Is possible to show that the linear operator $T(\varphi)(x) = \int_{V_x\cap M} \varphi(y)\text{d}y$ has spectral radius $>0$. Fix some $σ>2/(3\sqrt{3})$, let $M$ be the interval $[x_-,x_+]$, where $$x_- = \text{unique real root of 
 $x^3 + \sigma = x$}$$
and
$$x_+ = \text{unique real root of 
 $x^3 - \sigma = x$}.$$
Moreover, define the set $$V_x=\{z\in \mathbb{R};\ z = x^3+\omega \sigma,\ \mbox{for some $\omega\in[-1,1]$}\}.$$
Consider the Banach Space $\left(\mathcal C^0(M),\|\cdot\|_\infty\right)$. Where $$\mathcal C^0(M):=\{f:M\to\mathbb{R};f\ \text{is continuous}\}\ \mbox{and }\|\phi\|_\infty=\sup_{x\in M}|\phi|.$$
Now, define the continuous linear map
\begin{align*}
T: \mathcal C^0(M)&\to \mathcal{C}^0 (M)\\
\varphi&\mapsto \left(x\mapsto\int_{V_x\cap M}\varphi(y)\text{d} y\right).
\end{align*}

My Question: Is it possible to guarantee that the continuous linear operator $T$ has a positive spectral radius?

Remark: It is possible to show that $T$ is a compact operator.
 A: Yes, it always has a positive spectral radius. 
Note that for $x \in [-1/4, 1/4]$, since $\sigma > 2/3\sqrt{3}$, we have $x^3 + \sigma \geq -1/64 + \sigma > 1/4$ and similarly $x^3 - \sigma < -1/4$, so $[-1/4, 1/4] \subset V_x$. You can also check that $x_+ > 1$ and $x_- < -1$, so $[-1/4, 1/4] \subset M$ as well. Now let $\varphi = 1 \in \mathcal{C}^0(M)$ be the constant function with value $1$.
Proposition. For every $n$, $T^n(\varphi)$ is nonnegative and has $T^n(\varphi)(x) \geq 2^{-n}$ on $[-1/4, 1/4]$.
Proof: Clearly this holds for $n = 0$. Suppose now that the statement holds for some $n$. Then for any $x \in M$, $T^{n+1}(\varphi)(x) = \int_{V_x \cap M} T^n(\varphi)(y) \,dy \geq 0$ since $T^n(\varphi)$ is nonnegative, meaning $T^{n+1}(\varphi)$ is also nonnegative. Furthermore, for $x \in [-1/4, 1/4]$, we have $[-1/4, 1/4] \subset V_x \cap M$, so since $T^n(\varphi)$ is nonnegative everywhere on $M$ and $\geq 2^{-n}$ on $[-1/4, 1/4]$, it follows that
$$T^{n+1}(\varphi)(x) = \int_{V_x \cap M} T^n(\varphi)(y) \,dy \geq \int_{-1/4}^{1/4} T^n(\varphi)(y) \,dy \geq \int_{-1/4}^{1/4} 2^{-n} = 2^{-(n+1)}$$
so the statement also holds for $n+1$.
By this proposition, $||T^n(\varphi)|| \geq 2^{-n}$ for each $n$, hence since $||\varphi|| = 1$ it follows that $||T^n|| \geq 2^{-n}$ for each $n$. By the formula $\rho(T) = \lim_{n \to \infty} ||T^n||^{1/n}$, the spectral radius is at least $1/2$. Better lower bounds can be found by optimizing the interval beyond $[-1/4, 1/4]$.
