Is there a math symbol for $f(x) > g(x)$ except when $x=0$, in which case $f(x)=g(x)$?

Question

Title says it all, except that $x \in R^n$. For want of a better choice, I've specified the amsmath symbol $\gtrdot$ for this relation, but probably this actually means something quite different that google hasn't revealed?? If no such symbol exists, and my choice means something different, does anybody have an alternative suggestion? Thanks very much!

Answer 1

No, there isn't, there doesn't need to be.

When you write "$f(x) > g(x)$, except when $x = 0$, in which case $f(x) = g(x)$," that's plenty clear. Getting carried away with symbols might obscure rather than clarify your meaning in this instance.

If you're dealing with a whole bunch of functions related to each other in this manner, what you might need is some kind of terminology. I don't know what that terminology is, so I'll just "marklarable" for the sake of providing examples. Thus, in your paper, you might write "$f(x)$ is marklarable to $g(x)$, as is $\alpha(x)$ to $\beta(x)$."

Answer 2

You could define a Boolean function: $$D^+(f, g) = \delta^{f(0)}_{g(0)} | f(x) > g(x) \forall x \neq 0 |$$ Presumably everyone reading your paper knows what the Kronecker delta and the Iverson bracket are. It's much easier to look for a prior definition of a function than it is to look for a prior use of a symbol, as you yourself have already experienced.

For example, if $f(x) = 3x^2$ and $g(x) = \frac{x^2}{3}$, then $D^+(f, g) = 1$. But with the prime counting function $\pi_1(x)$ and the semiprime counting function $\pi_2(x)$, we'd have $D^+(\pi_1, \pi_2) = 0$ because we also have $\pi_1 = \pi_2$ for $-2 < x < 2$.

Then, later on, to use Bob's example, you could write $D^+(\alpha, \beta) = 1$.