What does it mean that a function is unbounded below in every neighborhood?

Question

In this paper Strong Convexity Does Not Imply Radial Unboundedness

In [3], Tapia gives this result showing that a strongly convex functional is either radially unbounded (and so minima-existence results for bounded domains may be applied to unbounded domains) or it is unbounded below in every neighborhood (in which case the functional has no local minima and is unsuited for optimization).

I don't quite understand about a function is unbounded below in every neighborhood, could someone help to clarify? Thanks!

Answer 1

It means just what it sounds like - that no matter how small you make the neighborhood around a point, you can always find some value in there where the function is extremely deep negative, no matter how small and no matter how deep you want to imagine it.

This can actually occur with a function on the real numbers, too. Such functions are not continuous. An example is

$$f(x) := \begin{cases} -q && \text{if $x = \frac{p}{q}$, rational in reduced form}\\ 0 && \text{otherwise} \end{cases}$$

Suppose $x = 3$. Note that $3 \approx 3 + 10^{-10000} = \frac{3 \times 10^{10000} + 1}{10^{10000}}$. This last fraction is in reduced form due to Bézout's lemma: taking $x = 1$ and $y = -3$ yields that $px + qy = 1$ (where $p$ and $q$ are the numerator and denominator of that fraction). Hence, the value of $f(x)$ at the point $3 + 10^{-10000}$ is $-10^{10000}$, and that will also work in general for $3 + 10^{-n}$ for any integer $n > 0$ by the same logic, e.g. we can even force $f$ to be deeper than negative Graham's number, in that same tiny neighborhood of $3$.

Your example does just the same thing, but in function spaces. Of course, to do that, we need some way to say that functions are/aren't "close together", but that's what the topology (which provides the neighborhoods) does. Then, we have a function of functions $F[f]$, and then it means that, for a given function and functions suitably "close" in that sense, we can do the same thing - no matter how close, we will always be able to find one that pounds the value into the ground as deep as we'd like it to.

Answer 2

The example in this paper is of the form

$$ \|x\|^2 - \ell(x),$$

where $\ell$ is a linear discontinuous functional. You will have to require your underlying space to be infinite-dimensional for this weirdness to occur.

$f$ is unbounded below in every neighborhood – presumably of a point $x_0$ – means that for every $\varepsilon>0$, we have $\inf f\big(B(x_0;\varepsilon)\big) = -\infty$, where $B(x_0;\varepsilon)$ is the ball entered at $x_0$ of radius $\varepsilon$.