3
$\begingroup$

For any bounded function $f: X \to \mathbb R$, not-necessarilly continuous, one can define for any $x$, the real functions $$ \limsup f(x) = \inf_{U\in\mathcal V_x} \sup_{u\in U} f(u) $$ and $$ \liminf f(x) = \sup_{U\in\mathcal V_x} \inf_{u\in U} f(u), $$ where $\mathcal V_x$ denotes the set of open neighborhoods of $x$. Then, we can measure the oscilation of $f$ in $x$ as $$o_f(x) = \limsup f(x)-\liminf f(x).$$ The greater the value of $o_f(x)$ the farther $f$ is from any function $g$ continuous at $x$.

The Katetov-Tong theorem guarantees that, provided $X$ is a normal space, there exists a continuous function $g: X \to \mathbb R$ such that $$\|f-g\|=\frac{1}{2}\sup_{x\in X} o_f(x),$$ where the norm considered is the usual supremum norm.

I would like to know whether it is possible to obtain an analogous result for functions $f: X \to \mathbb C$. The closest result I got from this relies on Michael's Selection Theorem. I was able to prove that, provided $X$ is a paracompact space, for any $f: X \to \mathbb C$, there exists a continuous $g: X\to \mathbb C$ such that $$ \|f-g\|=\frac{1}{2}\sup_{x\in X} o_f(x), $$ where $o_f(x)$ is defined as the diameter of the smallest ball containing all limit points of nets of the form $(f(x_i))_{i\in I}$ with $x_i\to x$. This definition coincides with the definition of $o_f$ for real functions.

I wonder if this result is valid not only for paracompact spaces but also for normal spaces, giving us a complex version of Katetov-Tong's theorem.

$\endgroup$
1
  • 3
    $\begingroup$ This gives a worse estimate, but you can just say that $f=(f_1,f_2)$, where $f_1$ and $f_2$ are real-valued, and then produce continuous $g_1$ and $g_2$ based on them. $\endgroup$
    – erz
    Commented Aug 3, 2020 at 9:57

0

You must log in to answer this question.

Browse other questions tagged .