TITLE: Is $f(x,y) = f(\mathbf{x})$ abuse of notation?
QUESTION [10 upvotes]: A scalar function $f(x,y)$ is often written as $f(\mathbf{x})$, where $\mathbf{x} = (x,y)$, but as far as I know, there is a difference between the scalar function inputs $(x,y)$ and the vector input $(x,y) = x\imath+y\jmath$. As I see it $f(\mathbf{x}) = f((x,y)) = f(x\imath+y\jmath) \neq f(x,y)$. Am I wrong, or is there a simple bijection between the two concepts?
Is it simply shorthand for $f': \mathbf{x} \mapsto f''(\imath\cdot\mathbf{x},\jmath\cdot\mathbf{x})$, s.t. $f'(\mathbf{x})= f''(x,y)$?
If it's of an relevance, I'm reading about scalar fields, and this definition came up:  

$\displaystyle\dfrac{\partial f}{\partial x}(x,y) = \lim_{h\to 0} \dfrac{f(x+h, y)-f(x,y)}{h} \overset{\color{green}{?}}{=} \dfrac{f(\mathbf{x}+h\imath)-f(\mathbf{x})}{h} = \dfrac{\partial f}{\partial\imath}(\mathbf{x})$

While it looks nice I'm just curious if it's correct. However I don't see how $f$ can be differentiated with respect to both $\imath$ (a vector) and $x$ (a scalar), unless it's actually two different functions $f'$ and $f''$...

REPLY [3 votes]: A viewpoint that hasn't been mentioned yet:
We can take $f\mathbf{x}$ and $f\langle x,y\rangle$ as the "correct" notations, and then stipulate that we can include round brackets wherever we want, as disambiguators. Under these conventions, the notations $$f(\mathbf{x}),\quad f\langle x,y\rangle$$ are perfectly acceptable.
This also gives us notation for "global" elements; if $x \in X$, then the corresponding function $1 \rightarrow X$ can be denoted $\langle x \rangle,$ since by definition, this is an element of $X^1$.