Question

2) Let $T: X \to X$ be an orbitally continuous mapping and $X$ be $T$-orbitally complete. If $T$ satisfies the following condition \\ $\min\{d(Tx, Ty), d(x, Tx), d(y, Ty)\} - \min\{d(x, Ty), d(y, Tx)\} \le kd(x, y)$ \\ for some $k < 1$ and for all $x, y \in X$, then $T$ has a fixed point in $X$.

          2) Let $T: X \to X$ be an orbitally continuous mapping and $X$ be $T$-orbitally complete. If $T$ satisfies the following condition \\
$\min\{d(Tx, Ty), d(x, Tx), d(y, Ty)\} - \min\{d(x, Ty), d(y, Tx)\} \le kd(x, y)$ \\
for some $k < 1$ and for all $x, y \in X$, then $T$ has a fixed point in $X$.

2) Let T: X → X be an orbitally continuous mapping and X be T-orbitally complete. If T satisfies the following condition

min{d(Tx, Ty), d(x, Tx), d(y, Ty)} - min{d(x, Ty), d(y, Tx)}≤ kd(x, y)

for some k < 1 and for all x, y ∈ X, then T has a fixed point in X.

Added by Paul G.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

11/21/2023

Step 1

Then, by the given condition, we have $$min\{d(T^nx, T^ny), d(T^{n-1}x, T^nx), d(T^{n-1}y, T^ny)\} - min\{d(T^{n-1}x, T^ny), d(T^{n-1}y, T^nx)\} \leq kd(T^{n-1}x, T^{n-1}y).$$ Show more…

Show all steps

Thanks for your feedback!

2) Let $T: X \rightarrow X$ be an orbitally continuous mapping and $X$ be $T$-orbitally complete. If $T$ satisfies the following condition $$min\{d(Tx, Ty), d(x, Tx), d(y, Ty)\} - min\{d(x, Ty), d(y, Tx)\} \leq kd(x, y)$$ for some $k < 1$ and for all $x, y \in X$, then $T$ has a fixed point in $X$.