Let $f: \mathbb{R} \rightarrow [0, B]$ and for every $\varepsilon > 0$, let $\varphi_{\varepsilon}(x) := \sup_{\{y: |x - y| < \varepsilon\}}f(y)$. Since for each fixed $x$, $\varphi_{\varepsilon_{1}}(x) \leq \varphi_{\varepsilon_{2}}(x)$ for $\varepsilon_{1} < \varepsilon_{2}$, it follows that $\lim_{\varepsilon \rightarrow 0}\varphi_{\varepsilon}(x)$ exists. Let $\varphi(x)$ be this limit. Why must $\varphi$ be upper semi-continuous?

  • $\begingroup$ What does it mean that $\varphi(x) < c$? What follows for $y$ close enough to $x$? $\endgroup$ – Daniel Fischer Jul 21 '14 at 18:31

Your Answer

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

Browse other questions tagged or ask your own question.