Consider the probability space $([0,1], \mathcal B([0,1]), \lambda(\mathrm{d} x) )$, where $\mathcal B([0,1])$ is the Borel $\sigma$-algebra and $\lambda$ the Lebesgue measgure on $[0,1]$.
Let $P,T:L^1([0,1]) \to L^1([0,1])$ operators satisfying the following properties:
For every $p\in [1,\infty],$ $$ \left.P\right|_{L^p[0,1]},\left.T\right|_{L^p[0,1]}: L^p [0,1] \to L^p[0,1],$$ is well defined and a bounded linear operator.
For every $p \in [1,\infty]$ $$\left(\left. T\right|_{L^p([0,1])}\right)^* = \left.P\right|_{L^{p^*}([0,1])},$$ where $\frac{1}{p} + \frac{1}{p^*} = 1 \ \text{and }{}^*\ \text{denotes the adjoint operator}. $
If $0 \leq f \in L^p$ then $0\leq T f $ and $0\leq Pf.$ I think this hypothesis is not relevant.
For every $p\in [1,\infty]$ let us define $$\sigma (T,p ) := \sigma (\left. T\right|_{L^p}) \ \text{and } \sigma (P,p ) := \sigma (\left. P\right|_{L^p}),$$ where $\sigma(T)$ denotes the spectrum of the operator $T$.
Question: Is it possible to conclude that $\sigma(T,1) = \sigma(T, p)$ or $\sigma(T,\infty) = \sigma(T,p)$, for all $p\in (1,\infty).$?
For a fixed sequence of positive numbers $\omega=\{\omega_n\}_{n=0}^\infty$ consider the weighted spaces $\ell^p(\omega_n)$ defined by $$\ell^p(\omega)=\left \{x=\{x_n\}_{n=0}^\infty \,:\, \sum_{n=0}^\infty |x_n|^p\omega_n <\infty \right \}$$ with norm $$\|x\|_p=\left (\sum_{n=0}^\infty |x_n|^p\omega_n \right)^{1\over p}.$$ The space $\ell^p(\omega)$ is isometrically isomorphic to the standard space $\ell^p$ by the mapping $$\ell^p(\omega) \ni \{x_n\}_{n=0}^\infty\longmapsto \{x_n\,\omega_n^{1/p}\}_{n=0}^\infty \in \ell^p$$ Let $\omega_n=2^{n}.$ Consider the shift operator $T:\ell^p(\omega)\to \ell^p(\omega),$ acting by $(Tx)_n=x_{n+1}.$ The operator $T$ preserves positivity of the sequences. The operator $T$ is bounded: $$\|Tx\|_p^p=\sum_{n=0}^\infty |x_{n+1}|^p2^n={1\over 2} \sum_{n=0}^\infty |x_{n+1}|^p2^{n+1}\le {1\over 2}\|x\|_p^p$$ hence $\|T\|_{p\to p}\le 2^{-1/p}.$ For $x=\delta_1$ we get $\|T\delta_1\|_p=\|\delta_0\|_p=2^{-1/p}\|\delta_1\|_p.$ Therefore $$\hspace{5cm}\|T\|_{p\to p}=2^{-1/p}\hspace{5cm} (1)$$ The adjoint operator $T^*$ acts by the formula $$(T^*x)(n)={\omega_{n-1}\over \omega_n}x_{n-1}={1\over 2} x_{n-1},$$ with convention $x_{-1}=\omega_{-1}=0.$
Let $\sigma(T,p)$ denote the spectrum of $T:\ell^p(\omega)\to \ell^p(\omega).$ We will show that $$\qquad\qquad\qquad \sigma(T,p)=\{z\in \mathbb{C}\,:\, |z|\le 2^{-1/p}\}\qquad \ \ \qquad \qquad \qquad\qquad (2)$$ The inclusion "$\subset $" follows from $(1).$ Let $|z|<2^{-1/p}.$ Then the sequence $v_z=\{z^n\}_{n=0}^\infty $ belongs to $ \ell^p(\omega)$ and $Tv_z=z\,v_z.$ Therefore the number $z$ is an eigevalue of $T:\ell^p(\omega)\to \ell^p(\omega).$ Hence $z\in \sigma(T,p),$ i.e. $$\{z\in \mathbb{C}\,:\, |z|< 2^{-1/p}\}\subset \sigma(T,p)$$ This concludes the proof of $(2).$