Does there exists an operator with these properties? Consider with $(\Omega,\Sigma,\mu)$ a $\sigma$-additive measure space. Is there a linear operator $P \neq 0$
$$P : L^1(\mu) \to L^1(\mu) $$
which fulfills
$$ \|Pf \| \leq \|f\|,$$
$$ f\geq0 \Rightarrow Pf \geq0 $$
such that there exists no $\alpha \in (0,1]$ with
$$\alpha \|f\| \leq \|Pf\| $$ for all $f\in L^1(\mu)$?
 A: If $\Omega\subset\mathbb{R}^n$ is bounded, then convolution operators will do:
They are compact (if you convolute with, say a standard mollifier), as a consequence of Kolmogorov-Riesz, thus do not fulfill $\alpha \|f\|\leq \|Pf\|$ (note that zero is in the spectrum of compact operators). Furthermore, if you convolute with something positive you'll get $f\geq 0\Rightarrow Pf\geq 0$
If $\Omega\subset\mathbb{R}^n$ is unbounded, first multiply with the characteristic function of the unit-ball.
A: Is this a simple example? Let $\Omega = [0,1)$ with the Lebesgue measure. Let $\chi_{[0,1/2)}$ be the characteristic function of the interval $[0,1/2)$. Set
$$
Pf = f\chi_{[0,1/2)}, \quad f \in L^1(\mu). 
$$
Clearly, $P : L^1(\mu) \to L^1(\mu)$, $\|Pf\| \leq \|f\|$ for all $f \in L^1(\mu)$ and $Pf \geq 0$ whenever $f \geq 0$. Also, for every $\alpha \gt 0$ we have 
$$\alpha \|\chi_{[1/2,1)}\| = \frac{\alpha}{2} \gt \|P\chi_{[1/2,1)}\| = 0$$. 
Or, if you want $P$ for which $0$ is not an eigenvalue define 
$$
Pf = fg, \quad f \in L^1(\mu), 
$$
where $g(t) = t$. Then, again $P : L^1(\mu) \to L^1(\mu)$, $\|Pf\| \leq \|f\|$ for all $f \in L^1(\mu)$ and $Pf \geq 0$ whenever $f \geq 0$. 
Let $\alpha \gt 0$. Then there exists $n \in {\mathbb Z}^+$ such that $\alpha \gt 1/(2n)$. Set $f = n\chi_{[0,1/n)}$ and we have
$$
\alpha = \alpha\|f\| > \|Pf\| = \|fg\| = \frac{1}{2n}.
$$
