Operatornorm bound by splitting operator

Question

Let ($M$, $\mu$) be a measure space. Consider the operator $T:L^{2}(M, \mu)\to L^{2}(M, \mu)$ with $$Tf(x)=\int_{M}{k(x, y)f(y)d\mu},$$ where $k(x, y)\geq 0$ and $k(x, y)=k(y, x)$ satisfying $$k(x, y)\leq h_{1}(x, y)+h_{2}(x, y)$$ where $h_{i}(x, y)=h_{i}(y, x)$ and $h_{i}(x, y)\geq 0$. Suppose that for the operator $H_{i}:L^{2}(M, \mu)\to L^{2}(M, \mu)$ with $$H_{i}f(x)=\int_{M}{h_{i}(x, y)f(y)d\mu},$$ we have $$||H_{1}||_{2\to 2}\leq \lambda_{1}$$ and $$||H_{2}||_{2\to 2}\leq \lambda_{2}.$$ Does it follow that $$||T||_{2\to 2}\leq C(\lambda_{1}+\lambda_{2})?$$ Thanks in advance!

Answer 1

You can even take $C = 1$ in this case.

First, I claim that $\|T\| = \sup\{\|Tf\|: \|f\| = 1, f \geq 0\}$. This is straightforward to see since by using the fact that $k$ is a non-negative kernel, we get that for $f \in L^2(M,\mu)$, $|Tf(x)| \leq T(|f|)(x)$ and $\||f|\| = \|f\|$. In particular, the same is true for $H_1,H_2$ and $H_1 + H_2$ also.

So suppose $\|f\| = 1$ and $f \geq 0$. Then \begin{align*} \|Tf\|^2 =& \int_M \bigg( \int_M k(x,y)f(y) \mu(dy) \bigg)^2 \mu(dx)\\ \leq& \int_M \bigg( \int_M h_1(x,y)f(y) + h_2(x,y)f(y) \mu(dy) \bigg)^2 \mu(dx) \\ =& \int_M (H_1f(x) + H_2f(x))^2 \mu(dx) \\=& \|H_1f + H_2f\|^2 \end{align*} where I use the fact that we can take $f$ non-negative to pass to the second line. Hence, by the triangle inequality, we can conclude that $\|Tf\| \leq \|H_1f\| + \|H_2f\|$ and hence that $\|T\| \leq \|H_1\| + \|H_2\|$ as desired.