Let $f:X\rightarrow \Bbb{R}\,\cup\,\{+\infty\}$ be a map where $X$ is a real normed space. Then, for an arbitrary $x_0\in X,$ I want to prove that the following always holds:

\begin{align}f(x_0)\geq \sup\limits_{V\in U(x_0)}\inf\limits_{x\in V}f(x) \end{align}


Let $x_0\in X,$ then $\exists\,V\in\,U(x_0)$ such that $f(x_0)\geq f(x),\;\forall\;x\in V$, where $U(x_0)$ is the set of all neighbourhoods of $x_0$. So,

\begin{align}f(x_0)\geq \inf\limits_{x\in V}f(x),\;\;\text{for some}\;\;V\in U(x_0), \end{align} which implies that \begin{align}f(x_0)\geq \sup\limits_{V\in U(x_0)}\inf\limits_{x\in V}f(x) \end{align}

Kindly check if I am correct or wrong. If it happens that I'm wrong, kindly give an alternative proof. Thanks

  • $\begingroup$ You need to give us a hint - what is $U(x_0)$? $\endgroup$ – David C. Ullrich Nov 24 '18 at 13:32
  • $\begingroup$ @David C. Ullrich: Sorry! $U(x_0)$ is the set of all neighbourhoods of $x_0$ $\endgroup$ – Omojola Micheal Nov 24 '18 at 13:34
  • $\begingroup$ @Mike Oh, so $X$ is a topological space? But why should that matter when $f$ is not necessarily continuos? $\endgroup$ – Hagen von Eitzen Nov 24 '18 at 13:35
  • $\begingroup$ @Hagen von Eitzen: Yes! Infact, $X$ is a real normed space. $\endgroup$ – Omojola Micheal Nov 24 '18 at 13:36

There is not the slightest reason why $V\in U(x_0)$ with $f(x_0)\ge f(x),\forall x\in V$ should exist.

However, if $V\in U(x_0)$ then $x_0\in V$ (here, I make a wild guess that $U(x_0)$ denotes some non-empty subset of the power set of $X$ and each of its elements contains $x_0$) and hence $\inf_{x\in V} f(x)\le f(x_0)$.


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