For every real function $f$, is there an everywhere continuous real function $g$ above $f$?

Question

This is a follow-up to my previous question, here: For every real function $f$, is there a strictly increasing real function $g$ above $f$?. Suppose $f$ is a function from reals to reals. It can be any kind of function, not necessarily continuous. Does there exist a real function $g$ such that $g$ is everywhere continuous and everywhere greater than or equal to $f$?

Answer 1

The problem here is that an arbitrary function $f$ can be incredibly complicated and pathological. There is no real way of getting control over such an arbitrary function, hence there is little hope of bounding such a function (even on some bounded interval). As an example, consider the function $f$, defined by the formula $$f(x) = \begin{cases} (-1)^q q & \text{if $x = p/q$ is rational, in lowest terms, and} \\ 0 & \text{if $x$ is irrational.} \end{cases}$$ This function is unbounded (both above and below) on any interval. As such, this function cannot be bounded by a continuous function, since every continuous function is bounded on any closed interval.

Answer 2

For the question as stated, there is a very simple counterexample: $$f(x)= \begin{cases} 1/x & \text{if }x\neq0\\ 0 & \text{if }x=0 \end{cases}$$ This function is unbounded on the interval $[0,1]$. Any continuous function on a closed interval achieves a maximum and a minimum, so it cannot dominate $f(x)$.

(One says that $g$ dominates $f$ on a set $S$, if $f(x)\leq g(x)$ for all $x\in S$.)

If you weaken the domination requirement, then examples become harder to construct. This function $f$ is not dominated by a continuous function on any open interval: $$f(x) = \begin{cases} q & \text{if }x = p/q \text{ is rational in lowest terms}\\ 0 & \text{if }x \text{ is irrational} \end{cases}$$ However, the identically-zero function dominates this with countably many exceptions.

Suppose we ask, is there a function $f$ such that no everywhere continuous dominates $f$, even allowing countably many exceptions? The answer is yes. We can construct it using the well-ordering theorem. Let $$\{r_\alpha : \alpha<\gamma\}$$ be a well-ordering of the interval $[0,1]$; here, $\gamma$ is the first ordinal with cardinality $2^{\aleph_0}$. Now, every ordinal is of the form $\lambda+n$ where $\lambda$ is a limit ordinal or 0, and $n$ is a natural number (aka finite ordinal). Define $$f(r_{\lambda+n})=n$$ and define $f$ arbitrarily outside the interval $[0,1]$.

Suppose $g$ is an everywhere continuous function (or just continuous on the interval $[0,1]$). Let $E$ be a countable set. We demand that $f(x)\leq g(x)$ only for $x\not\in E$. Since $E$ is countable, there is a limit ordinal $\lambda<\gamma$ such that all elements of $E\cap[a,b]$ have indices less than $\lambda$ in the enumeration. Say $M$ is the maximum of $g(x)$ on $[0,1]$, and let $n$ be greater than $M$. Then $f(r_{\lambda+n})>g(r_{\lambda+n})$, and $r_{\lambda+n}$ is not in $E$.

With a minor modification, $f$ is not dominated on any interval, even allowing countably many exceptions. For the initial enumeration $\{r_\alpha : \alpha<\gamma\}$, demand also that $|r_\lambda-r_{\lambda+n}|<1/n$; this is easily done using transfinite induction. For any interval $(a,b)\subset[0,1]$, we must have a limit ordinal $\lambda$ such that $r_\lambda\in(a,b)$ because $(a,b)$ has cardinality $2^{\aleph_0}$ and $\gamma$ is the first ordinal with cardinality $2^{\aleph_0}$. Also we can demand that $\lambda$ is larger than the indices of all the elements of $E$, as before. For $n$ large enough, $r_{\lambda+n}$ is also in $(a,b)$, not in $E$, and $f(r_{\lambda+n})>g(r_{\lambda+n})$ as before.

aschepler suggests in a comment a more concrete construction which accomplishes the same thing. On the interval $[0,1]$, let $f(x)=n$ if the longest string of consecutive zeros in the binary expansion of $x$ is $n$. If there is no maximum, then $f(x)=0$. Given any interval $(a,b)$, we pick a prefix so that all $x$'s with that prefix lie in $(a,b)$. (This is always possible because the finite-length binary expansion represent the dyadic rationals, which are dense in $[0,1]$.) There are clearly an uncountable number of $x$'s with that prefix for which $f(x)=n$, so at least one is not in $E$.