# Notation for function mapping between cases

Suppose I have a function $f$ which can take different mappings depending on how $f$ is parameterized. For example, I want to write something like this:

$$\label{eq4:bounds_ex1} f: \begin{cases} \mathbb{R}^n \rightarrow \mathbb{C}, & \text{if f is a function of }x \\ \mathbb{R}^{n+m} \rightarrow \mathbb{C}, & \text{if f is a function of }x\text{ and }y \end{cases}$$

Is there a correct (and neat) mathematical way to express this?

Thanks!

• If you mean there are ‘partial’ $f_1\colon A\to C$, $f_2\colon B\to C$, $A\cap B=\varnothing$, and $f = f_1\cup f_2$, then you can say $f\colon A\cup B\to C$. In your case, $f\colon\mathbb R^n\cup\mathbb R^{n+m}\to\mathbb C$. (But is all this indeed necessary?) Commented Dec 19, 2017 at 12:23
• Thanks for your response. Well, I want to be able to distinguish the mappings of the 'partial' functions. But wouldn't the union of $\mathbb{R}^n$ and $\mathbb{R}^{n+m}$ just be $\mathbb{R}^{n+m}$? Commented Dec 19, 2017 at 12:32
• As mere sets, no, $\mathbb R^n\cup\mathbb R^{n+m}$ consists of $n$- and $(n+m)$-tuples of reals. Also, this notation doesn’t imply you can’t ‘distinguish the mappings’, you can state $f(x_1,\ldots,x_n) = f_1(x_1,\ldots,x_n)$ and $f(x_1,\ldots,x_{n+m}) = f_2(x_1,\ldots,x_{n+m})$ or something alike, and these two statements (and $f\colon\mathbb R^n\cup\mathbb R^{n+m}\to\mathbb C$) would in fact count as a definition of $f$. But this is not an absolutely clear stating; maybe you’d better leave $f_1$ and $f_2$ separate and not united in $f$. Commented Dec 19, 2017 at 12:50

I think it is best to create $2$ different functions. Functions by definition have $1$ domain. Thus, a function cannot take elements from two different spaces. There are ways to embed $\mathbb{R}^n$ in $\mathbb{R}^{n+m}$, so that your function can be thought of as a function from $\mathbb{R}^{n+m}$ that "looks like" a function from $\mathbb{R}^n$ under certain conditions. However, I think it is much clearer to create $2$ functions $f_1$ and $f_2$ instead.