Proving that there exists only one non-negative eigenfuction for the following operator.

Question

Fix some constants $a>1$, $\sigma>0$ and consider $x_- = -\sigma/(\alpha-1)$, $x_+ = \sigma/(\alpha - 1)$, and $M=[x_-,x_+]$. Consider the Banach space $V = \left(\mathcal C^0 (M),|\cdot |_{\infty}\right)$ where $$\mathcal C^0(M) = \{f:M\to\mathbb R;\ f\ \text{is a continuous function}\}, $$ and $$\left|f\right|_\infty = \sup_{x\in M}\left|f(x)\right|. $$

Defining the linear operator \begin{align*} T:V&\to V\\ f&\mapsto\left(x\mapsto \frac{1}{2\sigma} \int_{\frac{1}{\alpha}\left(x-\sigma\right)}^{\frac{1}{\alpha}\left(x+\sigma\right)} f(y)\ \text d y\right), \end{align*} it is ''easy to see'' that $T$ is a continous and compact operator, moreover $r(T)>0$ (where $r(T)$ is the spectral radius of $T$).

Defining convex cone $K:= \mathcal C^0_+(M) = \{f\in V;\ f(x)\geq 0, \forall\ x\in M\}$, and noticing that $T(K)\subset K$ and $K-K= V$, by Krein-Rutman theorem there exists an eigenfunction $g\in K$, such that $$T(g) = r(T)g. $$

My Question: Is it possible to show that $g$ is the unique eigenfunction of $T$ lying in the cone $K$?

Comment: I think that the only non-negative eigenfunction of $T$ is the constant function $1$.

Answer 1

Suppose $g \in K$ is an eigenfunction of $T$, so $Tg = \lambda g$ for some $\lambda$.

Note first that for $x \in [x_-, x_+]$, we have $\alpha^{-1}(x-\sigma) \leq x \leq \alpha^{-1}(x+\sigma)$, with equality on the left iff $x = x_-$, and equality on the right iff $x = x_+$. In fact -- this will come up later -- defining $I(x) = [\alpha^{-1}(x-\sigma), \alpha^{-1}(x + \sigma)]$, we see that for any $x \in M$, $x$ is in the interior of $I(x)$ with respect to $M$: when $x = x_-$ or $x_+$ this is true because the sets $[x_-, y)$ and $(y, x_+]$ are open in $M$, and otherwise $I(x)$ contains an open interval around $x$. Then taking $x$ with $g(x) \neq 0$, we see that $g$ is nonnegative and not identically zero on $I(x) = [\alpha^{-1}(x-\sigma), \alpha^{-1}(x + \sigma)]$, so $(Tg)(x) \neq 0$ as well, and thus in particular $Tg \neq 0$, and $\lambda > 0$ (since $g, Tg \geq 0$).

Now define, for $f \in \mathcal{C}^0(M)$, the quantities $c(f) = \min_{x \in M} f(x)$ and $C(f) = \max_{x \in M} f(x)$. Then since $c(g) \leq g(y) \leq C(g)$ on $M$, we have $$\alpha^{-1}c(g) \leq \frac{1}{2\sigma} \int_{\alpha^{-1}(x-\sigma)}^{\alpha^{-1}(x+\sigma)} g(y) \,dy \leq \alpha^{-1}C(g)$$ for all $x \in M$, hence $\alpha^{-1}c(g) \leq (Tg)(x) \leq \alpha^{-1}C(g)$ on $M$, giving $c(Tg) \geq \alpha^{-1}c(g)$ and $C(Tg) \leq \alpha^{-1}C(g)$. Then because $g, Tg \geq 0$ and neither is identically zero, we have $C(g), C(Tg) > 0$ and $c(g), c(Tg) \geq 0$, hence $$\frac{c(Tg)}{C(Tg)} \geq \frac{c(Tg)}{\alpha^{-1}C(g)} \geq \frac{\alpha^{-1}c(g)}{\alpha^{-1}C(g)} = \frac{c(g)}{C(g)}$$ but on the other hand, $Tg = \lambda g$, so we have $c(Tg) = \lambda c(g)$ and $C(Tg) = \lambda C(g)$, meaning the LHS and RHS above are equal, and equality must hold in each of the above inequalities. In particular, equality in the second inequality means $c(Tg) = \alpha^{-1}c(g)$.

Let $N$ be the subset of $M$ where $g$ attains its minimum, i.e. the set of those $x \in M$ with $g(x) = c(g)$. Note $N$ is nonempty and closed. Because $Tg = \lambda g$, $N$ is also the set of those $x$ where $Tg$ attains its minimum, that is, where $(Tg)(x) = c(Tg) = \alpha^{-1}c(g)$. Then let $x_0 \in N$, so $(Tg)(x_0) = \alpha^{-1}c(g)$. Because $g(y) \geq c(g)$ on $I(x_0)$, we have $(Tg)(x_0) \geq \alpha^{-1} c(g)$ with equality holding iff $g(y) = c(g)$ everywhere on $I(x_0)$, so this last part must hold, i.e. we must have have $I(x_0) \subset N$. However, $x_0$ is in the interior of $I(x_0)$ (relative to $M$), so this means $x_0$ is in the interior of $N$ as well. Thus all points of $N$ are in its interior, hence $N$ is open in $M$, so since $N$ is also closed and $M$ is connected, it follows that $N = M$. As $N$ was the subset where $g$ attained its minimum, this means $g$ is constant.