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Let $(X,\mathcal{A}, \mu)$ be an measure space. Let $f$ be an extended complex-valued $\mathcal{A}-$measurable function on $X$ such that $|f|<\infty$ $\mu$-a.e. on $X$. Suppose that $fg\in L^1(X,\mathcal{A}, \mu)$ for every $g\in L^q(X,\mathcal{A}, \mu)$. Show that $f\in L^{p}(X,\mathcal{A}, \mu)$. $(p>1, q>1, \frac{1}{p}+\frac{1}{q}=1)$

A similar problem I asked $fg\in L^1$ for every $g\in L^1$ prove $f\in L^{\infty}$

but I cannot find similar construction of counterexample in this case $f\notin L^p$. Almost the same problem Given $f\notin L^p$ find $g\in L^q$ s.t. $fg\notin L^1$, the construction given in the answer require $\mu(f\geq t)$ to be finite, but that is not a condition of the problem (nor can we derive from $f\notin L^p$).

Looking for an answer using little or no functional analysis technique.

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  • What do you mean by "without using advanced functional analysis"? For example, do you know the theorem that says that the continuous dual of $L^p$ is isometrically isomorphic to $L^q$ (when $p < \infty$)? – User8128 Apr 06 '16 at 23:22
  • @User8128, that is something I'll learn in the near future but at the chapter the problem show up in the book (first introductory chapter of $L^p$ space in real analysis book), that concept is not introduced yet. –  Apr 06 '16 at 23:26
  • Edited the last sentence a little bit. –  Apr 06 '16 at 23:35
  • Wow, I've been thinking about it for the last 15 min or so and I can't come up with anything. I'll post my solution that uses dual spaces, but I can't really get around that. Sorry man. Maybe this will help you get some ideas that lead you in the right direction. – User8128 Apr 06 '16 at 23:46
  • @User8128 thanks for your answer, I post another solution I get from my real analysis professor in case you want to see. –  Apr 08 '16 at 07:35

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Suppose that $fg \in L^1$ for all $g \in L^q$ but that there is no $C \in (0, \infty)$ such that $\| f g \|_{1} \le C \| g \|_q$ for all $g \in L^q$. Then we can find a sequence $g_n \in L^q$ such that $\| g_n \|_q = 1$ and $\| fg \|_1 > 3^n$ for all $n$ (do you see why this follows?). Then put $$g = \sum^\infty_{n=1} \frac{\lvert g_n \rvert}{2^n}. $$ Then $$\| g \|_q \le \sum^\infty_{n=1} \frac{\| g_n \|_q}{2^n} = \sum^\infty_{n=1} \frac{\| g_n \|_q}{2^n} = 1,$$ so $g \in L^q$ but $$\|fg\|_1 = \int \sum^\infty_{n=1} \frac{\lvert fg_n \rvert}{2^n} = \sum^\infty_{n=1} \frac{\| f g\|_1}{2^n} > \sum^\infty_{n=1} \frac{3^n}{2^n} = \infty.$$ Thus we have contradicted our assumption and so there must exist some $C$ such that $$\| fg \|_1 \le C \| g \|_q.$$ This shows that $g \mapsto \| fg\|_1$ is a bounded linear functional on $L^q$. By duality, the norm of this functional (which is less than $C$ as defined above) is the $L^p$ norm of $f$. Hence $\| f \|_p \le C < \infty.$ Thus $f \in L^p$.

User8128
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Indeed a very difficult construction...

Here is the answer I get from Professor's office hour,

Use the following lemmas(results from earlier HW),

Lemma 1: Let $(X,\mathcal{A},\mu)$ be a $\sigma-$finite but not finite measure space we can find a disjoint sequence $\cup E_n=X$ and $\mu(E_n)\in [1,\infty)$ for every $n\in \mathbb{N}$.

This one is easy to prove.

Lemma 2: Same assumption on $(X,\mathcal{A},\mu)$, there is a measurable $f\notin L^1$ but in $L^p$ for every $p\in (1,\infty]$.

Define $f=\frac{1}{n\mu(E_n)}$,$E_n$ defined in lemma 1, easy to check the rest.

Back to the problem, suppose $||f||_p=\infty$, define a measure $\nu$ by $d\nu=|f|^pd\mu$ not hard to check $\nu$ is $\sigma$-finite but not finite. By lemma 2 there is an $h\notin L^1(X,\nu)$, $h\in L^r(X,\nu)$ for $r>1$. Define $g=h|f|^{p-1}$. We have $\int_X|g|^qd\mu=\int_X|h|^q|f|^{pq-q}d\mu=\int_X|h|^q|f|^{p}d\mu=\int_X |h|^qd\nu<\infty$ so $g\in L^q(X,\mu)$. On the other hand $\int_{X}|fg|d\mu=\int_X |h||f|^pd\mu=\int_X |h|d\nu=\infty$ so $fg\notin L^1(X,\mu)$.