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Let $E$ be a normed space. If $E'$ (dual of $E$) is separable then we know that the balls of $E''$ (bidual of $E$) are metrizable for the weak star topology on $E''$. If I have a function $f\colon E'' \to \mathbb{R}$ that is inferiorly semicontinuous with respect to the weak star topology. How can I conclude that $f$ has a minimum point? Furthermore, I know that an lower semicontinuous function over a compact reaches a minimum. My question is: Is it possible to somehow argue that these balls in $E''$ are compact in the weak star topology? Or is there another argument that I'm not seeing?

Thanks for the help!!!

AlphaMaths
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  • What is the definition of "inferiorly semicontinuous" – Evangelopoulos Foivos Mar 10 '25 at 19:20
  • @EvangelopoulosFoivos A function $f: X \to \mathbb{R} \cup {-\infty, +\infty}$ defined on a topological space $X$ is called lower semicontinuous at a point $x_0 \in X$ if $\liminf_{x \to x_0} f(x) \geq f(x_0)$. But for my specific case $f: E''\to \mathbb{R}$ being lower semicontinuous is equivalent to saying that if $(x_n)_{n\in \mathbb{N}}$ is a sequence of elements of $X''$, if $x_n$ converges weakly star to $x_0$ then $||x_0||\leq \liminf ||x_n||$. – AlphaMaths Mar 10 '25 at 19:27
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    @EvangelopoulosFoivos "inferiorly semicontinuous" is the same thing "lower semicontinuous" for me. – AlphaMaths Mar 10 '25 at 19:32
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    If $f$ is coercive, i.e. $f(x)\to \infty$ as $|x|\to\infty$ then $f$ attains its infimum, you can work as in https://math.stackexchange.com/questions/499326/prove-a-lower-semi-continuous-and-coercive-function-attains-its-infimum-and-is-b – Evangelopoulos Foivos Mar 10 '25 at 19:48

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