Is it possible to have $g\colon\Omega\to\Bbb C$ which defines an unbounded functional?

Question

Let $\Omega$ be an infinite space with a nontrivial measure $\mu$. We define $L^p$ spaces as usual, then for $1<p<\infty$ if $\frac1p+\frac1q=1$, then $(L^p)^*=L^q$. This is all pretty much a classical theorem. Fix $1<p<\infty$ for the discussion.

Suppose $g\colon\Omega\to\Bbb C$ is a function such that for every $f\in L^p$, the pointwise multiplication $gf\in L^1$. Is it necessarily the case that $g\in L^q$?

It is not hard to show that $f\mapsto\int_\Omega fg\ d\mu$ is a linear functional. So $g\in L^q$ if and only if this functional is continuous. It is even not hard to show that if $g$ is measurable and $\mu$ is $\sigma$-finite then $g\in L^q$.

Is it possible to have $g\notin L^q$ such that for every $f\in L^p$, $gf\in L^1$?

Why is this difficult? It is consistent with $\sf ZF+DC$ that every linear functional on a Banach space is continuous. In such case $g$ has to be in $L^q$, since the functional is indeed continuous. This means that if there is a counterexample it has to involve the axiom of choice.

We can even point further that such example has to involve sets without the Baire property since if every set of reals (equiv. complex numbers) has the Baire property, then every linear functional on a Banach space is continuous.

Answer 1

Let $X=\{0\}$, with the unique $\sigma$-algebra. Define $\mu$ by $\mu(X)=\infty$. Let $g(0)=1$. Then $gf\in L^1$ for every $f\in L^p$, simply because $f\in L^p$ implies $f=0$. But $g\notin L^q$.

Of course $g$ does nonetheless define a bounded linear functional on $L^p$, even though $g\notin L^q$ and $(L^p)^*=L^q$. So it's not quite true that "if this functional is continuous then $g\in L^q$"; the truth is that if this functional is continuous then there exists $h\in L^q$ which defines the same functional.

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So that answers the question you asked, but not the related question which you thought was equivalent but isn't:

Q2 If $gf\in L^1$ for every $f\in L^p$ then is the functional $f\mapsto\int gf$ continuous on $L^p$?

I think the answer is yes, it must be continuous, but I don't quite have a proof. The comments you make about set theory this and that of course really apply to the second question.

It's not clear to me what happens if we don't assume that $g$ is measurable, is the thing. If $g$ is measurable then it seems to me that yes, it does define a continuous linear functional:

If $E$ is $\sigma$-finite then everything is nice if we restrict everything to $E$, and in particular $\int_E|g|^q<\infty$. Since a countable union of $\sigma$-finite sets is $\sigma$-finite it follows that there exists $B<\infty$ so that $\int_E|g|^q\le B^q$ for every $\sigma$-finite $E$. Now if $f\in L^p$ there exists a $\sigma$-finite $E$ such that $f$ vanishes off $E$, and hence $|\int fg|\le B||f||_p.$

Edit: In fact this works without assuming that $g$ is measurable. Because if $E$ is $\sigma$-finite it's easy to see that $g\chi_E$ is measurable, and then the rest of the argument goes through. (PhoemueX's answer gives the proof I wanted...)

So the answer to Q2 is yes, period. Twice. Of course no answer to the original question is possible without a definition of "non-trivial".

(Yes, the $\sigma$-finite case is just like $\Bbb R^n$.)

Answer 2

I don't know much about the set theoretic subtleties you are talking about, but using ZFC, the functional is always continuous. To see this, consider the well-defined(!) linear map $$ \Phi : L^p \to L^1, g \mapsto fg. $$

It is easy to see that this map has a closed graph: If $g_n \to g$ in $L^p$ and $f g_n \to h$ in $L^1$, then switching to subsequences, we get the convergence almost everywhere. Note that this does not use measurability of $f$, only that of $g_n,g, f g_n$ and $h$, all of which hold by assumption. But this implies $h = fg =\Phi(g)$ a.e., so that the graph of $\Phi$ is closed.

Thus, $\Phi$ is bounded by the closed graph theorem, which easily yields the claim.

Note though (as David Ullrich pointed out) that this does not imply $f\in L^q$ in general.

Answer 3

No, $g$ need not be in $L^q$, even for what I suspect is a non-trivial measure.

Let $X$ be the disjoint union of uncountably many disjoint copies of $[0,1]$; call them $I_j$. Say $E$ is measurable if (i) $E\cap I_j$ is Lebesgue measurable for every $j$ and (ii) either the set of $j$ such that $E\cap I_j\ne\emptyset$ is countable or the set of $j$ such that $I_j\cap E\ne I_j$ is countable. Let $\mu(E)$ be the sum of the Lebesgue measures of $E\cap I_j$.

Choose $g=\chi_E$ where $E\cap I_j$ is a singleton for every $j$. Then $fg\in L^1$, in fact $fg$ is measurable and $fg=0$ almost everywhere, for every $f\in L^p$. But $g$ is not measurable.

(Not "hard", because the reasons you gave for this being "hard" apply to obtaining a discontinuous linear functional; the functional defined by $g$ is continuous.)

If this one is also hatefully trivial please define "non-trivial".