4
$\begingroup$

Let $p,q \in [1,+\infty\rangle$ and let $K : \Bbb{R}^2 \to \Bbb{R}$ be a measurable function such that the linear map $T : L^p(\Bbb{R}) \to L^q(\Bbb{R})$ given by$$(Tf)(x) = \int_{\Bbb{R}} K(x,y)f(y)\,dy, \quad \text{for a.e. $x \in \Bbb{R}$ and $f \in L^p(\Bbb{R})$}$$ is well-defined. Can we conclude that $T$ is bounded?

My attempt:

We shall use the Closed Graph Theorem to show that $T$ is bounded.

Since $T$ is well-defined, for every $f \in L^p(\Bbb{R})$ the function $Tf$ is a well-defined $L^q(\Bbb{R})$ function which is equal to the above integral for a.e. $x \in \Bbb{R}$. In particular, the integral $\int_{\Bbb{R}} K(x,\cdot)f(\cdot)$ exists for a.e. $x \in \Bbb{R}$.

We can conclude that for every $f \in L^p(\Bbb{R})$ there exists a set $N_f \subseteq \Bbb{R}$ of measure zero such that for all $x \in \Bbb{R}\setminus N_f$ holds $K(x,\cdot)f(\cdot) \in L^1(\Bbb{R})$.

Here comes the unclear part. Can we say that there exists a "global" set $N \subseteq \Bbb{R}$ measure zero such that for every $x \in \Bbb{R}\setminus N$ we have $$K(x,\cdot)f(\cdot) \in L^1(\Bbb{R}),\quad \text{ for all }f \in L^1(\Bbb{R})?$$

If so, we can proceed as in this answer:

For all $x \in \Bbb{R}\setminus N$ we define a linear map $A_x : L^p(\Bbb{R}) \to L^1(\Bbb{R})$ as $A_x(f) := K(x,\cdot)f(\cdot)$. It is easy to show that $A_x$ is bounded by the Closed Graph Theorem.

Now for all $x \in \Bbb{R}\setminus N$ we define the linear functional $l_x : L^p(\Bbb{R}) \to \Bbb{C}$ as $$l_x(f) := \int_{\Bbb{R}} K(x,\cdot)f(\cdot) = \int_{\Bbb{R}} A_x(f) = (Tf)(x), \quad f\in L^p(\Bbb{R})$$ Since $A_x$ is bounded, $l_x$ is bounded as well.

Now assume $f_n \xrightarrow{L^p} 0$ and $Tf_n \xrightarrow{L^q} f \in L^q(\Bbb{R})$ and we wish to show that $f = 0$. Since $l_x$ is continuous for all $x \in \Bbb{R}\setminus N$, for all such $x$ we have $$(Tf_n)(x) = l_x(f_n) \xrightarrow{n\to\infty} l_x(0) = 0$$ and hence $Tf_n \xrightarrow{\mathrm{a.e.}} 0$. From $Tf_n \xrightarrow{L^q} f$ by passing to a subsequence we conclude $f = 0$. Hence, $T$ is bounded by CGT.

So basically, my question boils down to the fact whether the quantifiers "for every" and "for almost every" commute. Are the statements

  • $$(\text{for every } f \in L^p(\Bbb{R}))(\text{for a.e. }x \in \Bbb{R}) \quad K(x,\cdot)f(\cdot)\in L^1(\Bbb{R})$$
  • $$(\text{for a.e. }x \in \Bbb{R})(\text{for every } f \in L^p(\Bbb{R})) \quad K(x,\cdot)f(\cdot)\in L^1(\Bbb{R})$$

equivalent?

$\endgroup$
2
  • $\begingroup$ Are $p$ and $q$ related by $\frac{1}{p}+\frac{1}{q}=1$ ? $\endgroup$ Commented Aug 9, 2021 at 22:38
  • $\begingroup$ @DannyPak-KeungChan No, I usually denote the conjugate exponent of $p$ by $p'$. $\endgroup$ Commented Aug 10, 2021 at 9:48

2 Answers 2

5
$\begingroup$

The equivalence is false. In fact, if the second assertion holds for every $f\in L^p(\mathbb R)$ it follows that $K(x,\cdot)\in L_{p'}(\mathbb R)$ with $\frac1p+\frac1{p'}=1$ which is not true, in general if $T\colon L^p\to L^q$.

A more concrete counterexample is $K(x,y)=g(x-y)$ with $g\in L^1(\mathbb R)$ where the second assertion is false, in general, but the first holds with $p=1$. (In fact, you can pick $q=1$ by Fubini-Tonelli.)

However, the operator $T$ is automatically bounded - you can find the proof in Banach's classical monograph, BTW. The proof goes as follows:

Let $f_n\to0$ in $L^p$ and $Tf_n\to g$ in $L^q$. As you observe, one has to show that $g=0$. Passing to a subsequence, you can assume that $Tf_n\to g$ almost everywhere. Passing to a further subsequence, you can assume that $f_n\to0$ a.e., and moreover, that there is a function $h\in L^p$ such that $\lvert f_n(y)\rvert\le h(y)$ for almost every $y$. (The latter assertion is a little bit tricky to show and is usually proved implicitly in the proof of the completeness of $L^p$; in Rudin's "Real and complex analysis" you will find the proof.)

Now it suffices to observe that for the particular function $h$ the function $K(x,\cdot)h(\cdot)$ is integrable for almost all $x$. Hence, for almost all $x$ you can use Lebesgue's dominated convergence theorem to show that $Tf_n(x)\to0$.

$\endgroup$
5
  • $\begingroup$ Amazing answer, thanks a lot. So, would you say that the approach in the linked question is wrong then? It basically assumes that the second assertion holds, when the problem from Conway says "for $f \in L^1(\mu)$ and a.e. $x \in X$ holds $k(x,\cdot)f(\cdot) \in L^1(\mu)$", which is ambiguous. $\endgroup$ Commented Aug 10, 2021 at 9:39
  • 1
    $\begingroup$ It seems that the answer in the link you gave, although correct in showing the statement that $k(x,\cdot)\in L_q(dy)$ for almost all $x$,, is not used correctly by the OP. The approach suggested by MartinVäth allows you to apply dominated convergence to prove that $\int_X\int_Yk(x,y) f_n(y)\,dy\,\nu(dx)\xrightarrow{n\rightarrow\infty}0$. $\endgroup$
    – Mittens
    Commented Aug 10, 2021 at 17:10
  • $\begingroup$ @OliverDiaz How is it correct that for a.e. $x \in X$ holds $k(x,\cdot) \in L^q(\mu)$? The counterexample by Martin Väth shows that for $p=1$ and $g \in L^1(\Bbb{R})\setminus L^\infty(\Bbb{R})$ the linear map $$K : L^1(\Bbb{R}) \to L^1(\Bbb{R}), \quad (Tf)(x) = \int_{\Bbb{R}} g(x-y)f(y)\,dy, \quad x \in \Bbb{R}, f \in L^1(\Bbb{R})$$ is well defined (and bounded) but $y \mapsto g(x-y)$ is not in $L^\infty(\Bbb{R})$ for any $x \in \Bbb{R}$. $\endgroup$ Commented Aug 10, 2021 at 19:49
  • 1
    $\begingroup$ To clarify a bit: The difficulty of integral operators is always that the “exceptional” set of $Tf$ (in the sense that $Tf$ can be undefined or very large on this set) is dependent on $f$ in general. In general, you get the sufficient estimate $\lVert T\rVert\le\Bigl(\int\Bigl(\int\lvert k(x,y)\rvert^{p'}\,dy\Bigr)^{q/p'}dx\Bigr)^{1/q}$ which you can easily prove by Hölder by the “pointwise” estimate, but the finiteness of the right-hand side is far from being necessary for $T\colon L^p\to L^q$. Even $Sf(x)=\int\lvert k(x,y)\rvert f(y)\,dy$ need not satisfy $S\colon L^p\to L^q$ $\endgroup$ Commented Aug 10, 2021 at 20:57
  • $\begingroup$ @mechanodroid: one may have $K(x,\cdot)\in L_p(\nu)$, for all $f\in L_p(\nu)$ $\mu$-a.s. in $X$, in which case $K(x,\cdot)\in L_q(\nu)$ for $\mu$-a.s. in $X$. This however does not mean that $x\mapsto \int_Y|K(x,y)|^q\,\nu(dy)$ is integrable or in any $L_r(\mu)$ $\endgroup$
    – Mittens
    Commented Aug 10, 2021 at 21:47
1
$\begingroup$

This is just to provide a few more details to the answer provided ny @MartinVäth.

  • Assumptions: $(X,\mathscr{F},\mu)$ and $(Y,\mathscr{G},\nu)$ are $\sigma$--finite measure spaces, and $p,q\geq1$; $K:(X\times Y,\mathscr{F}\otimes\mathscr{G})\rightarrow\mathbb{R}$ $$Tf(x):=\int_Yf(y)K(x,y)\,\nu(dy)\in L_q(\mu)$$ whenever $f\in L_p(\nu)$.

  • As suggested by the OP, we apply the closed graph theorem to show that $T:L_p(\nu)\rightarrow T_q(\mu)$ is bounded.

  • It suffices to show that for any sequence $\{f_n:n\in\mathbb{N}\}\subset L_p(\nu)$ such that $(f_n,Tf_n)\xrightarrow{n\rightarrow\infty}(0,g)$ in $L_p(\nu)\times L_q(\mu)$, we have that $g=0$.

  • Claim: There is a set $A\subset Y$ with $\nu(A)=0$ and $h\in L_p(\nu)$ such that on $X\setminus A$ and along a subsequence $n'$ $|f_n'|\leq h$ and $f_{n'}\xrightarrow{n\rightarrow\infty}0$.
    Proof: Take a subsequence $n_k$ such that $\|f_n-f_{n_k}\|_p<2^{-k}$ for all $n\geq n_k$, and set $g=\sum_k|g_{n_k}-g_{n_{k+1}}|$. Then $0\leq g<\infty$ $\nu$-a.s., and $f'=f_{n_1}+\sum^\infty_{j=1}f_{n_{j+1}}-f_{n_j}$ converges absolutely $\nu$-a.s. For $n\geq n_k$ $$\|f'-f_n\|_p\leq\|f'-f_{n_k}\|_p+\|f_n-f_{n_k}\|_p\leq \Big\|\sum^\infty_{j=k}(f_{n_{j+1}}-f_{n_j})\Big\|_p+2^{-k}\leq 2^{-k+1}\xrightarrow{k\rightarrow\infty}0$$ Hence $f=f'$ and $|f_{n_k}|\leq|f|+g=h$ $\nu$-a.s. Let $A$ be the exceptional set in $Y$.
    By a similar argument, there is subsequence $k'$ along which $$Tf_{n_{k'}}\xrightarrow{k'\rightarrow\infty}g\qquad\text{$\mu$--a.s.}$$ Since $$\int_X|Th(x)|^q\,\mu(dx)=\int_X\Big|\int_YK(x,y) h(y)\,\nu(dy)\Big|^q\,\mu(dx)<\infty$$ there is a set $B\subset X$ with $\mu(B)=0$ such that $|K(x,\cdot)\, h(\cdot)|\in L_1(\nu)$ for all $x\in X\setminus B$. On $(X\setminus B)\times(Y\setminus A)$ we have $$|K(x,y) f_n(x)|\leq |K(x,y) h(x)|$$ and $$K(x,y)f_n(x)\xrightarrow{n\rightarrow\infty}0$$ By dominated convergence, for all $y\in X\setminus B$, $$\lim_n\int_X K(x,y) f_n(x)\,d\mu(dx)=0$$ and $$|\int_XK(x,y) f_n(x)\,d\mu(dx)|\leq\int_X|K(x,y) h(x)|\,\mu(dx)\in L_q(\nu)$$ Another application of dominated convergence implies that $$ \lim_n\int_Y\Big|\int_XK(x,y)\,f(x)\,\mu(dx)\Big|^p\,\mu(dy)=0 $$ That is, $Kf_n\xrightarrow{n\rightarrow\infty}0$ in $L_q(\nu)$. This implies that $g=0$ $\mu$-a.s. The closed graph theorem implies that $T$ is a bounded operator, i.e., there is $c>0$ such that $\|Tf\|_{L_q(\nu)}\leq c\|f\|_{L_p(\mu)}$ for all $f\in L_p(\mu)$. $\Box$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .