# How to properly describe or define symmetric (even) function in $\mathbb{R}^n$

I've recently been thinking how to properly describe symmetric (even) function in any $\mathbb{R}^n$ with $n\in\mathbb{N}^*$.

I know that in the case of a real scalar function an even function is defined by: $$f\mbox{ is an even function} \Longleftrightarrow \forall x\in D, -x\in D,\ \quad f(x)=f(-x)$$

Where D is the definition domain of the function $f$

Then, i've been thinking of a generalization of this property for bigger spaces $\mathbb{R}^n,\ n>1$

For example, in the $\mathbb{R}^3$ case (2 variables scalar function), i came up with this idea ( i might be wrong)

\begin{align*}f\mbox{ is an even function}\Longleftrightarrow& \forall(x,y)\in D,\ \left\{\begin{array}{l} (-x,-y)\in D\\ (-x,y)\in D\\ (x,-y)\in D \end{array}\right.\\& \mbox{and}\ f(x,y)=f(-x,-y)=f(-x,y)=f(x,-y) \end{align*}

The extension of the idea to any $\mathbb{R}^n,\ n>3$ is formulated as follows:

We begin by defining the matrix we can call the symmetry matrix $S_n\in M_{n\times(2^n-1)}(\mathbb{R})$.

For $n=2,3$: $$S^2=\left(\begin{matrix}-x_1 & -x_1 & x_1 \\ -x_2& x_2&-x_2\end{matrix}\right)\quad S^3=\left(\begin{matrix}-x_1&-x_1&-x_1&-x_1&x_1&x_1&x_1\\ -x_2&-x_2&x_2&x_2&-x_2&-x_2&x_2\\ -x_3&x_3&-x_3&x_3&-x_3&x_3&-x_3 \end{matrix}\right)$$ Then, we can say that $f$ is an even function if and only if: $$\forall (x_1,\cdots,x_n)\in D,\ \forall j\in[1,2^n-1],\ S_{i=1,\cdots,n;j}^n\in D\ \ \mbox{and}\ \ f((S_{i=1,\cdots,n;j}^n)^T)=f(x_1,\cdots,x_n)$$

Questions:

1. I would like to know if this approach to extend the even notion in any finite dimension space is correct?

2.Where are my mistakes, if i made any?

• Do you know what is a symmetric polynomial? A symmetric function is an extension of this idea Nov 28, 2017 at 15:58
• The concept of symmetry you mentioned is about interchanging variables, which is not the same i described in the question. I am talking about extending the notion of 'even function' (its graph is symmetric with respect to the y axis) Nov 28, 2017 at 16:02
• I see. So maybe that the better extension is to asymmetry with respect to the origin, as $f(\vec x)=f(-\vec x)$ ? Nov 28, 2017 at 16:06
• The standard way to extend the definition of even for functions of many variables, is that the function takes the same value at $x$ as it does at $-x$, where $x \in \mathbb{R}^n$. It is not the only way how one can extend the concept from a function of 1 variable to a function of many variables, but it is the common standard way. You have indeed found another one, which could be useful in some instances indeed. Dec 15, 2017 at 22:06
• More generally, if $\Gamma$ is for instance a finite subgroup of $O(n)$, where $O(n)$ is the group generated by rotations and reflections in $\mathbb{R}^n$, you can define $f$ to be $\Gamma$-invariant, if $f$ has the same value at each of point in a $\Gamma$-orbit. Dec 15, 2017 at 22:09

To me, the most obvious definition would be the following: Let $f:\mathbb{R}^n \to \mathbb{R}^m$. We can think of the input being a vector, or a column matrix of size $n \times 1$. So we are looking at $f(X)=Y$, where $X=\begin{pmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \\ \end{pmatrix}$

and $Y=\begin{pmatrix} y_{1} \\ y_{2} \\ \vdots \\ y_{m} \\ \end{pmatrix}$

Of course, scalar multiplication is well defined on matrices, so the most logical way to extend the definition would be to let $F(X)=F(-X)$. Any other definition seems to require a more requirements and hence would need a different word than even I think.

E.g. $f:\mathbb{R}^2 \to \mathbb{R}$ given by $f(x,y)=xy$ would be even under this definition, as $f(-(x,y))=f(-x,-y)=(-x)(-y)=xy$. Hope this helps.

I understand what are you aiming at, and I would go in exactly the same way from one-variable to two-variable case, and further to more-than-two-variables cases.

In one variable case, evenness means that the function takes equal values at the points symmetric with respect to the origin, and they are points of a line segment (one-dimensional cube with opposite vertices symmetric with respect to the origin).

In two variable case, evenness would mean that the function takes equal values at $2^2=4$ points that are vertices of a square (two-dimensional cube) that has opposite vertices symmetric to the origin and has the origin as its intersection of diagonals (as its "center")).

More generally, evenness in the case of a function with $n$ variables would mean that the function of $n$ real variables (that is, of a $n$-tuple) takes equal values at $2^n$ points of a hypercube that has opposite vertices symmetric with respect to the origin and has origin as its "center".

You now only have to prove that for every point in $\mathbb R^n$ there are $2^n-1$ points in $\mathbb R^n$ that are such that those $2^n$ points uniquely determine one and only one hypercube that has an origin as its "center" and has opposite vertices symmetric with respect to the origin.

This is all non-rigorous and imprecise but it shows you that you only need to do some job with hypercubes to successfully extend the notion of evenness (in at least one way, maybe the most natural of all possible ways) to functions of $n$ variables.

• Well first thanks for the answer. It is indeed a very good approach to my proposition, but this is a pure geometrical extension of the evenness notion. And i think that an analytical approach would be more direct and useful in this topic Dec 15, 2017 at 18:32
• @hamzaboulahia There is nothing analytical needed to extend successfully that notion. We only need some kind of symmetry that hypercubes centered at origin have.
– user480281
Dec 19, 2017 at 22:09

I think the most natural definition is as follows: A function $f:\mathbb{R}^n\to\mathbb{R}^m$ is called even if $f(-x)=f(x)$, for all $x\in\mathbb{R}^n$ and called odd if $f(-x)=-f(x)$, for all $x\in\mathbb{R}^n$.

This is essentially defining evenness or oddness with respect to the origin. One nice property that stays preserved is that any function $f:\mathbb{R}^n\to\mathbb{R}^m$ can be written as a sum of an even and an odd function as follows. $$f(x)=\frac{1}{2}\big(f(x)+f(-x)\big)+\frac{1}{2}\big(f(x)-f(-x)\big),\quad x\in\mathbb{R}^n.$$