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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 ?

Proof of the theorem

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2 Answers 2

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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.

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  • $\begingroup$ I have a small question related to your answer when ๐œ† -> 1 why we have ๐œ†๐‘ฅ¯+(1โˆ’๐œ†)๐‘งโˆˆ๐ต(๐‘ฅ¯,๐œ–)? $\endgroup$
    – Ben10
    Oct 31, 2020 at 17:58
  • $\begingroup$ because $ \bar{x} \in B(\bar{x}, \epsilon)$ @Ben10 $\endgroup$
    – Red shoes
    Nov 2, 2020 at 1:43
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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.

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  • $\begingroup$ 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 $\endgroup$ Jan 19, 2018 at 15:07
  • $\begingroup$ That' not a correct way of thinking of proof! @rapidracim $\endgroup$
    – Red shoes
    Jan 22, 2018 at 3:11
  • $\begingroup$ when you take limit, at most you have $\leq$ not $<$ $\endgroup$
    – Red shoes
    Jan 22, 2018 at 3:18
  • $\begingroup$ @Redshoes i disagree, the limit is inside $f$; if it was outside, would be a different story $\endgroup$
    – gt6989b
    Jan 22, 2018 at 3:50

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