TITLE: Uniform semi-continuity
QUESTION [10 upvotes]: Background
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 $X_\delta^\epsilon = X$.

REPLY [7 votes]: 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.