Let $(X,\mathbf{X},\mu)$ be a measure space. Define the linear bounded operator $$G(f)=\int_X gfd\mu,\,\forall f\in L_1(\mathbf{X},\mu)$$ where $g\in L_\infty(\mathbf{X},\mu)$. Take $E_c=\{x\in X:|g(x)|>c\|G\| \}$ with $c>1$. And define $$f_c(x)=\chi_{E_c\cap\{x\in X:g(x)\geq 0\}}(x)-\chi_{E_c\cap \{x\in X:g(x)<0\}}(x).$$ Prove that $$c\mu(E_c)\|G\|\leq G(f_c)\leq \|G\|\mu(E_c)$$ where $\|G\|$ is the operator norm of $G$.
My question is: In order to have the inequality above one needs to have $f_c\in L_1$ for some $c>1$. But I can find how to prove this. Any suggestions?
Right, notice that $|f_c(x)| = \chi_{E_c}(x)$, so $f \in L^1$ if and only if $\mu(E_c)< \infty$. It is not clear to me how to show this directly, but maybe we can go around it. Let $\{A_n\}$ be an increasing sequence of sets with finite measure such that $\cup_n A_n = \mathbf{X}$. Define $f_{c,n}:=f_c \cdot \chi_{A_n}$. Now, we clearly have that $f_{n,c}$ is in $L^1$, so we can use it to evaluate $G$. Hence, $$ G(f_{c,n}) = \int |f(x)| \chi_{A_n\cap E_c}(x) dx. $$ We can then use: the upper bound $|G(f_{n,c}| \le \|G\|\|f_{n,c}\|_{L^1}$ and $\|f_{n,c}\|_{L^1}=\|G\|\mu(E_c \cap A_n)$, to get $$ G(f_{n,c}) \le \mu(E_c \cap A_n). $$ We can now use the Monotone Convergence Theorem to get that the inequality above holds as $n\to \infty$.
The strategy for the lower bound is the same. Hope that this helps.