36

Let $f$ be a measurable function on a measure space $X$ and suppose that $fg \in L^1$ for all $g\in L^q$. Must $f$ be in $L^p$, for $p$ the conjugate of $q$? If we assume that $\|fg\|_1 \leq C\|g\|_q$ for some constant $C$, this follows from the Riesz Representation theorem. But what if we aren't given that such a $C$ exists?

Guy Fsone
  • 25,237
user15464
  • 12,092

3 Answers3

42

So that $L^p$ and $L^q$ are duals of each other, let $1\lt p,q\lt\infty$, as well as $\frac1p+\frac1q=1$.

Furthermore, so that we can apply the Riesz Representation Theorem, we should assume that we are working in a measure space where the measure is countably additive on a regular, locally compact Hausdorf space.

Without these conditions, the counterexample given by Danny Pak-Keung Chan and David C. Ullrich in Danny's answer shows that the answer is no.

Suppose that $fg\in L^1$, but there is no $C$ so that $\|fg\|_{L^1}\le C\|g\|_{L^q}$. Without loss of generality, we can assume all functions are positive. Suppose we have a sequence of $L^q$ functions $\{g_k:\|g_k\|_{L^q}=1\}$ where $\int|fg_k|\;\mathrm{d}x>3^k$. Set $g=\sum\limits_{k=1}^\infty2^{-k}g_k$. $\|g\|_{L^q}\le1$ yet $fg\not\in L^1$. Thus, there must be a $C$ so that $\|fg\|_{L^1}\le C\|g\|_{L^q}$. Then, as you say, apply the Riesz Representation Theorem.

robjohn
  • 353,833
  • Since Didier said one should let oneself be inspired by this question: note that a very similar gliding hump argument is used by Sokal's article containing a proof of the uniform boundedness principle, posted in a comment there. This ties in nicely with Davide's suggestion... – t.b. Sep 03 '11 at 09:23
  • @Theo: Thanks for the references. Reading his paper, my answer is quite similar to Alan Sokal's proof of the Uniform Boundedness Principle. – robjohn Sep 03 '11 at 13:10
  • @robjohn: do you mind elaborating on your proof a bit. the part where you showed that $|fg|{L^1} \leq C |g|{L^q}$? Thanks – Kuku May 07 '12 at 22:08
  • @kuku: This is an indirect argument. I started by assuming there was no such $C$ so that $|fg|{L^1}\le C|g|{L^q}$ for all $g\in L^q$. Using that assumption, I derived a contradiction, which shows that the assumption was false. Thus, there must be a $C$ so that $|fg|{L^1}\le C|g|{L^q}$. – robjohn May 07 '12 at 22:37
  • sorry if I wasn't clear before. I was wondering if you could expand on the derivation of the contradiction. for example why $g=\sum\limits_{k=1}^\infty2^{-k}g_k|g|_{L^q}$ is less than one and why it implies that $fg\notin L^1$. – Kuku May 07 '12 at 22:47
  • 3
    @kuku: Since $|g_k|{L^q}=1$, $$ \begin{align} |g|{L^q} &\le\sum_{k=1}^\infty 2^{-k}|g_k|{L^q}\ &=1 \end{align} $$ Since $\int fg_k;\mathrm{d}x>3^k$ (we were assuming all functions were non-negative) $$ \begin{align} \int fg;\mathrm{d}x &=\sum{k=1}^\infty2^{-k}\int fg_k;\mathrm{d}x\ &>\sum_{k=1}^\infty2^{-k}3^k \end{align} $$ which diverges. – robjohn May 07 '12 at 23:17
  • @robjohn How does Riesz representation theorem solve the problem? – Idonknow Apr 13 '19 at 15:43
  • @Idonknow: The Riesz Representation Theorem ensures there is an $f\in L^p$ for the linear functional on $L^q$ given by $|fg|_1$. – robjohn Apr 13 '19 at 16:18
  • @robjohn I am sorry. I still do not see how RRT implies that $f\in L^p$ whenever we have $fg\in L^1$ for all $g\in L^q.$ – Idonknow Apr 14 '19 at 02:55
  • 3
    @Idonknow: We have shown that $T:g\mapsto\int fg,\mathrm{d}x$ is a bounded linear functional from $L^q$ to $\mathbb{R}$. Riesz says that there is an $h\in L^p$ so that $\int fg,\mathrm{d}x=\int hg,\mathrm{d}x$ for all $g\in L^q$. To show that $f=h$ a.e., assume not. Then, for some $0\lt\epsilon\le1$, $|f-h|\ge\epsilon$ on a set $E$ where $\epsilon\le m(E)\le1$. Set $$g(x)= \frac{|f(x)-h(x)|}{f(x)-h(x)} [x\in E]\in L^q$$ Then $0=\int(f-h)g,\mathrm{d}x\ge\epsilon^2$. Contradiction. – robjohn Apr 14 '19 at 09:41
  • Someone asked me just now, years later, what the error was with this answer. I inserted a detailed explanation into Danny's answer below, if you're curious... – David C. Ullrich May 06 '20 at 14:18
  • @DavidC.Ullrich: the question said that the implication was true if there were such a $C$, but what happens if there is no such $C$ given. Since this holds when a $C$ is given, is not the measure space in the counterexample excluded? – robjohn May 06 '20 at 15:23
  • ??? No, the statement that it holds if there exists $C$ is not an assumption, it's an assertion. Doesn't indicate anything that I can see, it's simply an error. – David C. Ullrich May 06 '20 at 16:02
  • It is stated in the question: "If we assume that $|fg|_1 \leq C|g|_q$ for some constant $C$, this $\left[,fg\in L^1,\forall g\in L^q\implies f\in L^p,\right]$ follows from the Riesz Representation theorem." I took this statement as a given (which limits the measure spaces considered). The quoted statement does not hold for the measure space given in the counterexample. So, yes, without limiting the measure spaces, the question is not well-posed. – robjohn May 06 '20 at 17:14
  • @DavidC.Ullrich: Does the caveat I've added to my answer help some? – robjohn May 06 '20 at 17:27
  • Not to be repetitive, but I don't follow what you're saying at all. In particular I don't see how "If we assume that there exists $C$ such that etc then this follows from RRT" limits the measure spaces in question. If we assume that then the result does _not_follow from RRT; that statement is simply false. – David C. Ullrich May 06 '20 at 22:02
  • Does the addition help? Whatever. It's your post. Reading "All the answer above does is relieve the need for a C to be given, but it does require the statement above to be true." it took me a minute to figure out various things were referring to – David C. Ullrich May 06 '20 at 22:03
  • In any case, it still looks at the start like you're saying the answer is yes. I don't know why you don't want to simply say the answer is no, since that's what it is,after all. Or to be more fair, "no without $\sigma$_finiteness, even for $1<p<\infty$". – David C. Ullrich May 06 '20 at 22:05
  • @DavidC.Ullrich: Okay. I have expressed the assumptions I made. – robjohn May 06 '20 at 22:51
  • Fabulous. In fact the counterexample was completely given by Danny before I started typing - I just added a bit of explanation regarding the nature of the error in the wrong answers. (In particular the problem is not with $(L^q)^*=L^p$.) – David C. Ullrich May 06 '20 at 23:31
  • Okay. I have corrected the attribution. I had it correctly in my aborted caveat and messed it up when adding the preamble. – robjohn May 06 '20 at 23:40
10

Oops. Someone asked me a question in a comment to this answer. So I sort of assumed it was an Answer from me, and replied to the comment with an edit. Oops, no it's an Answer from someone else. Sorry, didn't mean to revise his post. Although I didn't change anything, just added more content, so I'm tentatively leaving this as it is.

Original Answer

The answer is NO. All the proofs presented in above are invalid. Counter-example: Consider the measure space $(X,\mathcal{M},\mu)$, where $X=\{x_0\}$, $\mathcal{M}= \{\emptyset,X\}$, and $\mu(\emptyset)=0$, $\mu(X)=\infty$. Let $p,q\in(1,\infty)$ be such that $\frac{1}{p}+\frac{1}{q}=1$. In this case, $L^q=\{0\}$. Therefore, if we define $f:X\rightarrow\mathbb{R}$ by $f(x_0)=1$, then $fg\in L^1$ for all $g\in L^q$. However, $f\notin L^p$.

(Third-party) Edit

Just now I've been asked what's the problem with the accepted answer. It seems maybe worthwhile explaining, since it's actually a bit more subtle than I realized at first.

With apologies lest I misrepresent what someone had in mind; nobody included details for the wrong part, so to explain where the error is I have to start by inserting my own version of the details:

So assume that $fg\in L^1$ for all $g\in L^p$. Just to give a name to what we've correctly proved above:

Lemma. $||fg||_1\le C||g||_p$.

At this point everyone says that the lemma plus RRT imply $f\in L^p$. The wrong argument is this: Define $\Lambda g=\int fg$. Then $\Lambda\in (L^q)^*$. But RRT says $(L^q)^*=L^p$, hence $f\in L^p$.

The problem arises only at the very last step. At first I assumed the problem was with $(L^q)^*=L^p$ if the measure is not $\sigma$-finite, but that's not quite it:

Assume in addition that $1<p<\infty$. Then it is true that $(L^q)^*=L^p$ for any measure. But that doesn't show $f\in L^p$. Again we need to insert details: RRT does show that there exists a unique $F\in L^p$ with $\int Fg=\int fg$ for all $g\in L^q$. And now

Error. If $\int Fg=\int fg$ for all $g\in L^q$ then $f=F$ almost everywhere.

That's easy in the $\sigma$-finite case, but as we see in the example at the top, in general there need not be enough $L^q$ functions to separate points of $L^p$.

Community
  • 1
Danny Pak-Keung Chan
  • 12,923
  • The parts showing that $||fg||_1\le C||g||_q$ are valid; what's invalid is the inference (in the OP and elsewhere) that this implies $f\in L^q$. – David C. Ullrich Jul 18 '19 at 13:07
  • @DavidC.Ullrich Where is the mistake in robjohns proof? – 2132123 May 05 '20 at 21:47
  • @2132123 That's a better question than I thought it was at first. See edit above... – David C. Ullrich May 06 '20 at 13:59
  • @DavidC.Ullrich Thank you for giving more information a year later... one more question if I may. Robjohn uses a single function to show $F=f$ , $g(x)=\frac{|f(x)−h(x)|}{f(x)−h(x)}[x∈E]∈L^q$ where $E$ is a finite subset of a set on where $|f-g|>\epsilon$. His error was assuming such non empty $E$ exists, correct? As in Danny's case that set would have to be empty. – 2132123 May 10 '20 at 04:07
  • @2132123 Not non-empty and finite; the problem was assuming there exists $E$ with strictly positive finite measure. – David C. Ullrich May 10 '20 at 11:19
  • Interesting, but could we find a non-trivial example, say with the 1-dimensional Hausdorff measure on R^2 ? – LLD Feb 28 '21 at 12:07
7

I would like to add another answer to this question. Consider the linear functional $T:L^q\to\mathbb{C}$ defined by $$Tg=\int gf.$$

It is sufficient to prove that $T$ is bounded. To this end, assume that $g_n\in L^q$ is such that $\|g_n\|_q\to 0$.

We can extract a subsequence of $g_n$ (not relabeled) such that $$g_n(x)\to 0,\ |g_n(x)|\le h(x),\ \mbox{a.e.},\tag{1}$$

where $h\in L^q$ (this partial converse of Lebesgue theorem, can be found, for example, in Rudin's book "Real and Complex Analysis", Theorem 3.12, or in Brezis book "Functional Analysis", Theorem 4.9). It follows from $(1)$ that $$g_n(x)f(x)\to 0,\ |g_n(x) f(x)|\leq h(x)|f(x)|,\ \mbox{a.e}.\tag{2}$$ Note that by hypothesis, $h|f|\in L^1$, therefore, we can apply Lebesgue theorem to conclude that $$Tg_n\to 0.\tag{3}$$

As every subsequence of $g_n$, has a subsequence which satisfies $(1)$, we conclude that $(3)$ is true for the whole sequence.

Tomás
  • 22,985
  • I don't understand: are you trying to show that $f\in L^q$?. – robjohn Jun 13 '14 at 20:05
  • You've shown that $T$ is continuous. Therefore, it is bounded. Thus, you can use Riesz to get an $\tilde{f}\in L^p$ so that $\int\tilde{f}g,\mathrm{d}x=Tg=\int fg,\mathrm{d}x$ for all $g\in L^q$. Then you can show that $f=\tilde{f}$ a.e. so that $f\in L^p$. However, the proof of $(1)$ seems quite non-trivial. – robjohn Jun 13 '14 at 23:36
  • @robjohn, I would say that non-trivial is too much. If you take a look in the proof of $(1)$, you will see that all amounts to prove that, any subsequence $g_{n_i}$ of $g_n$, which satisfies $$|g_{n_i+1}-g_{n_i}|_q<2^{-i},$$ satisfies $(1)$ and it can be proved only by using the basics of measure theory. – Tomás Jun 13 '14 at 23:52
  • Yes, we can choose a sub-sequence $|g_n|q\le2^{-n}$ and let $h(x)=\sum\limits{n=1}^\infty|g_n(x)|$. Then $h\in L^q$ and $|g_n(x)|\le h(x)$. How do we arrange it so that $\lim\limits_{n\to\infty}g_n(x)=0$ for all (or even almost all) $x$? That is the part that seems less than trivial. – robjohn Jun 14 '14 at 00:52
  • @robjohn, do you think my answer is totally useless and deserves to be deleted? If so, I will have no problem in deleting it. – Tomás Jun 14 '14 at 01:00
  • I don't think it is useless, I am just trying to understand how it works. I don't have my copy of Rudin available, so I can't look up the proof of $(1)$. I would also like to understand how we can extend from specially chosen sub-sequences vanishing to the whole sequence vanishing. – robjohn Jun 14 '14 at 02:47
  • Ok @robjohn. To prove that $h=0$ a.e., it is sufficient to prove that $|h|_q=0$. It can be done by means of Fatou's lemma, indeed, note that $|h|_q\le |h-g_n|_q+|g_n|_q$. On the other hand, to extend the result to the whole sequence, we can use the following topological lemma: Let $X$ be a metric space and $x\in X$. Assume that $(x_n)\subset X$ and that every subsequence of $x_n$ converges to $x$. Then, the whole sequence does converge to $x$. – Tomás Jun 14 '14 at 11:22
  • Sorry @robjohn, again my notation is confused. My $h$ in the last comment is defined by $h(x)=\lim g_n(x)$, which is different of yoru $h$. Note that my $h$ is well defined, because $\sum_{n=1}^\infty |g_n(x)|\in L^q$, which implies in particular that $\sum_{n=1}^\infty |g_n(x)|<\infty$ a.e. – Tomás Jun 14 '14 at 12:29
  • Indeed, $|h|q=0\implies h=0\text{ a.e.}$ However, $\lim\limits{n\to\infty}|g_n|q=0$ does not imply that $\lim\limits{n\to\infty}g_n(x)$ even exists anywhere, so we need more justification why we can pick a subsequence so that $g_n(x)\to0$. – robjohn Jun 16 '14 at 22:01
  • You are right @robjohn, however, we have more than it. In fact, we have that $\sum_{n=1}^\infty |g_n(x)|\in L^q$, which implies, in particular, that this sum is finite everywhere. On the other hand, note that $$g_n(x)=\sum_{i=1}^n g_i(x)-\sum_{i=1}^{n-1}g_i(x)$$ – Tomás Jun 17 '14 at 01:38
  • @robjohn To show that $f=\widetilde{f}$, we may nend the whole space is $\sigma$-finite. – Xiang Yu Nov 12 '15 at 10:16
  • @XiangYu: Since $\int\tilde{f}g,\mathrm{d}x=Tg=\int fg,\mathrm{d}x$ for all $g\in L^q$, we have that $|\tilde{f}-f|_p=0$, which means that $\tilde{f}=f$ (a.e.) – robjohn Nov 12 '15 at 15:31
  • @robjohn How can we conclude $|\widetilde{f}-f|_{L^p}=0$ from $\int \widetilde{f}g=\int fg$? – Xiang Yu Nov 12 '15 at 15:41
  • @robjohn If $E$ is a measurable set with finite measure, then $1_E\in L^q$. Let $g=1_E$, we have $\int\widetilde{f}1_E=\int f1_E$ for all measurable set $\mu(E)<\infty$. If the whole space $X$ is $\sigma$-finite, then using proof by contradiction, we can show that $\widetilde{f}=f$ a.e. – Xiang Yu Nov 12 '15 at 15:47
  • Let $\overline{g}=(\tilde{f}-f),|\tilde{f}-f|^{p-2}$, then $0=\int(\tilde{f}-f)g,\mathrm{d}x=|\tilde{f}-f|_p^p$ – robjohn Nov 12 '15 at 16:10
  • All the proofs are invalid. I find a simple counter-example. – Danny Pak-Keung Chan May 12 '19 at 17:49