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Let $X$ be some space (eg. vector space or Banach space).

When is it true that: for any $\epsilon >0$ small, there exists an $f \in X$ such that $$(1+\epsilon) \inf_{g \in X} I(g) \geq I(f)?$$ Here $I:X \to \mathbb{R}$.

EasyStarter
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1 Answers1

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This is true whenever $$\inf_{g \in X}I(g) > 0 \ \ \ \mbox{ or } \ \ \ \min_{g \in X}I(g) = 0$$ to see this, simply apply the definition of $\inf$.

On the contrary, if $\inf_{g \in X}I(g) < 0$, you cannot find such an $f$ because otherwise $$I(f) \le (1+ \epsilon) \inf_{g \in X}I(g) < \inf_{g \in X}I(g) \le I(f)$$ a contradiction.

Crostul
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  • Thanks. So $f$ depends on $\epsilon$. Can we say that $f(\epsilon) \to \text{arginf}_g I(g)$ as $\epsilon \to 0$, or something similar? – EasyStarter Jul 18 '15 at 14:31
  • well, arginf doesn't exist necessarily, so maybe something similar holds? – EasyStarter Jul 18 '15 at 14:32
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    You need some hypothesis on $I$ (e.g. continuity) to say something like this. But I am not so sure. Actually I am not familiar with arginf at all. – Crostul Jul 18 '15 at 14:42