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

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


  • $\begingroup$ Are $p$ and $q$ related by $\frac{1}{p}+\frac{1}{q}=1$ ? $\endgroup$ Aug 9 '21 at 22:38
  • $\begingroup$ @DannyPak-KeungChan No, I usually denote the conjugate exponent of $p$ by $p'$. $\endgroup$ Aug 10 '21 at 9:48

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 $Tf_n(x)\to0$.

  • $\begingroup$ 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. $\endgroup$ Aug 10 '21 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 $\int_X\int_Yk(x,y) f_n(y)\,dy\,\nu(dx)\xrightarrow{n\rightarrow\infty}0$. $\endgroup$ Aug 10 '21 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 $x \in \Bbb{R}$. $\endgroup$ Aug 10 '21 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 $S\colon L^p\to L^q$ $\endgroup$ Aug 10 '21 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 $L_r(\mu)$ $\endgroup$ Aug 10 '21 at 21:47

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 $f\in L_p(\mu)$. $\Box$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.