebesgue measure. Define $F:x\mapsto1+\int\limits_0^xf^p$ and $g=f^{p-1}/F$.

Then $g^q=f^p/F^q$ and $f^p=F'$ hence $$ (q-1)\int_{\mathbb R^+}g^q=(q-1)\int_{\mathbb R^+}\frac{F'}{F^q}=\left[\frac{-1}{F^{q-1}}\right]_0^{+\infty}\leqslant1. $$ Likewise $fg=f^p/F=F'/F$ hence $$ \int_{\mathbb R^+}fg=\left[\log F\right]_0^{+\infty}, $$ which is infinite since $F(x)\to\infty$ when $x\to+\infty$. Thus $g$ is in $L^q$ and $fg$ is not in $L^1$.

The case of functions defined on a general space $E$ can be deduced from this one. Namely, one considers the function $g:x\mapsto f(x)^{p-1}/h(f(x))$ where, for every nonnegative $t$, $$ h(t)=1+\int\limits_Ef^p(y)\,[f(y)\geqslant t]\,\mathrm d\mu(y). $$ Then, if $\mu(f\geqslant t)$ is finite for every positive $t$, $fg$ is not in $L^1$ and $g$ is in $L^q$.

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Related to this. –  Did Dec 11 '11 at 10:49
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+1. Another related thread establishing the contrapositive of the statement in the title. –  t.b. Dec 11 '11 at 13:39
    
@leo, sorry but I have no idea what your last comment is about nor how it is related to the rest of this page. –  Did Dec 13 '11 at 8:22
    
@leo, do you sometimes answer the questions they ask you in comments? –  Did Dec 16 '11 at 22:27

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