Help showing a linear functional is bounded

Question

Let $(X,M,\mu)$ be a $\sigma$-finite measure space, and $k:X\times X \rightarrow \mathbb{C}$ (or $\mathbb{R}$) be $X\times X$ measurable. Suppose there are measurable functions $h,g:X\rightarrow(0,\infty)$ and constants $c_1,c_2>0$ such that $$\int_X |k(x,y)|g^q(y)d\mu(y)\leq c_1^qh^q(x)$$ a.e. and $$\int_X |k(x,y)|h^p(x)d\mu(x)\leq c_2^pg^p(y)$$ a.e. Show that $(Tf)(x)=\int_X k(x,y)f(y)d\mu(y)$ defines a bounded linear operator $T:L^p(X)\rightarrow L^p(X)$ with $||T||\leq c_1c_2.$

I can show that $T$ is linear with not trouble, but I don't see how to get it is bounded. Any help would be most appriciated.

Your domain is a Banach space. So you can apply the closed graph theorem to prove a linear transformation is bounded. — GEdgar, Dec 28 '18 at 21:44
@GEdgar I doubt that's helpful. I especially doubt it's helpful for showing $||T|| \le c_1c_2$. — mathworker21, Dec 28 '18 at 21:47

score 1 · Accepted Answer · answered Dec 28 '18 at 23:41

Here I'm assuming $1 < p < \infty$, and $q$ is conjugate to $p$.

Fix $f\in L^p(X)$. Note

$$\lvert Tf(x)\rvert \le \int_X \lvert k(x,y)\rvert \lvert f(y)\rvert\, d\mu(y) = \int_X \lvert k(x,y)\rvert^{1/q} \left(\frac{g(y)}{h(x)}\right)\cdot\lvert k(x,y)\rvert^{1/p}\left(\frac{h(x)}{g(y)}\right)\lvert f(y)\rvert\, d\mu(y)$$ and by Hölder's inequality, the latter integral is bounded by $$\left[\int_X \lvert k(x,y)\rvert \left(\frac{g(y)}{h(x)}\right)^{q}\, d\mu(y)\right]^{1/q} \left[\int_X \lvert k(x,y)\rvert \left(\frac{h(x)}{g(y)}\right)^p\, \lvert f(y)\rvert^p d\mu(y)\right]^{1/p}$$ Thus $$\lvert Tf(x)\rvert \le c_1\left[\int_X \lvert k(x,y)\rvert \left(\frac{h(x)}{g(y)}\right)^p\lvert f(y)\rvert^p\, d\mu(y)\right]^{1/p}\quad \text{a.e.}$$ which implies $$\|Tf\|_p \le c_1\left[\int_X\int_X\lvert k(x,y)\rvert \left(\frac{h(x)}{g(y)}\right)^p\lvert f(y)\rvert^p\, d\mu(y)\, d\mu(x)\right]^{1/p}$$ By Fubini and the hypothesis, $$\int_X\int_X\lvert k(x,y)\rvert \left(\frac{h(x)}{g(y)}\right)^p\lvert f(y)\rvert^p\, d\mu(y)\, d\mu(x) =\int_X\left[\int_X\lvert k(x,y)\rvert \left(\frac{h(x)}{g(y)}\right)^p\, d\mu(x)\,\right]\lvert f(y)\rvert^p d\mu(y) \le c_2^p\|f\|_p$$ Consequently, $$\left[\int_X\int_X\lvert k(x,y)\rvert \left(\frac{h(x)}{g(y)}\right)^p\lvert f(y)\rvert^p\, d\mu(y)\, d\mu(x)\right]^{1/p} \le c_2\|f\|_p$$ yielding $\|Tf\|_p \le c_1c_2\|f\|_p$. As $f$ was arbitrary, $T$ is bounded on $L^p(X)$ with $\|T\| \le c_1 c_2$.

Help showing a linear functional is bounded

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