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

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 ?

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.

• I have a small question related to your answer when 𝜆 -> 1 why we have 𝜆𝑥¯+(1−𝜆)𝑧∈𝐵(𝑥¯,𝜖)? Oct 31, 2020 at 17:58
• because $\bar{x} \in B(\bar{x}, \epsilon)$ @Ben10 Nov 2, 2020 at 1:43

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.

• that's what I thought he meant. but I feel like there's something wrong somewhere but it's probably just to me. thanks for the clarification btw Jan 19, 2018 at 15:07
• That' not a correct way of thinking of proof! @rapidracim Jan 22, 2018 at 3:11
• when you take limit, at most you have $\leq$ not $<$ Jan 22, 2018 at 3:18
• @Redshoes i disagree, the limit is inside $f$; if it was outside, would be a different story Jan 22, 2018 at 3:50