Let $K \in M_+(R_+^2), f \in M_+(R_+)$. Consider operator

$$ (T_k)(x)=\int_{R_+}K(x,y)f(y)dy, \quad y\in R_+. $$

Denote by $f^*(t)=\inf\{\lambda>0: \alpha x \in R_+: \mu_f(y)>\lambda\}$ the non-increasing rearrangement of $f$. Here $\mu_f(y)=\{\alpha x\in R_+: |f(x)|>y\}$.

Let $\Phi(x)=\int_0^x \phi(y)\,dy$, $x \in \mathbb{R}_+$, be an N-function, and let $u$ be locally integrable on $\mathbb{R}_+$. Consider the gauge norm $$ \rho_{\Phi,u}(f)=\inf\{\lambda>0: \int_{\mathbb{R}_+}\Phi\left(\frac{|f(x)|}{\lambda}\right)u(x)\,dx\leq 1\}, $$where $f \in M_+(R_+)$.

I am trying to find an example of such $u_1, u_2$ when Kantorovich conditions would not be true, but the following inequality would hold: $$ \rho_{\Phi_1,u_1}(T_Kf^*)\leq \rho_{\Phi_2,u_2}(f^*) $$

Let $K \in M_+(R_+^2), f \in M_+(R_+)$. Consider operator

$$ (T_k)(x)=\int_{R_+}K(x,y)f(y)dy, \quad y\in R_+. $$

Denote by $f^*(t)=\inf\{\lambda>0: \alpha x \in R_+: \mu_f(y)>\lambda\}$ the non-increasing rearrangement of $f$. Here $\mu_f(y)=\{\alpha x\in R_+: |f(x)|>y\}$.

Let $\Phi(x)=\int_0^x \phi(y)\,dy$, $x \in \mathbb{R}_+$, be an N-function, and let $u$ be locally integrable on $\mathbb{R}_+$. Consider the gauge norm $$ \rho_{\Phi,u}(f)=\inf\{\lambda>0: \int_{\mathbb{R}_+}\Phi\left(\frac{|f(x)|}{\lambda}\right)u(x)\,dx\leq 1\}, $$where $f \in M_+(R_+)$.

I am trying to find an example of such $u_1, u_2$ when Kantorovich conditions (stated that the $l_q$ norm of the kernel is finite) would not be true, but the following inequality would hold: $$ \rho_{\Phi_1,u_1}(T_Kf^*)\leq \rho_{\Phi_2,u_2}(f^*) $$

Let $K \in M_+(R_+^2), f \in M_+(R_+)$. Consider operator

$$ (T_k)(x)=\int_{R_+}K(x,y)f(y)dy, \quad y\in R_+. $$

Denote by $f^*(t)=\inf\{\lambda>0: \alpha x \in R_+: \mu_f(y)>\lambda\}$ the non-increasing rearrangement of $f$. Here $\mu_f(y)=\{\alpha x\in R_+: |f(x)|>y\}$.

Let $\Phi(x)=\int_0^x \phi(y)\,dy$, $x \in \mathbb{R}_+$, be an N-function, and let $u$ be locally inferable on $\mathbb{R}_+$. Consider the gauge norm $$ \rho_{\Phi,u}(f)=\inf\{\lambda>0: \int_{\mathbb{R}_+}\Phi\left(\frac{|f(x)|}{\lambda}\right)u(x)\,dx\leq 1\}, $$where $f \in M_+(R_+)$.

I am trying to find an example of such $u_1, u_2$ when Kantorovich conditions would not be true, but the following inequality would hold: $$ \rho_{\Phi_1,u_1}(T_Kf^*)\leq \rho_{\Phi_2,u_2}(f^*) $$

Let $K \in M_+(R_+^2), f \in M_+(R_+)$. Consider operator

$$ (T_k)(x)=\int_{R_+}K(x,y)f(y)dy, \quad y\in R_+. $$

Denote by $f^*(t)=\inf\{\lambda>0: \alpha x \in R_+: \mu_f(y)>\lambda\}$ the non-increasing rearrangement of $f$. Here $\mu_f(y)=\{\alpha x\in R_+: |f(x)|>y\}$.

Let $\Phi(x)=\int_0^x \phi(y)\,dy$, $x \in \mathbb{R}_+$, be an N-function, and let $u$ be locally integrable on $\mathbb{R}_+$. Consider the gauge norm $$ \rho_{\Phi,u}(f)=\inf\{\lambda>0: \int_{\mathbb{R}_+}\Phi\left(\frac{|f(x)|}{\lambda}\right)u(x)\,dx\leq 1\}, $$where $f \in M_+(R_+)$.

I am trying to find an example of such $u_1, u_2$ when Kantorovich conditions would not be true, but the following inequality would hold: $$ \rho_{\Phi_1,u_1}(T_Kf^*)\leq \rho_{\Phi_2,u_2}(f^*) $$