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Simple question, fully expressed in the Title line. Is the dot within the parenthesis intended to mean, "any possible function"?

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    $\begingroup$ Just to avoid a dummy variable here. $\endgroup$
    – MonkeyKing
    May 17 '15 at 16:17
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    $\begingroup$ Unless $\cdot$ denotes a variable or a constant, I would believe $f(\cdot)$ means $f : x \mapsto f(x)$, although $f=f(\cdot$), so it's sort of pointless. $\endgroup$ May 17 '15 at 16:18
  • $\begingroup$ And how do you get the dot right in the middle with latex? $\endgroup$ May 17 '15 at 16:20
  • $\begingroup$ @toni Try \cdot $\endgroup$ May 17 '15 at 16:25
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    $\begingroup$ Just $f(\cdot)$ is rather pointless. The notation is more useful when you want to refer to a function which is defined in some more complicated fashion in terms of some other function. For instance $f=g(x,\cdot)$ is the same as $f(y)=g(x,y)$. Another important place is when you have a functional operator like convolution. For instance, $f(x,\cdot) * g(y,\cdot)$ means "convolve $f$ and $g$ in their second arguments, with the first arguments fixed as $x$ and $y$ respectively". It might be written as $f(x,z) *_z g(y,z)$. $\endgroup$
    – Ian
    May 17 '15 at 16:29
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Usually we use it to avoid writing more letters $x$, $y$, etc. One example I see a lot: let $B: V \times V \to W$ be a bilinear form, and fix $y \in V$. When we write $B(\cdot, y)$, we mean the map $$V \ni x \mapsto B(x,y) \in W,$$ so we don't write this extra $x$ if we don't need to. If we're going to write $f(\cdot)$ just like this, as in the title question, then there isn't much advantage - just talk about the function $f$ and be done with it. The advantage I see is where you want to simplify the writing of some function that uses another one "in the background", like the example with the bilinear form I gave above.

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    $\begingroup$ When you write a function, you usually write $f:A \to B$, no? Instead of $A$ and $B$, I used $V$ and $W$. Also, $V \times V = \{ (x,y) \mid x,y \in V \}$ is the cartesian product that you have probably seen in some point of your life. And I used $B$ instead of $f$. So $B : V \times V \to W$ is a function $B$, from the set $V \times V$ to the set $W$. You can safely ignore the words "bilinear form" here - it was just for giving context to the explanation, but not needed at all. $\endgroup$
    – Ivo Terek
    May 17 '15 at 16:45
  • $\begingroup$ Does the "dot" have a name? $\endgroup$
    – Mel
    Jul 4 '17 at 3:37
  • $\begingroup$ Hmmm... "dot"? :-P $\endgroup$
    – Ivo Terek
    Jul 5 '17 at 15:09

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