Linear map between $L^p$ and $L^q$.

Question

Let $h: \mathbb{R} \rightarrow \mathbb{R}$ be a real measurable function. If $hf \in L^q(\mathbb{R})$ for all $f \in L^p(\mathbb{R})$ then what can we say about $h$?

What I have:

(i) As $L^q, L^p$ are both Banach spaces, by closed graph theorem with the fact that $hf_n \xrightarrow{L^q} y$, and $f_n \xrightarrow {L^p} f $ implies existence of a.e. convergent subsequence, we deduce the linear map $$T_h: L^p \rightarrow L^q, \quad , f \mapsto fh$$ is bounded.

(ii) From (i), I can deduce if $p'$ were the conjugate of $p$, and $q=1$, then $h \in L^{p'}$:

Take simple functions $\varphi_n \rightarrow h$, such that $|\varphi_n| \le |h|$ and has compact support. Let $f_n:= \frac{|\varphi_n|^{p'-1} \overline{sgn \, h} }{||\varphi_n||^{p'-1}_{p'}}$ so $||f_n||_p = 1 $. By Fatou's and boundedness of $T_h$, \begin{align*} ||h||_{p'} & = \liminf ||\varphi_n||_{p'} = \liminf \int |f_n \varphi_n | \\ & \le \liminf \int f_n h = ||T_h f_n||_1 \le || T_h || \end{align*}

I was told that one can deduce more about the properties of $h$. What is there?

Answer 1

The god-awful mess below is really nothing but a lot of trivial details. If there's a point to it all it's this:

If $p,q,r$ are such that $h\in L^r$ implies $hf\in L^q$ for every $f\in L^p$ then this fact "must" follow from Holder's inequality. And the converse, that if $h$ has that property then $h\in L^r$ for a certain $r$, "must" follow from what's sometimes called the reverse Holder inequality, aka $(L^p)^*=L^{p'}$.

The cases $q=1$ and $q=p$ are very easy:

$hf\in L^1$ for every $f\in L^p$ if and only if $h\in L^{p'}$.

<p>$hf\in L^p$ for every $f\in L^p$ if and only if $h\in L^\infty$.</p>

This looks like the condition $h\in L^r$ should have something do with it, where $$\frac1r+\frac1p=\frac1q.$$

Let's see. Yes, $T_h$ must be bounded, so $$\int|h|^q|f|^q\le c||f||_p^q.$$

Assume $p\ge q$. Then $f\in L^p$ implies $|f|^q\in L^{p/q}$. And in fact $$\{|f|^q:f\in L^p\}=\{|\phi|:\phi\in L^{p/q}\},$$so we have $|h|^q\phi\in L^1$ for every $\phi\in L^{p/q}$, which implies that $|h|^q\in L^{(p/q)'}$. Let's see, $$\frac qp+\frac{p-q}p=1,$$so $(p/q)'=p/(p-q)$. So $|h|^q\in L^{(p/q)'}$ says $h\in L^r$, where $r=pq/(p-q)$. Yes, that's the same as $1/r+1/p=1/q$, cool.

We've proved the harder half of this:

Suppose $p\ge q$ and $1/r+1/p=1/q$. Then $hf\in L^q$ for every $f\in L^p$ if and only if $h\in L^r$.

The other half must be just Holder's inequality. Suppose $h\in L^r$. Note that $q/p+q/r=1$, so $p/q$ and $r/q$ are conjugate exponents. Hence $$\int|hf|^q\le\left(\int|h|^r\right)^{q/r}\left(\int|f|^p\right)^{q/p}.$$

It's easy to see that $p\ge q$ is essential:

If $p<q$ and $hf\in L^q$ for every $f\in L^p$ then $h=0$ almost everywhere.

If $h$ does not vanish almost everywhere there exists $E$ with $m(E)>0$ such that $|h|\ge c>0$ on $E$. There exists $f\in L^p$ such that $f$ vanishes off $E$ and $f\notin L^q$.

Details on that: There are at least two ways to see there is such an $f$. (i) Note that $E$ is measure-space-isomorphic to $(0,m(E))$, so wlog $E$ is $(0,a)$; now let $f(t)=t^\alpha$ for a suitable $\alpha<0$. (ii) Choose disjoint sets $E_j\subset E$ with $m(E_j)>0$ and let $f=\sum a_j\chi_{E_j}$ for a suitable sequence $(a_j)$.

And so there you are:

$hf\in L^q$ for every $f\in L^p$ if and only if (i) $h=0$ almost everywhere or (ii) $p\ge q$ and $h\in L^r$, where $1/p+1/r=1/q$.