A nowhere locally bounded function? Hi everyone: Let $O$ be an open set of $\mathbb{R}^m $. Is there a real-valued function $f(x)$ defined on $O$ and an open subset $V$ of $O$ such that $f$  is not locally bounded at any point of $V$?
Thanks in advance.
 A: Sure, there are loads of ways to do this.


*

*Let $f$ be the function that writes out the decimal expansion of each co-ordinate, moves the decimal point to after the last $5$ (if any), and sums the results. So for example, $(0.0353, 0.7776523) \mapsto 35.3 + 77765.23$.

*Similarly, but instead write each rational co-ordinate as $p/q$ in lowest terms, and map it to $q$.

*Identify a countable family of pairwise-disjoint dense subsets (e.g. rationals and cosets of the rationals) and map the first to $1$, the second to $2$, the third to $3$, and so on.

A: Let's just look at this for $\mathbb{R}$, as you can just copy the function to each other $\mathbb{R}$ in $\mathbb{R}^m$. 
Then I encourage you to look into the history of Cauchy's function equation:
$$ f(x + y) = f(x) + f(y).$$
There are two types of functions that satisfy this functional equation: linear functions $f: x \mapsto \lambda x$, and pathological, nowhere continuous, dense additive functions that are difficult to describe. 
It's easy to show that any such $f$ satisfying this equation satisfies $f(0) = 0$, $f(-x) = -f(x)$, and $f(nx) = nf(x)$ for each integer $n$. (In fact, $f(rx) = rf(x)$ for each rational $r$). 
If we assume the axiom of choice, there is a basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$. Then we are free to define our function however we like on each element of this basis (such a basis is called a Hamel basis, and the function we are creating is sometimes called a Hamel Function). 
For each $z \in \mathbb{R}$, there is a unique set of basis elements $x_1, \ldots, x_n$ such that $z = \sum r_i x_i$ with $r_i \in \mathbb{Q}$. The behaviour of $z$ under $f$ will be based upon how $f$ handles the $x_i$. If $f(1) = \alpha$ (meaning that $f(r) = \alpha r$ for every rational $r$), then as long as $f(x_i) \neq \alpha x_i$ for at least one $x_i$, then the resulting function $f$ is everywhere dense, nowhere continuous, and not locally bounded. 
