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Continuity of functions of several variables
Before we consider continuity of functions of several variables, let's recall how the issue is handled in Calc1 as there is a subtle difference.
Problem. For the function $$f(x) = \sqrt{x}$$
discuss continuity.
- at $x = 0$, $f$ is not continuous. However it is right-continuous at $\{0 \}$ as $x = 0$ is the end point of $D(f)$.
- at $x > 0$, $f$ is continuous.
- for $x < 0$, f is undefined.
Thus $f$ is continuous on on $( 0, \infty )$ or continuous in the extended sense on $[ 0, \infty )$.
Recall the definition. A function $$f \colon {\bf R} \rightarrow {\bf R}$$
is continuous at point $a \in {\bf R}$ if for any $\epsilon > 0$ there is a $\delta > 0$ such that $| x - a | < \delta \rightarrow |f(x)-f(a)| < \epsilon$.
Note that the definition implicitly assumes that $x, a \in$ $D(f)$.
How does the definition change in the $n$-dimensional case?
Definition. A function $$f \colon {\bf R}^n \rightarrow {\bf R}, f(\vec{x}) = y,$$ is continuous at point $\vec{a} \in D(f)$ if for any $\epsilon > 0$ there is a $\delta > 0$ such that $|| \vec{x} - \vec{a} || < \delta$ and $\vec{x} \in D(f) \rightarrow || f(\vec{x}) - f(\vec{a}) || < \epsilon$.
First, we have replaced the absolute value with the norm and unidimensional $x$ with multidimensional $\vec{x}$. This simply reflect the way we measure "closeness" in ${\bf R}$ versus in ${\bf R}^n$ (see Continuous functions).
Second, $x$ is required to be in the domain of $f$. This makes a difference. With the new definition, $\sqrt{x}$ is continuous on $[ 0, \infty )$ (verify). Thus one-side continuity is no longer an issue.
Now for ${\rm dim} > 1$.
Example. Consider $$f( x, y ) = ( x + y )^{\frac{1}{2}},$$ $$D(f) = { ( x, y ): x + y ≥ 0 }.$$
Without actually verifying the definition, we see that the function is continuous on the whole domain - the half-plane including the border.
Example. Let $$f(x) = \left\{ \begin{array}{ll} 0 &: x \leq 0 \\ 1 &: x \geq 0 \end{array} \right. .$$
Is $f$ continuous? Yes! It is surprising, but since the domain is "disconnected", the function can't be (unlike the second example on the right).
Example. Let $$f(x) = c$$ a constant function. Then $$|| f(x) - f(a) || = || c - c || = 0 < \epsilon,$$
so $f$ is continuous.
Theorem. Let $$f \colon {\bf R}^n \rightarrow {\bf R}$$ $$f_k(x) = x_k,$$ be the projection, where $$x = ( x_1, \ldots , x_n ).$$
Then $f$ is continuous.
(For example, if $n = 3$, $f_2( 1, 5, 7 ) = 5$.)
Proof. Consider $$f_k(x) = a_k,$$ where $a = ( a_1, ..., a_n )$.
In order to show this, we have to find $\delta$ for a given $\epsilon > 0$. We hence want to show that for $$|| x - a || < \delta$$ it follows that $$| f_k(x) - f_k(a) | = | x_k - a_k | < \epsilon.$$
We rewrite $$((x_1 - a_1)^2 + (x_2-a_2)^2 + \ldots)^{\frac{1}{2}} < \delta,$$ and $$((x_1-a_1)^2+(x_2-a_2)^2+ \ldots)^{\frac{1}{2}} \geq ((x_k-a_k)^2)^{\frac{1}{2}} = |x_k-a_k| < \epsilon$$
Choose $\delta = \epsilon$. Then if $$((x_1-a_2)^2+(x_2-a_2)^2+ \ldots)^{\frac{1}{2}} < \delta = \epsilon$$ then $$| x_k - a_k | < \epsilon.$$
A more topological discussion follows.
The continuity of function $f$ of one variable at point $x=a$ can be defined in terms of closeness (proximity):
This is the usual definition of continuity of a real valued function of single variable $f$ at point $x=a$:
The second part can be easily rewritten for a function several variables with the norm replacing the absolute value:
Definition 1. Given a function $f \colon {\bf R}^n \rightarrow {\bf R}^m$, it is continuous at $x=a$ if
This definition still follows the proximity idea above.
However, this is not the way the continuity of a function of $n$ variables is defined.
Definition 2. A function $f \colon {\bf R}^n \rightarrow {\bf R}^m$ is represented as a combination of $m$ functions of $n$ variables: $$f(x) = (f_1(x), f_2(x), \ldots , f_m(x)), {\rm \hspace{3pt} where \hspace{3pt}} x = (x_1, x_2, \ldots , x_n).$$
Then $f$ is called continuous if each of the functions $$f_1, f_2, ..., f_m \colon {\bf R}^n → {\bf R}$$ is continuous. Now, a function $g \colon {\bf R}^n \rightarrow {\bf R}$ is called continuous at $x=a$ if it is continuous with respect to each of these variables:
- for each combination $a_1, a_2, \ldots, a_n$ and each $k$, the function $h \colon {\bf R}^n \rightarrow {\bf R}$ given by
$$h(u) = g(a_1, a_2, ..., a_{k-1}, u, a_{k+1}, ..., a_n)$$ is continuous at $u=a_k$ as a function of single variable.
For example, define $f \colon {\bf R}^2 \rightarrow {\bf R}^2$ by $$f(x,y)=({\rm sin}(xy), x+y).$$
It is continuous because for each $a,b$ the functions $${\rm sin}(ay), {\rm sin}(xb), a+y, {\rm \hspace{3pt} and \hspace{3pt}} x+b$$ are continuous functions of $x$ or $y$ respectively.
This means, in particular, that if the graph of ${\rm sin}(xy)$ is cut by a plane parallel to the $xz$-plane or the $yz$-plane, the result is a continuous curve. Meanwhile, Definition $1$ suggests that this graph is a continuous surface. One would expect that the latter implies the former, but, as it turns out, these two definitions are equivalent.
To illustrate let's show that for a function $f \colon {\bf R} \rightarrow {\bf R}^2$ (a parametric curve) Definition $2$ implies Definition $1$. The idea of the proof comes from the purely topological approach to continuity (see Continuous functions). The definition can be re-written one more time with: $${\rm if \hspace{3pt}} x \in B(a,\delta), {\rm \hspace{3pt} then \hspace{3pt}} f(x) \in B(f(a),\epsilon),$$ where $B(p,d)= \{u \colon ||u-p|| < d\}.$
Now we use the observation that no matter how small a circle is, one can always inscribe a square inside: $${\rm if \hspace{3pt}} q < \frac{\epsilon}{\sqrt{2}}, {\rm \hspace{3pt} then \hspace{3pt}} \{(x,y) \colon |x-a| < q, |y-b| < q \} \subset B((a,b), \epsilon) = \{(x,y) \colon ||(x,y)-(a,b)|| < \epsilon \}.$$
Also, for the converse, it suffices to consider the opposite situation. Expressed in other words, these two collections form two bases of the same topology on ${\bf R}^2$, see Introduction to point-set topology.
Suppose a is a real number. Given $\epsilon >0$, we need to prove that there is $\delta >0$ such that $${\rm if \hspace{3pt}} |x - a| < \delta, {\rm \hspace{3pt} then \hspace{3pt}} || f(x) - f(a) || < \epsilon.$$
Suppose $f = (f_1, f_2)$. For $i=1,2$, since $f_i$ is continuous at $a$, there is $\delta_i$ such that $${\rm if \hspace{3pt}} |x - a| < \delta_i {\rm \hspace{3pt} then \hspace{3pt}} | f_i(x) - f_i(a) | < \epsilon' = \frac{\epsilon}{\sqrt{2}}.$$
Let $$\delta = {\rm min}\{ \delta_1, \delta_2 \}.$$ Then $${\rm if \hspace{3pt}} |x - a| < \delta {\rm then \hspace{3pt}} | f_i(x) - f_i(a) | < \epsilon' = \frac{\epsilon}{\sqrt{2}}.$$
As explained above, we need these inequalities to conclude that $$(f_1(x) - f_1(a))^2 + (f_2(x) - f_2(a))^2 < \epsilon^2.$$ The right-hand side is the square of the norm, hence $${\rm if \hspace{3pt}} |x - a| < \delta {\rm \hspace{3pt} then \hspace{3pt}} || f(x) - f(a) || < \epsilon.$$ $\blacksquare$
