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 $f^{-1}(-\infty,c).$

  • $\begingroup$ $f$ is usc. iff $-f$ is lsc. $\endgroup$ – copper.hat Apr 10 '17 at 2:59
  • $\begingroup$ @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 $c$. $\endgroup$ – Idonknow Apr 10 '17 at 3:07
  • $\begingroup$ 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 $f_k = - \phi_k$. $\endgroup$ – copper.hat Apr 10 '17 at 3:10

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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.