Let us assume that we have two integral operators, $L_{K_{1}}$ and $L_{K_{2}}$ with respective kernels $K_1(x,y) , K_2(x,y)\in L^{1}(\mathbb{R}^{2})\cap L^{2}(\mathbb{R}^{2})$. . Also, assume that the operators $L_{K_1}$ and $L_{K_{2}}$ are both positive operators, i.e. in this case $\forall$ $\phi \in L^{2}(\mathbb{R}^{2})$ $\langle \phi, L_{K_{i}} \phi\rangle \geq 0$.
I would like to take the trace distance of these operators. The trace distance is definde as follows.
$$\| A- B\|_{1} :=\frac{1}{2}Tr|A-B| =\frac{1}{2}Tr\sqrt{(A-B)^{\dagger}(A-B)}.$$
I know that using lidskii's theorem one can take the trace of an operator simply by integrating along the diagonal. i.e. $$Tr(L_{K_1}) = \int dx K_{1}(x,x)dx. $$
I am wondering if the follwoing is true.
$$\|L_{K_1}-L_{K_2}\|_{1} = \frac{1}{2}Tr|L_{K_1}-L_{K_2}| = \frac{1}{2}\int |K_{1}(x,x)-K_{2}(x,x)| dx$$ I suspect that this is not true but I can not prove this hypothesis.
The symbol $\chi_E$ will denote the indicator function of the set $E.$
Let $$K_1(x,y)=\chi_{[0,1]}(x)\chi_{[0,1]}(y), \qquad K_2(x,y)=\chi_{[0,2]}(x)\chi_{[0,2]}(y)$$ Then $K_2\ge K_1$ and $${1\over 2}{\rm Tr}(L_{K_2}-L_{K_1})={1\over 2}\int |K_2(x,x)-K_1(x,x)|\,dx={1\over 2}$$ The operator $L_{K_2}-L_{K_1}$ is not positive definite as for $f=\chi_{[1,2]}-\chi_{[0,1]}$ we have $$L_{K_2}f=0,\qquad L_{K_1}f=-\chi_{[0,1]},\qquad (L_{K_2}-L_{K_1})f=\chi_{[0,1]}$$ hence $$\langle (L_{K_2}-L_{K_1})f,f\rangle =-1$$ Therefore $${1\over 2}{\rm Tr}|L_{K_2}-L_{K_1}|>{1\over 2}{\rm Tr}[L_{K_2}-L_{K_1}]={1\over 2}$$
Remark I have used a general fact that if a trace class operator $T$ is not positive then ${\rm Tr}\,|T|>|{\rm Tr}\,T|.$ The proof is easy, especially if $T$ is self-adjoint like in the example above.