I am working on Suzuki contractions in non-triangular metric spaces and I want to know if there are certain conditions under which such maps become orbitally continuous. For reference, the formal definitions which we are dealing with in this question are as follows:
Definition (Non-triangular metric spaces): Let $X$ be a non-empty set and $d: X \times X \rightarrow \left[ 0, \infty \right)$ be a map. Then, $\left( X, d \right)$ is called a non-triangular metric space if
- $d(x, x) = 0$ for every $x \in X$.
- $d(x, y) = d(y, x)$ for every $x, y \in X$.
- If $(x_n)$ is a sequence in $X$ such that $\lim\limits_{n \rightarrow \infty} d \left( x_n, x \right) = 0$ and $\lim\limits_{n \rightarrow \infty} d \left( x_n, y \right) = 0$ for some $x, y \in X$, then $x = y$.
All in all, non-triangular metric spaces are similar to the usual metric spaces, except for the fact that we cannot use the triangle inequality. However, the definition does ensure the uniqueness of the limits of sequences.
Definition (Suzuki Contractions): Let $\left( X, d \right)$ be a non-triangular metric space and $T: X \rightarrow X$ be a map. Then, $T$ is called a Suzuki contraction if there is some $\lambda \in \left[ 0, 1 \right)$ such that for all $x \neq y \in X$, we have $$\dfrac{1}{2} d \left( x, Tx \right) < d \left( x, y \right) \Rightarrow d \left( Tx, Ty \right) \leq \lambda d \left( x, y \right).$$
Definition (Orbitally Continuous): Let $\left( X, d \right)$ be a non-triangular metric space. A map $T: X \rightarrow X$ is orbitally continuous if for some subsequence $\left( n_k \right)$ for $\mathbb{N}$, $T^{n_k} x_0 \rightarrow x$ implies $T \left( T^{n_k} x_0 \right) \rightarrow T \left( x_0 \right)$.
Here, the definition of convergence is similar to that used in the case of usual metric spaces. For reference, I mention it here:
Definition (Convergence): Let $\left( X, d \right)$ be a non-triangular metric space and $\left( x_n \right)$ be a sequence in $X$. We say that $x_n \rightarrow x$ for some $x \in X$ if $\lim\limits_{n \rightarrow \infty} d \left( x_n, x \right) = 0$.
My question is, under what conditions can we guarantee that a Suzuki contraction on a non-triangular metric space is Orbitally continuous? I have tried to get the orbital continuity by proving that Suzuki contractions are asymptotically regular for each $x \in X$. However, I am stuck in getting the premise for the definition of Suzuki contractions.
In particular, I would like some conditions on $T$ or $X$ so that we get
$$\dfrac{1}{2} d \left( T^{n_k} x_0, T \left( T^{n_k} x_0 \right) \right) < d \left( T^{n_k} x_0, x \right)$$ for all $k \geq k_0$ (for some $k_0 \in \mathbb{N}$).
Any help will be appreciable! Thank you!