# For bounded real valued function $f$ show that $\omega_f$ is upper continuous

Let $$f:X\to\Bbb R$$ be bounded, where $$X$$ is a metric space. Let the function

$$\omega_f(x):=\inf_\epsilon \omega_f(x,\epsilon)=\lim_{\epsilon\to 0^+}\omega_f(x,\epsilon)\tag{1}$$

where $$\omega_f(x,\epsilon):=\sup \{|f(z)-f(y)|:z,y\in \Bbb B(x,\epsilon)\}\tag{2}$$ And a function is upper continuous when

$$f(a)\ge \limsup f(x_n),\text{ whenever }(x_n)\to a\tag{3}$$

for all $$a\in{\rm dom}(f)$$. An equivalent definition for $$(3)$$ is: for any $$\epsilon>0$$ exists some $$\delta>0$$ such that

$$x\in\Bbb B(a,\delta)\implies f(x)-f(a)<\epsilon\tag{4}$$

Check this proof please. For this proof I will use a functional definition of limit superior:

$$\limsup_{x\to a}f(x)=\inf_\epsilon\sup f(\Bbb B(a,\epsilon)\setminus\{a\})=\lim_{\epsilon\to 0^+}\sup f(\Bbb B(a,\epsilon)\setminus\{a\})\tag{5}$$

noticing that $$(5)$$ imply the sequential definition of $$\limsup$$ for any sequence $$(x_n)\to a$$, in the same way that the functional definition of limit imply the sequential definition of limit for any sequence that converges to the limit point of the domain.

Writing $$(5)$$ in an $$\epsilon{-}\delta$$ form we can says that if $$L=\limsup_{x\to a}f(x)$$ then for any $$\epsilon>0$$ exists a $$\delta>0$$ such that

$$|\sup f(\Bbb B(a,\delta)\setminus\{a\})-L|<\epsilon\tag{6}$$

Hence from $$(6)$$ and $$(4)$$ we can write

$$\sup\omega_f(\Bbb B(a,\delta)\setminus\{a\})<\epsilon+L\tag{7}$$

And from $$(5)$$ we can write $$L$$ as

$$L=\inf_\delta\sup \omega_f(\Bbb B(a,\delta)\setminus\{a\})\tag{8}$$

Now, using the definitions above, observe that

\begin{align}L&\le \inf_\delta\sup \omega_f(\Bbb B(a,\delta)),&\text{from }(8)\\&=\inf_\delta\sup\{\omega_f(x):x\in\Bbb B(a,\delta)\}\\&\le\inf_\delta\sup\{|f(y)-f(z)|:y,z\in\Bbb B(a,\delta)\},&\text{from }(1)\text{ and }(2)\\&=\omega_f(a)\end{align}

Then from above and $$(7)$$ we finally have that

$$\bbox[border:2px solid gold,8pt]{\forall\epsilon>0,\exists\delta>0:\sup\omega_f(\Bbb B(a,\delta)\setminus\{a\})<\epsilon+\omega_f(a)}\tag{9}$$

what is equivalent to $$(4)$$, that is

$$\forall\epsilon>0,\exists\delta>0:x\in\Bbb B(a,\delta)\implies\omega_f(x)<\epsilon+\omega_f(a)$$

• You are applying the distance $d_X$ to real values ($f(x),f(y)$) above? – copper.hat Nov 20 '16 at 8:19

We want to show that $\omega_f(a)\ge\varlimsup \omega_f(x_n)$ for any sequence $(x_n)$ in $X$ that converge to $a$. First some notation $$\operatorname{diam}(X):=\sup\{d(x,y):x,y\in X\}\implies \omega_f(x)=\lim_{\delta\to 0^+}\operatorname{diam}[f(\Bbb B(x,\delta))]\tag1$$ Hence clearly if $A\subset X$ then $\operatorname{diam}(A)\le\operatorname{diam}(X)$. Let a sequence $(x_n)\to a$ and set $$\epsilon_n:=\sup\{|x_k-a|:k\ge n\}+1/n\tag2$$ Then we find that \begin{align*}\varlimsup \omega_f(x_n)&=\lim_{n\to\infty}\sup_{k\ge n}\omega_f(x_k)\\ &=\lim_{n\to\infty}\sup_{k\ge n}\lim_{\delta\to 0^+}\operatorname{diam}[f(\Bbb B(x_k,\delta))]\\ &\le\lim_{n\to\infty}\operatorname{diam}[f(\Bbb B(a,\epsilon_n))]\\ &=\omega_f(a)\end{align*}\tag3 where we used the fact that for enough small $\delta>0$ we have that $\Bbb B(x_k,\delta)\subset\Bbb B(a,\epsilon_n)$ because $x_k\in\Bbb B(a,\epsilon_n)$ and $\Bbb B(a,\epsilon_n)$ is open.$\Box$