Let $f:\mathbb{R} \to \mathbb{R}$ be a function, what does $f(\cdot)$ mean usually? Is it another way of writing this function, or is it a real number?
2 Answers
It is another way of writing the function, emphasising that the value of $f$ at, for instance, $5$ is written as $f(5)$, and not $f5$ or $5f$ or $(5)f$ or $f|_5$ or anything else. The dot is just a placeholder.
Some would write this as $f(x)$ instead of $f(\cdot)$, but this is a slightly different emphasis again. The notation $f(x)$ tends to be associated to a specific description of $f$, for instance $f(x) = 4x-3$.

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The meaning of the symbol $f(\,)$ can be somewhat ambiguous depending upon the context.
For example, if $f(x)=x^3$ then in the equation $y=f(x)$ the meaning of $f(x)$ is $x^3$.
However, in the equation $y=f(2)$ the meaning of $f(2)$ is the number obtained by cubing 2.
So in the former equation, it represents a mathematical expression and in the latter it represents a number.

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1So the symbol $f(\cdot)$ refers to the function $f$, and the symbol $f(x)$ refers to the number that $f$ would output if we put $x$ in it? Is this the reason why people write $f(\cdot)$ instead of $f(x)$? – user56834 Aug 31 '17 at 16:27
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Personally, I have never found it necessary to use the notation $f(\cdot)$. If I wish to refer to the function alone, I would simply write $f$. The expression $f(x)$ can simultaneously stand for a mathematical expression in $x$ such as, for example the expression $x^3$ as well as the number resulting from performing the operations denoted by that expression. If, instead of a variable, one has a constant such as $x=2$ then $f(2)$ means the number resulting from performing the operations. – John Wayland Bales Aug 31 '17 at 16:42
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So if we define $f(x)=x^3$ then $f(2)$ stands for the number $2^3=8$. Now the expression $f(\cdot)$ could merely be a way of indicating that $f$ is a function of a single variable without expressing how it is defined. – John Wayland Bales Aug 31 '17 at 16:42
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: The function goes from $\Bbb R$ to $\Bbb R$, and $x$ is mapped to $y$. I agree that the difference is subtle, but it's definitely established. – Arthur Jan 25 '17 at 23:40