i am stuck as to how the author jumps from the 2 inequalities to two equalities in the following theorem.
For context i state theorem 1.11: Let X be a uniformly convex Banach space and $C \subset X$ which is nonvoid,closed,convex and bounded.If $f: C\rightarrow C$ is a non expansive map then f has a fixed point.
I understand how the inequalities are derived but not how these inequalities imply the equalities.
Presumably,i would like to show that $||f(x_t)-f(x_0)||<t||x_1-x_0||$ is false and so the equality has to occur. Any hint?
You have $$||x_1 - x_0|| = ||f(x_t) - x_0 + x_1 - f(x_t)||\leq ||f(x_t) - x_0|| + ||f(x_t) - x_1||$$ by triangle inequality. Hence $$||x_1 - x_0|| \leq ||f(x_t) - x_0|| + ||f(x_t) - x_1||\leq $$ $$ \leq t||x_1 - x_0|| + (1-t)||x_1 - x_0|| = ||x_1 - x_0||$$ Thus we have $=$ everywhere instead of $\leq$.