any local minimum of a convex function is a global minimum over a convex set

Question

in this proof I can't see the contradiction that the author of this proof is talking about when $\lambda \to 1$

is it just the fact that $f(\overline{x}) < f(\overline{x}) $ is non-sense or something else ?

Answer 1

For a $ 0 < \lambda < 1 $ close enough to $1$, one has $$ \lambda \bar{x} + (1-\lambda)z \in B(\bar{x}, \epsilon)$$

while we have $f \left( \lambda \bar{x} + (1-\lambda)z\right) < f(\bar{x}) $

which is contradicting with $\bar{x}$ being local minimum.

Answer 2

As $\lambda \to 1$ you have, as he says, $$ \lim_{\lambda \to 1} f \left( \lambda \bar{x} + (1-\lambda)z\right) = f\left(\bar{x}\right) $$ so the inequality becomes $$ f\left(\bar{x}\right) < f\left(\bar{x}\right), $$ an obvious contradiction.