# Using “$\cdot$” as variable placeholder

I've occasionally come across the use of "$\cdot$" as placeholder for a variable, most recently in a paper on radial basis functions, which were defined as

$$s(\cdot) = p(\cdot) + \sum_{i=1}^N \lambda_i \phi(|\cdot - x_i|).$$

What's the benefit of writing "$s(\cdot)$" rather than "$s(x)$", which would be easier to parse (IMO)?

• It's simply cool. – pppqqq Sep 25 '13 at 20:16
• I agree that in this case, using $x$ rather than "$\cdot$" would look much nicer. – Tyler Holden Sep 25 '13 at 20:23
• $s(\cdot)$ is an application while $s(x)$ is its value in $x$. – Pece Sep 25 '13 at 20:25

I've seen this used to curry a two(or more)-variable function. If $f = f(x,y)$ is a function of two variables, then it's nice to be able to write something like $f(x,-)$ to indicate the one variable function $f_x(y) := f(x,y)$.
As $f(x)$ is not a function but rather the value of the function $f$ evaluated at $x$, one should never use $f(x)$ to denote a function. So what one does to denote a function given by an expression is to write either $x\mapsto f(x)$, which is somewhat cumbersome, or use a notation like $f(\cdot)$ for this.
• Why not use just $f$ to refer to the function instead of $f(\cdot)$? – seek Jul 20 '15 at 16:12
The example that first comes to my mind is when the map you are defining has an output which still takes an input. For example, if $V$ is a finite dimensional vector space over a field $F$ let $V^*$ be its dual. Let $\langle \cdot,\cdot \rangle: V^* \otimes V \to F$ be the natural pairing (this was not the intended use of cdot, but there you go!), so that $\langle f,v \rangle = f(v)$. There is a canonical map $^\sharp: V \to V^{**}$ given by $v^\sharp = \langle \cdot, v \rangle$, that is, $v^\sharp(f) = \langle f,v \rangle = f(v)$. In this instance, if I omitted the cdot I would have to define $v^\sharp$ by inputting an element of $V^*$.
In this particular instance, I can see no particular benefit, but this notation has some uses. For example, assume that $f$ is a function of two variables, and we want to compute the Laplace transform with respect to the second variable. Then $$\mathcal L(f(x,\cdot))$$ is a convenient way to write it.