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Suppose $X$ is a metrizable space. A real-valued function $f:X \rightarrow \mathbb{R}$ is upper semicontinous if for any real number $c$, its preimage $f^{-1}(-\infty,c)$ is open in $X$.

In this post, we see that every lower semicontinuous can be expressed as thesupremum of increasing continuous functions, where the sequence of continuous functions is defined as $f_k(x) = \inf \{ f(y) + k d(x,y): y \in X \}$.

I am curious as to how would one show that an upper semicontinuous function $f$ can be expressed as the infimum of non-increasing continuous functions that converge pointwise, by using the definition of upper semicontinuity above, which is $f^{-1}(-\infty,c).$

Michael Greinecker
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Idonknow
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  • $f$ is usc. iff $-f$ is lsc. – copper.hat Apr 10 '17 at 02:59
  • @copper.hat: Yes, I know that. But still I do not see how the fact can help to us to construct a sequence of continuous functions from $f^{-1}(-\infty,c)$ for any real number $c$. – Idonknow Apr 10 '17 at 03:07
  • Why not use the same general form as above (with appropriate sign and $\sup,\inf$ adjustments? Use the above to construct some $\phi_k$ that is the $k$th approximation of $-f$. Then let $f_k = - \phi_k$. – copper.hat Apr 10 '17 at 03:10

1 Answers1

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If $f$ is usc. then $-f$ is lsc.

Let $\phi_k(x) = \inf_y (-f(y) + k d(x,y)) $, and $f_k(x) = - \phi_k(x) = -\inf_y (-f(y) + k d(x,y))$.

Rearranging gives $f_k(x) = \sup_y (f(y) - k d(x,y))$.

copper.hat
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