# If an integral operator $T : L^p(\Bbb{R}) \to L^q(\Bbb{R})$ is well-defined, it is bounded

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?

• Are $p$ and $q$ related by $\frac{1}{p}+\frac{1}{q}=1$ ? Commented Aug 9, 2021 at 22:38
• @DannyPak-KeungChan No, I usually denote the conjugate exponent of $p$ by $p'$. Commented Aug 10, 2021 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 $$Tf_n(x)\to0$$.

• 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. Commented Aug 10, 2021 at 9:39
• 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 $\int_X\int_Yk(x,y) f_n(y)\,dy\,\nu(dx)\xrightarrow{n\rightarrow\infty}0$. Commented Aug 10, 2021 at 17:10
• @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 $x \in \Bbb{R}$. Commented Aug 10, 2021 at 19:49
• 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 $S\colon L^p\to L^q$ Commented Aug 10, 2021 at 20:57
• @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 $L_r(\mu)$ Commented Aug 10, 2021 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 $$f\in L_p(\mu)$$. $$\Box$$