Maps from Rn to Rm

Continuous maps

So far, we have considered multivariate functions where the domain is (a subset of) ${\mathbb{R}}^n$ ($n\in \mathbb{N}$) and the codomain is $\mathbb{R}$. We can generalize the codomain to ${\mathbb{R}}^m$ for any $m\in \mathbb{N}$. To do this, first consider the following $m$ real-valued functions on $S(\subset {\mathbb{R}}^n)$:

$$f_1(x_1,x_2,\cdots ,x_n),f_2(x_1,x_2,\cdots ,x_n),\cdots ,f_m(x_1,x_2,\cdots ,x_n).$$

Now, for each $a=(a_1,a_2,\cdots ,a_n)\in S$, we can assign a point in ${\mathbb{R}}^m$

$$(f_1(a),f_2(a),\cdots ,f_m(a))=(f_1(a_1,a_2,\cdots ,a_n),f_2(a_1,a_2,\cdots ,a_n),\cdots ,f_m(a_1,a_2,\cdots ,a_n)).$$

By this correspondence, we can define a map from $S$ to ${\mathbb{R}}^m$ by

$$a=(a_1,a_2,\cdots ,a_n)\mapsto (f_1(a),f_2(a),\cdots ,f_m(a)).$$

We often write

$$F:S\to {\mathbb{R}}^m$$

and

$$F(a)=(f_1(a),f_2(a),\cdots ,f_m(a)).$$

Example. When $n=1$ and $m=2$, a map from the closed interval $[0,1]$ to ${\mathbb{R}}^2$, $t\mapsto (f(t),g(t))$, where $f(t)$ and $g(t)$ are continuous functions of $t$, is a path in ${\mathbb{R}}^2$. □

Example. Let $f(u,v),g(u,v),h(u,v)$ be continuous (real-valued) functions on an open region $U$ in ${\mathbb{R}}^2$. Then the mapping $U\to {\mathbb{R}}^3$ defined by $(u,v)\mapsto (f(u,v),g(u,v),h(u,v))$ represents a continuous surface in ${\mathbb{R}}^3$, parameterized by $u$ and $v$. □

Definition (Continuous map)

Let $S\subset {\mathbb{R}}^n$. The map $F:S\to {\mathbb{R}}^m$, $x=(x_1,x_2,\cdots ,x_n)\mapsto F(x)=(f_1(x),f_2(x),\cdots ,f_m(x))$, is said to be continuous at $a=(a_1,a_2,\cdots ,a_n)$ if the following condition holds:

• For any $\epsilon >0$, there exists a $\delta >0$ such that, for all $x\in S$, if $x\in N_{\delta }(a)$ then $F(x)\in N_{\epsilon }(F(a))$.

In a logical form,

$$\mathrm{\forall }\epsilon >0\mathrm{\exists }\delta >0\mathrm{\forall }x\in S[x\in N_{\delta }(a)⟹F(x)\in N_{\epsilon }(F(a))].$$

Note that "$x\in N_{\delta }(a)$" can be also written as "$\Vert x-a\Vert <\delta$" according to the definition of the open ball. Similarly, "$F(x)\in N_{\epsilon }(F(a))$" is equivalent to $\Vert F(x)-F(a)\Vert <\epsilon$. Using these notations, the above condition can be written as:

• For any $\epsilon >0$, there exists a $\delta >0$ such that, for all $x\in S$, if $\Vert x-a\Vert <\delta$ then $\Vert F(x)-F(a)\Vert <\epsilon$.

In this way, the definition of continuity looks very similar to that of a univariate (one-variable) function $f:S(\subset \mathbb{R})\to \mathbb{R}$:

• For any $\epsilon >0$, there exists a $\delta >0$ such that, for all $x\in S$, if $|x-a|<\delta$ then $|f(x)-f(a)|<\epsilon$.

Theorem (Continuity of a map)

Let $S\subset {\mathbb{R}}^n$. The map $F:S\to {\mathbb{R}}^m$ defined by $F(x)=(f_1(x),f_2(x),\cdots ,f_m(x))$ is continuous at $a=(a_1,a_2,\cdots ,a_n)$ if and only if all $f_i(x_1,x_2,\cdots ,x_n)$, $i=1,2,\cdots ,m$, are continuous at $a$.

Proof. ($\Rightarrow$, only if) Suppose $F(x)$ is continuous at $a$. Then, for any $\epsilon >0$, there exists a $\delta >0$ such that for all $x\in S$, if $x\in N_{\delta }(a)$ then $F(x)\in N_{\epsilon }(F(a))$.

Now, if $x\in N_{\delta }(a)$, then

$$|f_i(x)-f_i(a)|\le \sqrt{\sum _{j=1}^m{\{f_j(x)-f_j(a)\}}^2}=d(F(x),F(a))<\epsilon .$$

Thus, $f_i(x)$ is continuous at $a$.

($\Leftarrow$, if) Suppose each $f_i(x),i=1,2,\cdots ,m$ is continuous at $a$. Then, for any $\epsilon >0$, there exists a ${\delta }_i$ such that, if $x\in N_{{\delta }_i}(a)$, then $|f_i(x)-f_i(a)|<\frac{\epsilon }{\sqrt{m}}$. Let $\delta =min\{{\delta }_1,{\delta }_2,\cdots ,{\delta }_m\}$. Then, if $x\in N_{\delta }(a)$, then

$$d(F(x),F(a))=\sqrt{\sum _{j=1}^m{\{f_j(x)-f_j(a)\}}^2}<\sqrt{m{\left(\frac{\epsilon }{\sqrt{m}}\right)}^2}=\epsilon .$$

Therefore, $F(x)$ is continuous at $a$. ■

Map composition

Let $S\subset {\mathbb{R}}^n$ and $T\subset {\mathbb{R}}^m$. Suppose we have the maps $F:S\to {\mathbb{R}}^m$ and $G:T\to {\mathbb{R}}^l$. If the image of $S$ by $F$, $F(S)=\{F(x)|x\in S\}$ is contained in $T$, that is, $F(S)\subset T$, then, we can define a new map from $S$ to ${\mathbb{R}}^l$ by composing $F$ and $G$. That is, by

$$S\stackrel{F}{⟶}T\stackrel{G}{⟶}{\mathbb{R}}^l$$

we can define

$$G\circ F:S\to {\mathbb{R}}^l.$$

Let's break down this composition into pieces. First, for $$x=(x_1,x_2,\cdots ,x_n)\in S\subset {\mathbb{R}}^n,$$

we have $$F(x)=(f_1(x),f_2(x),\cdots ,f_m(x))\in {\mathbb{R}}^m.$$

Next, for $y=F(x)$, we have

$$G(y)=(g_1(y),g_2(y),\cdots ,g_l(y))\in {\mathbb{R}}^l.$$

Composing these, $$\begin{array}{rcl}(G\circ F)(x)& =& (g_1(f_1(x),f_2(x),\cdots ,f_m(x)),\\ & & g_2(f_1(x),f_2(x),\cdots ,f_m(x)),\\ & & ⋮\\ & & g_l(f_1(x),f_2(x),\cdots ,f_m(x))).\end{array}$$

Example. Let $F(x)=(f_1(x_1,x_2),f_2(x_1,x_2))$ and $G(x)=(g_1(x_1,x_2),g_2(x_1,x_2))$ be both maps $F,G:{\mathbb{R}}^2\to {\mathbb{R}}^2$. Then, we can compose them to have $G\circ F$. That is,

$$\begin{array}{rcl}(G\circ F)(x)& =& (G\circ F)(x_1,x_2)\\ & =& (g_1(f_1(x),f_2(x)),g_2(f_1(x),f_2(x)))\\ & =& (g_1(f_1(x_1,x_2),f_2(x_1,x_2)),g_2(f_1(x_1,x_2),f_2(x_1,x_2)))\end{array}$$ □

Theorem (Continuity of a composed map)

Let $F:S\to {\mathbb{R}}^m$ and $G:T\to {\mathbb{R}}^l$ be maps, where $S\subset {\mathbb{R}}^n$ and $T\subset {\mathbb{R}}^m$. Suppose $F(S)\subset T$. If $F$ and $G$ are continuous, then their composition $G\circ F$ is also continuous.

Proof. Suppose $F:S\to {\mathbb{R}}^m$ is a continuous map, $g:T\to \mathbb{R}$ is a continuous function, and $F(S)\subset T$. By the above theorem, it suffices to show that the function 

$$h(x)=g(F(x))=g(f_1(x),f_2(x),\cdots ,f_m(x)),x=(x_1,x_2,\cdots ,x_n),$$

is continuous on $S$. Pick an arbitrary point $a=(a_1,a_2,\cdots ,a_n)\in S$ and let $b=F(a)=(f_1(a),f_2(a),\cdots ,f_m(a))$. 

As $g$ is continuous, for any $\epsilon >0$, there exists a ${\delta }^{\prime }>0$ such that, if $y\in N_{{\delta }^{\prime }}(b)$, then $|g(y)-g(b)|<\epsilon$.

As $F(x)$ is continuous, there exists a $\delta >0$ such that if $x\in N_{\delta }(a)$, then $F(x)\in N_{{\delta }^{\prime }}(F(a))=N_{{\delta }^{\prime }}(b)$.

Thus, with $y=F(x)$, if $x\in N_{\delta }(a)$, then $|g(F(x))-g(F(a))|=|g(y)-g(b)|<\epsilon$. This shows that $g(F(x))$ is continuous at the point $a$, but this point is an arbitrary point in $S$. Therefore, $g(F(x))$ is continuous on $S$. ■