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Vector Calculus An Introduction

Daniel Daners

Section 3.3 Continuous Functions

Using the notion of convergence from Section 3.1 we now define what we mean by a continuous function of several variables. Note that the definition looks the same as in case of a function of one variable. Before we give a definition of continuity we introduce limits.
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Definition 3.20. Limits of functions.

Suppose that \(\vect f\) is a function defined on a set \(D\subset\mathbb R^N\) with values in \(\mathbb R^k\text{.}\) If \(\vect x_0\) is in the closure of \(D\) we write
\begin{equation*} \lim_{\vect x\to\vect x_0}\vect f(\vect x) =\vect y_0 \end{equation*}
if for every \(\varepsilon\gt 0\) there exists \(\delta\gt 0\) such that \(\|\vect f(\vect x)-\vect y_0\|\lt \varepsilon\) for all \(\vect x\in D\) for which \(0\lt \|\vect x-\vect x_0\|\lt \delta\text{.}\) We also write
\begin{align*} \vect f(\vect x)\amp\to\vect y_0\text{ as }\vect x\to\vect x_0 \amp \text{or}\amp\amp \vect f(\vect x)\amp\xrightarrow{\vect x\to\vect x_0}\vect y_0, \end{align*}
and say that \(\vect f(\vect x)\) converges to \(\vect y_0\) as \(\vect x\) tends to \(\vect x_0\text{,}\) or \(\vect y_0\) is the limit of \(\vect f(\vect x)\) as \(\vect x\) tends to \(\vect x_0\text{.}\)
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As for functions of one variable there is a characterisation of limits involving sequences.
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We omit the proof of the above proposition and state some properties of limits. Compare with functions of one variable!
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We now define what we mean by a continuous function. As with the definition of limits, compare it to what you know about continuity of functions of one variable.
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Definition 3.23. Continuous function.

Suppose that \(\vect f\) is a function defined on a set \(D\subset\mathbb R^N\) with values in \(\mathbb R^k\text{.}\) We say that \(f\) is continuous at a point \(\vect x_0\in D\) if
\begin{equation*} \lim_{\vect x\to\vect x_0}\vect f(\vect x) =\vect f(\vect x_0)\text{.} \end{equation*}
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We say that \(f\) is continuous if \(f\) is continuous at every point in its domain \(D\text{.}\)
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Depending on the situation the following characterisations of continuity are useful.
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The proof of the above facts follow from Proposition 3.21 and the definition of limits as discussed in Section 3.1. Next we give a few examples of continuous and non-continuous functions.
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Example 3.25.

Consider \(f(x,y):=x^2-y^2\) defined on \(\mathbb R^2\text{.}\) We claim that \(f\) is continuous. To see this pick a point \((x,y)\in\mathbb R^N\text{.}\) If \((x_n,y_n)\) is an arbitrary sequence we need to show that \(f(x_n,y_n)=x_n^2-y_n^2\to x^2-y^2=f(x,y)\text{.}\) This is quite clear since \((x_n,y_n)\to (x,y)\) means that \(x_n\to x\) and \(y_n\to y\text{.}\) Hence by the continuity of the square function and addition we have \(x_n^2-y_n^2\to x^2-y^2\text{,}\) showing that \(f\) is continuous.
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Example 3.26.

Show that \(f\) given by \(f(x,y,z,t):=(x^2+y^2+z^2)e^{-t}\) is a continuous function on \(\mathbb R^4\text{.}\)
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Solution.
If \((x_n,y_n,z_n,t_n)\to(x,y,z,t)\text{,}\) then in particular \(x_n\to x\text{,}\) \(y_n\to y\text{,}\) \(z_n\to z\) and \(t_n\to t\text{.}\) As \(f\) consists of products and sums of the four variables as in the previous example, and the exponential function is continuous we have indeed \(f(x_n,y_n,z_n,t_n)\to f(x,y,z,t)\text{.}\) Hence \(f\) is continuous.
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Example 3.27.

Determine the domain of definition of \(f(x,y):=x/y\text{,}\) and determine whether \(f\) is continuous on its domain.
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Solution.
The natural domain of \(f\) is \(D=\{(x,y)\in\mathbb R^2\mid y\neq 0\}\text{.}\) Geometrically this is the plane with the \(x\)-axis removed. If \(y\neq 0\) and \((x_n,y_n)\to(x,y)\) then clearly \(x_n/y_n\to x/y\text{,}\) and thus \(f\) is continuous on its domain.
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Example 3.28.

Define the function \(f\) on \(\mathbb R^2\) by setting
\begin{equation*} f(s,t):= \begin{cases} \dfrac{st^2}{s^2+t^2} \amp \text{if $(s,t)\neq (0,0)$,} \\ 0 \amp \text{if $(s,t)=(0,0)$.} \end{cases} \end{equation*}
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Determine whether \(f\) is continuous or not.
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Solution.
Along the same lines as in the previous examples we can show that \(f\) is continuous at every point \((s,t)\neq (0,0)\text{.}\) The difficulty is at \((0,0)\text{.}\) We need to show that
\begin{equation} \Bigl|\frac{s_nt_n^2}{s_n^2+t_n^2}-0\Bigr| =\Bigl|\frac{s_nt_n^2}{s_n^2+t_n^2}\Bigr| \xrightarrow{n\to\infty}0\tag{3.3} \end{equation}
whenever \((s_n,t_n)\xrightarrow{n\to\infty}(0,0)\text{.}\) If \(a,b\) are non-negative real numbers, then clearly \(0\leq(a-b)^2=a^2+b^2-2ab\text{.}\) This implies that
\begin{equation*} ab\leq\frac{1}{2}\bigl(a^2+b^2)\text{.} \end{equation*}
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Applying the inequality to estimate \(|s_n||t_n|\) we get
\begin{equation*} 0\leq\Bigl|\frac{s_nt_n^2}{s_n^2+t_n^2}\Bigr| =|s_n||t_n|\frac{|t_n|}{s_n^2+t_n^2} \leq\frac{1}{2}\bigl(s_n^2+t_n^2)\frac{|t_n|}{s_n^2+t_n^2} =\frac{1}{2}|t_n|\text{.} \end{equation*}
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If \((s_n,t_n)\xrightarrow{n\to\infty}(0,0)\) then in particular \(|t_n|\to 0\text{,}\) and thus by the “squeezing lemma”(3.3) holds as \((s_n,t_n)\xrightarrow{n\to\infty}(0,0)\text{.}\) This shows that \(f\) is also continuous at \((0,0)\text{,}\) and therefore continuous everywhere. The graph of \(f\) is shown in Figure 3.29.
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Figure 3.29. A function continuous at \((0,0)\text{.}\)
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Example 3.30.

Define the function \(f\) on \(\mathbb R^2\) by setting
\begin{equation} f(x,y):= \begin{cases} \dfrac{xy^2}{x^2+y^4} \amp \text{if $(x,y)\neq (0,0)$,} \\ 0 \amp \text{if $(x,y)=(0,0)$.} \end{cases}\tag{3.4} \end{equation}
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Determine whether \(f\) is continuous or not.
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Solution.
Along the same lines as in the previous examples we can show that \(f\) is continuous at every point \((x,y)\neq (0,0)\text{.}\) The difficulty is at \((0,0)\text{.}\) We need to check whether
\begin{equation*} \Bigl|\frac{x_ny_n^2}{x_n^2+y_n^4}-0\Bigr| =\Bigl|\frac{y_ny_n^2}{x_n^2+y_n^4}\Bigr| \xrightarrow{n\to\infty}0 \end{equation*}
for every sequence with \((x_n,y_n)\to (0,0)\text{.}\) If we find one particular sequence for which the limit is not zero then the function is not continuous. Let \((y_n)\) be a sequence converging to zero and set \(x_n:=y_n^2\text{.}\) Then \((x_n,y_n)=(y_n^2,y_n)\to (0,0)\text{,}\) and
\begin{equation*} f(x_n,y_n)=f(y_n^2,y_n) =\frac{y_n^2y_n^2}{y_n^4+y_n^4} =\frac{1}{2} \end{equation*}
for all \(n\in\mathbb N\text{.}\) Hence \(f(x_n,y_n)\to 1/2\neq 0=f(0,0)\) for that particular sequence, and so \(f\) is not continuous at \((0,0)\text{.}\)
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Remark 3.31.

1. One could try and make the function (3.4) into a continuous one by modifying it at \((0,0)\text{.}\) This does not work as for instance if \(y_n\to 0\) then \(f(0,y_n)=0\to 0\text{,}\) but \(f(y_n^2,y_n)=0\to 1/2\text{,}\) so the limits are different for the sequence \((0,y_n)\) and \((y_n^2,y_n)\text{.}\)
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2. When checking whether \(f\) is continuous at \((0,0)\) we cannot take separate limits, that is first the limit as \(x\to 0\) and then the limit as \(y\to 0\text{.}\) For the function (3.4) we get
\begin{equation*} \lim_{x\to 0}\frac{xy^2}{x^2+y^4}=0, \end{equation*}
and thus
\begin{equation*} \lim_{y\to 0}\lim_{x\to 0}\frac{xy^2}{x^2+y^4}=0\text{,} \end{equation*}
but as we know the limit is not zero for all sequences.
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3. The example shows even more. To show continuity, we could try to see whether it is sufficient to take limits along radial straight lines. This means we just consider sequences of the form \((tx,ty)\) as \(t\to 0\) for every \((x,y)\neq (0,0)\text{.}\) If we consider such a sequence we get
\begin{equation*} \lim_{t\to 0}f(tx,ty) =\frac{t^3xy^2}{t^2x^2+t^4y^4} =\frac{txy^2}{x^2+t^2y^4} =0 \end{equation*}
no matter whether \(x=0\) or not. Hence we really need to check convergence for every sequence! From the graph given in Figure 3.32 one can clearly see this very well.
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Figure 3.32. A function discontinuous at \((0,0)\text{.}\)
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Along the curve \((y^2,y)\) the graph has a ridge (valley) of constant height (depth) going into \((0,0)\text{,}\) but never completely closing up. (Due to the insufficient resolution of the plot near \((0,0)\) there is quite a gap between the ridges. In reality, however, the gap is only one point!) Compare the graph to the graph of the similar function from Example 3.28 in Figure 3.29.
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Example 3.33.

Define \(\vect f(x,t)=(x^2+t,e^{-x}\cos(t))\) for \(x,t\in\mathbb R^2\text{.}\) The function \(\vect f\) is a vector valued function of two variables. To check whether it is continuous we need to determine whether every component function is continuous. The two component functions are given by \(f_1(x,t)=x^2+t\) and \(f_2(x,t)=e^{-x}\cos(t)\text{.}\) It is easily checked that they both are continuous, and thus \(\vect f\) is continuous.
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Example 3.34.

Is \((1/x,x)\) continuous at \(x=0\text{?}\)
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Solution.
\(1/x\) is not defined at \(x=0\text{,}\) so \(0\) is not in the domain of definition of the given vector valued function. Hence it does not make sense to talk about continuity of \((1/x,x)\) at \(0\text{.}\)
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