Operator in function spaces

Question

Suppose I define the integral Operator $T:L^{\phi}(X)\to L^{\psi}(X)$ by $(Tf)(x)=\int_{X}K(x,y)f(y)d\mu$

Answer 1

Lets show that $Tf\in L^\Psi$ without using the assumption that $\Psi(x)\leq C \Phi(x)$. First note that any $f\in L^\Phi$ is also in $L^1$, this can be seen as follows:

Let $f\in L^\Phi$. Since $\lim_{x\to\infty}\frac{\Phi(x)}x=\infty$ and $\Phi(x)$ is convex there is some $r>0$ so that $\Phi(x)>x$ for all $x>r$. Denote with $X_+ = \{ x\in X\mid |f(x)|>r\}$, then $$\int_X|f| = \int_{X^+}|f|+\int_{X\setminus X^+}|f| \leq \int_{X^+}|\Phi(f)|+|X|\cdot r<\infty.$$

With this in hand let $|K|:=\sup_{x,y}|K(x,y)|$, then $|Tf (x)| \leq |K| \int |f|$ and so: $$\int_X \Psi(|Tf|)\leq |X|\, \Psi(|K|\int_X|f|)<\infty.$$