# Uniform semi-continuity

It is a standard and important fact in basic calculus/real analysis that a continuous function on a compact metric space is in fact uniformly continuous. That is, suppose $(X,d)$ is a compact metric space and $f\colon X \to\mathbb R$ is such that for every $x\in X$ and $\varepsilon>0$ there exists $\delta>0$ such that $d(x,y)<\delta$ implies $|f(x)-f(y)|<\varepsilon$. Then in fact, such a $\delta$ can be chosen independently of $x^. ## Question Does a similar statement hold regarding semi-continuous functions? For concreteness, let's consider upper semi-continuous functions, so suppose$(X,d)$is compact and$f\colon X \to\mathbb R$has the property that for every$x\in X$and$\varepsilon >0$there exists$\delta >0$such that$d(x,y)<\delta$implies$f(y) < f(x)+\varepsilon$. (Note the asymmetry of$x$and$y$in this definition.) Then is it true that$\delta=\delta(\varepsilon)$can be chosen independently of$x$? ## Reformulation Given$\delta, \epsilon > 0$, consider the set $$X_\delta^\epsilon := \lbrace x\in X \mid f(y) < f(x) + \epsilon \text{ for every } y\in B(x,\delta) \rbrace.$$ Then$f$is upper semi-continuous if and only if$\displaystyle\bigcup_{\delta>0} X_\delta^\epsilon = X$for every$\epsilon > 0$, and$f$is uniformly upper semi-continuous if and only if this union stabilises -- that is, if for every$\epsilon > 0$there exists$\delta>0$such that$X_\delta^\epsilon = X$. - ## 2 Answers$f(x)=0 \ (x\le0)$,$f(x)=-1/x \ (x\gt0)$is upper semi-continuous on$[-1;1]$— but not uniformly. - Of course... thanks for the example. Do you know what happens if we require the function f to be bounded below? (In the context that motivated the question, I'm considering such functions.) – Vaughn Climenhaga Aug 9 '10 at 14:09 Turns out boundedness doesn't help (see my answer). In retrospect this all seems quite obvious... maybe this is why I shouldn't post questions in the wee hours of the morning. – Vaughn Climenhaga Aug 9 '10 at 15:10 A little further thought reveals the following: uniform semi-continuity implies uniform continuity. Thus the answer to my question is a resounding "no", since any function that is upper semi-continuous but not continuous cannot be uniformly upper semi-continuous. Proof. Let$f$be uniformly upper semi-continuous. Then for every$\epsilon>0$there exists$\delta>0$such that for every$x\in X$, we have$f(y) < f(x) + \epsilon$whenever$y\in B(x,\delta)$. However, since this statement holds for every$x$, it also holds with$x$and$y$reversed; in the language of the original post, both$x$and$y$are contained in the set$X_\delta^\epsilon = X$. Since$y$is in this set and$x\in B(y,\delta)$, we also have$f(x) < f(y) + \epsilon$, and thus$|f(x) - f(y)| < \epsilon\$. But this is just the definition of uniform continuity.

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ah! (I should have thought about this instead of constructing artificial counter-examples) – Grigory M Aug 9 '10 at 15:22