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This is an exercise on Stein's Real Analysis.

Suppose $w$ is a measurable function with $0 < w < \infty$ almost everywhere, and $K$ a measurable function on $\mathbb R^{2d}$ that satisfies

(i) $\displaystyle\int |K(x,y)|w(y) dy \leq Aw(x)$ for almost every $x \in \mathbb R^d$,

(ii) $\displaystyle\int |K(x,y)|w(x) dx \leq Aw(y)$ for almost every $y \in \mathbb R^d$.

Show the integral operator $T \colon L^2(\mathbb R^d) \to L^2(\mathbb R^d)$ defined by $\displaystyle Tf(x) = \int K(x,y)f(y) dy$ is bounded with $||T|| \leq A$.

I have completed this exercise by first showing that

$\displaystyle \Big( \int |K(x,y)||f(y)| dy \Big)^2 \leq A w(x) \int |K(x,y)||f(y)|^2w(y)^{-1}dy$,

and then use this inequality to argue that

$\displaystyle ||Tf||^2 \leq \int Aw(x) \int |K(x,y)||f(y)|^2w(y)^{-1} dydx = \int A|f(y)|^2 w(y)^{-1} \int |K(x,y)|w(x) dxdy \leq A^2 ||f||^2$.

Now I want use a similar argument to show the following statement in $L^p$:

Let $(X,\mu)$ and $(Y,\nu)$ be two finite measure spaces, and $K(x,y)$ measurable on $X \times Y$. Let $1 \leq p \leq \infty$, and let

$\displaystyle A = \sup_x \int_Y |K(x,y)|d\nu$, $\displaystyle B = \sup_y \int_X |K(x,y)|d\mu$.

Define $T \colon L^p(Y) \to L^p(X)$ by

$\displaystyle Tf(x) = \int_Y K(x,y) f(y) d\nu$.

If $A,B < \infty$, then $||T|| \leq \max\{A,B\}$.

The function $w(x)$ in the previous exercise just becomes $w=1$ here. I use Holder's inequality to say that

$\displaystyle \Big( \int_Y |K(x,y)||f(y)| d\nu \Big)^p \leq \int_Y |K(x,y)| d\nu \cdot \Big( \int_Y |K(x,y)||f(y)|^q d\nu \Big)^{p/q} \leq A\Big( \int_Y |K(x,y)||f(y)|^q d\nu \Big)^{p/q}$.

In the next step, I would like to use Fubini's theorem to interchange the order of integration, just as in the previous exercise. But the exponent on the integral is preventing me from doing that. Could someone help me complete the argument here? I apologize that this might be a tedious computation.

P. Factor
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1 Answers1

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Here, I'll only deal with the case $1 < p < \infty$. Let $\|f\|_p = 1$. By Hölder's inequality,

$$\int_Y \lvert K(x,y)\rvert \lvert f(y)\rvert \, d\nu(y) = \int_Y \lvert K(x,y)\rvert^{1/q} \lvert K(x,y)\rvert^{1/p}\lvert f(y)\rvert\, d\nu(y) \le A^{1/q}\left(\int_Y \lvert K(x,y)\rvert\lvert f(y)\rvert^p\, d\nu(y)\right)^{1/p}$$

Hence

\begin{align}\|Tf\|_p &\le A^{1/q}\left(\int_X\int_Y \lvert K(x,y)\rvert \lvert f(y)\rvert^p\, d\nu(y)\, d\mu(x)\right)^{1/p}\\ &= A^{1/q}\left(\int_Y \lvert f(y)\rvert^p \left[\int_X \lvert K(x,y)\rvert\, d\mu(x)\right] d\nu(y)\right)^{1/p}\\ &\le A^{1/q}B^{1/p}\|f\|_p = A^{1/q}B^{1/p} \end{align}

Now since $0 \le A,B \le \max\{A,B\}$, we have $A^{1/q}B^{1/p} \le \max\{A,B\}^{1/q + 1/p} = \max\{A,B\}$. Thus $\|Tf\|_p \le \max\{A,B\}$. Since $f$ was arbitrary, it follows that $\|T\| \le \max\{A,B\}$.

kobe
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