Maps from \(\mathbb{R}^n\) to \(\mathbb{R}^m\)

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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 \(\varepsilon > 0\), there exists a \(\delta > 0\) such that, for all \(x \in S\), if \(x \in N_{\delta}(a)\) then \(F(x) \in N_{\varepsilon}(F(a))\).
In a logical form,
\[\forall \varepsilon > 0 ~ \exists \delta > 0 ~ \forall x \in S \left[ x \in N_{\delta}(a) \implies F(x) \in N_{\varepsilon}(F(a))\right].\]

Note that "\(x \in N_{\delta}(a)\)" can be also written as "\(\|x - a\| < \delta\)" according to the definition of the open ball. Similarly, "\(F(x) \in N_{\varepsilon}(F(a))\)" is equivalent to \(\|F(x) - F(a)\| < \varepsilon\). Using these notations, the above condition can be written as:
  • For any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that, for all \(x \in S\), if \(\|x - a\| < \delta\) then \(\|F(x) - F(a)\| < \varepsilon\).

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

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

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

\[|f_i(x) - f_i(a)| \leq \sqrt{\sum_{j=1}^{m}\left\{f_j(x) - f_j(a)\right\}^2} = d(F(x), F(a)) < \varepsilon.\]

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 \(\varepsilon > 0\), there exists a \(\delta_i\) such that, if \(x \in N_{\delta_i}(a)\), then \(|f_i(x) - f_i(a)| < \frac{\varepsilon}{\sqrt{m}}\). Let \(\delta = \min\{\delta_1, \delta_2, \cdots, \delta_m\}\). Then, if \(x \in N_{\delta}(a)\), then

\[d(F(x), F(a)) = \sqrt{\sum_{j=1}^{m}\left\{f_j(x) - f_j(a)\right\}^2} < \sqrt{m\left(\frac{\varepsilon}{\sqrt{m}}\right)^2} = \varepsilon.\]

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

\[S \stackrel{F}{\longrightarrow} T \stackrel{G}{\longrightarrow} \mathbb{R}^l\]

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, \[\begin{eqnarray}(G\circ F)(x) &=&(g_1(f_1(x),f_2(x),\cdots,f_m(x)),\\ &&~~g_2(f_1(x),f_2(x),\cdots,f_m(x)),\\ &&~~~~~\vdots\\ && ~~g_l(f_1(x),f_2(x),\cdots,f_m(x))).\end{eqnarray}\]

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,

\[\begin{eqnarray} (G\circ F)(x) &=& (G\circ F)(x_1, x_2)\\ &=& (g_1(f_1(x),f_2(x)), g_2(f_1(x),f_2(x)))\\ &=&(g_1(f_1(x_1,x_2),f_2(x_1, x_2)), g_2(f_1(x_1,x_2),f_2(x_1,x_2))) \end{eqnarray}\] □

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 \(\varepsilon > 0\), there exists a \(\delta' > 0\) such that, if \(y \in N_{\delta'}(b)\), then \(|g(y) - g(b)| < \varepsilon\).

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

Thus, with \(y = F(x)\), if \(x \in N_{\delta}(a)\), then \(|g(F(x)) - g(F(a))| = |g(y) - g(b)| < \varepsilon\). 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 \(S\). ■




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