I have a slight confusion with the notation $f(x)(y)$.

Does $f(x)(y)$ mean the same as $f(x,y)$?

Or does it mean something else?

Particularly I saw the notation $f(x)(y)$ in the following definition for Eulerian derivative of a functional $J: \Omega \rightarrow J(\Omega)$ to the direction of the vector field $V$:

$$dJ(\Omega;V) = \lim_{t\rightarrow 0}\frac{J(\Omega_t)-J(\Omega)}{t}$$

where $\Omega_t = T_t(V)(\Omega).$

  • $\begingroup$ If i read $f(x)(y)$, I would assume that $f$ is a map, which maps into a space of functions. $\endgroup$ – MooS Mar 22 '17 at 12:02

Let $X$, $Y$ and $Z$ be sets. Suppose that you are given a map $f \colon X \to Z^Y$, where $Z^Y$ is the set of all maps $Y \to Z$. Then, for any $x \in X$, it is legitimate to denote by $f(x)$ the map that sends $y \in Y$ to $f(x)(y) \in Z$.

This is, in particular, the case of derivatives, where at each differentiability point (x) of a function $J$ you want to attach a (linear) map $dJ(x)$ that acts again on points. Anyway, the derivative of $J$ at $x$ along $y$ is often denoted by $dJ(x;y)$, especially on old textbooks.

  • $\begingroup$ So it's a map from one set $X$ to functions between two sets $Y \rightarrow Z$? $\endgroup$ – mavavilj Mar 22 '17 at 12:11
  • $\begingroup$ @mavavilj Yes, this should be the standard interpretation. $\endgroup$ – Siminore Mar 22 '17 at 12:11

They almost mean the same thing: functions given sets $X,Y,Z$, there is a natural correspondence between functions $X \times Y \to Z$ and functions $X \to Z^Y$, where $Z^Y$ is the set of functions from $Y$ to $Z$. The translation between such kinds of functions is known as (un)currying, and is particularly important to computer scientists.

Specifically, given a function $f : X \to Z^Y$, we can define a new function $\widehat f : X \times Y \to Z$ defined by $\widehat f(x,y) = f(x)(y)$. Note that for each $x \in X$, $f(x)$ is a function from $Y$ to $Z$.

Conversely, given a function $g : X \times Y \to Z$, we can define a new function $\widetilde{g} : X \to Z^Y$ defined by letting $\widetilde{g}(x)$ be the function $Y \to Z$ defined by $\widetilde{g}(x)(y) = g(x,y)$.

This correspondence arises for the same reason that, for statements $p,q,r$, the implication $p \wedge q \Rightarrow r$ holds if and only if the implication $p \Rightarrow (q \Rightarrow r)$ holds. The underlying reason is that $Z^Y$ and $q \Rightarrow r$ are both examples of exponential objects.

This may have been more detail than you were looking for, but I hope it's helpful nonetheless.


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