Positivelinear operator on $L^p$-spaces Suppose $1<p<\infty$.
A linear operator $T \colon L^p(\Omega)\to L^p(\Omega)$ is positive if $f \geq 0$ imply $T(f)\geq 0$ (where $\Omega$ is a measure space).
1) Does there exist a positive operator $T \colon L^p(\Omega)\to L^p(\Omega)$ which is not bounded?
2) Does there exist a positive bounded operator $T \colon L^p(\Omega)\to L^p(\Omega)$ which does not induce a bounded operator on $L^\infty(\Omega)$?
 A: 1) No, every such operator is bounded. For simplicity, I assume that all functions are real-valued. 
If $T$ is not bounded, there is a sequence $f_n$ with $\Vert f_n \Vert_p \leq 2^{-n}$ and $\Vert Tf_n\Vert_p \geq 2^n$ (why?). 
Then Fatou's Lemma (or monotone convergence) shows $f :=\sum_n |f_n|\in L^p$ with $\Vert f\Vert_p \leq 1$. 
But $|f_n|\leq |f|=f$, which yields (why?) $|Tf_n|\leq Tf$. But for $n\to\infty$, this contradicts the fact that $Tf\in L^p$, because $\Vert Tf_n\Vert_p \to \infty$. 
2) There is such an example, consider $p=2$, $\Omega =(1,\infty)$ with Lebesgue measure and
$$
Tf := \int_1^\infty f(x) \cdot x^{-3/4}\,dx \cdot \chi_{(1,2)},
$$
where $\chi_A$ is the indicator function of the set $A$. 
I leave it to you to verify the details. The main observation is $x^{-3/4}\in L^2\setminus L^1$. 
There is an analogous counterexample in the case $\mu(\Omega) < \infty$. Take $\Omega = (0,1)$ with the usual Lebesgue measure and
$$
Tf := \int_0^1 f(x) \, dx \cdot x^{-1/2}.
$$
Then $T : L^1 \to L^1$ is well-defined and bounded (because of $x^{-1/2} \in L^1$) and a positive operator, but $Tf : L^\infty \to L^\infty$ is not even well-defined, because $x^{-1/2}$ is not (essentially) bounded.
