4.30 Limits and continuity on functions of 2 variables.

Topology in \(R^d\)

Continuity, differentiability and integrability of functions on \(R\) were defined using the notion of distance in \(R\), which is given by the absolute value.

Now we define the same notions for functions in \(R^2\) but using instead different notions of distance: the metric given in the past lecture.

This gives us another way to define neighborhoods of points

Definition

An open ball of radius \(r \ge 0\) centered at \(a \in R^d\) is the set \(B(a,r) = \{b \in R^d | d(a, b) < r\}\)

A closed ball of radius \(r \ge 0\) centered at \(a \in R^d\) is the set \(\overline{B(a,r)} = \{b \in R^{d} | d(a, b) \le r\}\)

Examples

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Before defining limits and continuity for functions on two variables, we introduce some general notions that are valid in any metric space.

Definitions

Let \(S \subseteq R^d\) be a non-empty subset.

1. Interior: A point \(x \in S\) is called an interior point if there exists \(r > 0\) such that \(B(x,r) \subset S\).

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   The subsets of interior points is denoted by \(S^{\circ}= \{x \in S | x\text{ is interior}\}\)

2. Open set: We say that \(S\) is open if \(S = S^\circ\); that is, if all points of \(S\) are interior points.

   Example. \(B(x, r)\) is open \(\forall x \in R^d, r > 0\).

   By definition, we need to prove all the points of \(B(x,r)\) is interior

   Proof

   Let's choose arbitrary point in the ball, call it \(y\), then it's clearly \(d(y,x)<r\) and let \(d=d(y,x)\).

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   The let \(s = \min\{\frac{r-d}{2}, \frac{d}{2}\}\),NTP: \(y\) is interior, by definition, NTP: \(B(y,s)\subseteq B(x,r)\)

   If \(z \in B(y,s) \Rightarrow d(y,z)<s\), we need \(d(x,z)<r\)

   By triangle ineq., \(d(x,z) \leq d(x,y) + d(y,z)\)

   Then \(d(x,z)<d+s\), since \(s<\frac{r-d}{2}\), then \(d(x,z)<\frac{r+d}{2}\)

   And since \(d<r\), then \(d(x,z)<r\). Thus \(B(y,s)\subseteq B(x,r)\) and \(y\) is interior

   Then \(B(x, r)\) is open

3. Closed sets: We say that \(S\) is closed if \(R^{d}- S\) is open. The set \(R^{d} - S\) is called the complement of \(S\).

   Example. Closed balls \(\overline{B(x,r)}\) are closed \(\forall x \in R^d, r > 0\). (Exercise)

   Proof

   Let's choose arbitrary point outside the ball, call it \(y\), then it's clearly \(d(y,x)>r\) and let \(d=d(y,x)\).

   The let \(s=\min\{\frac{d-r}{2},\frac{d}{2}\}\),NTP: \(y\) is interior, by definition, NTP: \(B(y,s)\subseteq R^{d}-B(x,r)\)

   If \(z \in B(y,s) \Rightarrow d(y,z)<s\), we need \(d(x,z)>r\)

   By triangle ineq., \(d(x,y)\leq d(x,z)+d(z,y)\)

   Then \(d(x,z)\geq d\left(x,y\right)-d\left(z,y\right)\Rightarrow d(x,z)> d-s\), since \(s<\frac{d-r}{2}\), then \(d(x,z)>\frac{r+d}{2}\)

   And since \(d>r\), then \(d(x,z)>r\). Thus \(B(y,s)\subseteq R^{d}-B(x,r)\) and \(y\) is interior

   Then \(R^{d}-B(x,r)\) is open

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4. Boundary: A point \(x\) is a boundary point of \(S\) if for every \(r > 0\), the ball \(B(x, r)\) contains one point from and one point of \(S\) and one point of the complement \(R^{d} - S\).

   Equivalently, \(x \notin S^\circ\) and \(x \notin (R^d−S)^\circ\)

   Note that \(x\) is not necessarily a point of \(S\).

   For example, for \(a, x \in R^d\) and \(r > 0\) with \(||x − a|| = r\), then \(x\) is a boundary point of \(B(a,r)\)

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   The set of boundary points of \(S\) is denoted by \(\partial S\) and it is called the boundary of \(S\).

5. Closure: A point \(x \in R^d\) is a limit point of \(S\) if for every \(r > 0\), the ball \(B(x, r)\) contains a point from \(S\) that is not \(x\).

   The set given by the union of \(S\) and all limit points of \(S\) is called the closure of \(S\) and it is denoted by \(\bar{S}\), that is \(\bar S=S\cup\{\text{limits points of }S\}\)

   Observe that all the interior points are limit points, but limits points can be interior points or boundary points

Exercises

1. Prove that \(\bar{S}= S \cup \partial S\).

   We know \(\bar S=S\cup\{\text{limits points of }S\}\), then NTP \(S\cup\{\text{limits points of }S\}=S\cup \partial S\)

   \(\subseteq\)) Take \(x\in S\), then \(x\in S\cup\{\text{limits points of }S\}\wedge s\in S\cup\partial S\)​

   Take \(x\in\{\text{limits points of }S\}\), then \(x\) is a limit point of \(S\).

   Then we have two cases. If \(x\in S^\circ\), then \(x\in S\).

   If \(x\notin S^\circ\), since \(x\) is a limit point, by definition \(B(x, r)\) contains a point \(y\) from \(S\) that is not \(x\).

   Then \(\forall r,B(x,r)\nsubseteq R^d-S\), then \(x\notin (R^d-S)^\circ\). Then \(x\) is a boundary point

   Thus \(S\cup\{\text{limits points of }S\}\subseteq S\cup\partial S\)

   \(\supseteq\)) It's trivial to take \(x\in S\). Then let's take \(x\in \partial S\), then \(x \notin S^\circ\) and \(x \notin (R^d−S)^\circ\)​

   Then by definition \(\forall r,B(x,r)\nsubseteq R^{d}-S\)​

   Then \(\exists x\neq z\in B(x,r)\) s.t. \(z\) is not a point in \(R^{d}-S\)

   Then \(z\) is a point in \(S\) and \(z\in B(x,r)\) and \(x\neq z\)​

   Then for every \(r > 0\), the ball \(B(x, r)\) contains a point from \(S\) that is not \(x\).

   Thus \(x\) is a limit point of \(S\), then \(S\cup\{\text{limits points of }S\}\supseteq S\cup\partial S\)​

2. \(S\) is closed if and only if \(\bar{S} = S\).

   Since we know \(\bar{S}= S \cup \partial S\), then it's equivalent to \(S\) is closed if and only if \(S\cup \partial S= S\).

   Then we need to prove \(S\) is closed if and only if \(\partial S\subseteq S\)​

   \(\Rightarrow\)) Since \(S\) is closed, then \(R^d-S\) is open, then \(R^{d}-S=(R^{d}-S)^{\circ}\)​

   Take \(x\in \partial S\), then \(x \notin S^\circ\) and \(x \notin (R^d−S)^\circ\), then \(x \notin S^\circ\) and \(x\notin R^{d}-S\)​

   Then \(x\notin S^{\circ}\) and \(x\in S\), then \(\partial S\subseteq S\)​

   \(\Leftarrow\)) We know \(\partial S\subseteq S\), if we take \(x\in \partial S\), then \(x \notin S^\circ\) and \(x \notin (R^d−S)^\circ\) and \(x\in S\)

   then to prove \(S\) is closed, we need to prove \(R^d-S\) is open

   Then we need to prove \(R^{d}-S=(R^{d}-S)^{\circ}\). Then we prove this by \(\subseteq\) and \(\supseteq\)​

   \(\subseteq\)) Take \(x\in R^d-S\), then \(x\notin S\), then \(x\notin \partial S\), then \(x \in S^{\circ}\) or \(x \in (R^{d}−S)^{\circ}\)

   If \(x \in (R^{d}−S)^{\circ}\), we are done. If \(x\in S^{\circ}\subseteq S\) contradiction. Thus \(x\) must in \((R^{d}-S)^{\circ}\)

   \(\supseteq\)) It's clearly \(R^{d}-S\supseteq(R^{d}-S)^{\circ}\) by definition

   Thus we finish the proof

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1. Bounded: A set \(S\) is bounded if there exists \(r > 0\), such that \(S \subseteq B(x,r)\).

2. Connected: A set \(S\) is disconnected if it is the disjoint union of two non-empty open sets \(S = U_1 \cup U_2, U_1 \ne \emptyset \ne U_2\) open sets and \(U_1 \cap U_2 = \emptyset\).

   A set \(S\) is connected if it is not disconnected.

Remark

Actually, in \(D \subset R^d\) connected imply connected by path (or path connected): for any pair of points \(p, q \in D\), there exists a continuous function \(f: [0,1] \to D\) (a path) such that \(f(0) = p\), \(f(1) = q\) and \(f(t)\in D,\forall t\in[0,1]\)

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We say that a set \(D \subset R^d\) is a domain if it is open and connected.

Example. Open balls \(B(x, r)\) are domains \(\forall x \in R^d, r > 0\). (Exercise)

Proof

Since we've proved open balls are open, then we just need to prove it's connected

Consider \(B(x,r)=\{y\in\R^d:\|y-x\|<r\}\), we are going to prove it's path connected

Let \(p,q\in B(x,r)\), then \(\|p-x\|<r\) and \(\|q-x\|<r\).

Consider \(f:[0,1]\to\mathbb{R}^{d},f(t)=(1-t)p+tq\), \(f(0) = p\), \(f(1) = q\)

Take any \(t\in[0,1]\), then \(\|f(t)-x\|=\|(1-t)(p-x)+t(q-x)\|\le(1-t)\|p-x\|+t\|q-x\|<(1-t)r+tr=r\)​

Thus \(f(t)\in B(x,r)\) for all \(t\). Thus \(B(x,r)\) is connected

Definition

A sequence of points \((p_n)_{n \ge 0}\) in \(R^d\) converges to a limit point \(p \in R^d\) if for every \(\varepsilon > 0\), there exists \(N \in \mathbb{N}\) such that \(p_n \in B(p, \varepsilon)\) for all \(n \ge N\).

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Theorem

A sequence of points \((p_n)_{n \ge 0}\) with \(p_{n}= (p_{n,1}, p_{n,2}, \ldots, p_{n,d}) \in R^{d}\) converges to a limit point \(p = (p_{1}, p_{2}, \ldots, p_{d}) \in R^{d}\) if and only if the sequences \((p_{n,i})_{n \ge 0} \subseteq R\) converge to \(p_i \in R\) for all \(1 \le i \le d\).

Proof. Exercise.

\(\Rightarrow\)) Since the hypothesis, then \(\forall\varepsilon > 0\), \(\exists N \in \mathbb{N}\), \(\forall n \ge N\) such that \(p_n \in B(p, \varepsilon)\)

Then \(\forall\varepsilon > 0\), \(\exists N \in \mathbb{N}\), \(\forall n \ge N\) such that \(||p_{n}-p||<\varepsilon\Rightarrow||(p_{n,1},p_{n,2},\ldots,p_{n,d})-(p_{1},p_{2} ,\ldots,p_{d})||<\varepsilon\Rightarrow||(p_{n,1}-p_{1}),...,(p_{n,d}-p_{d})||<\varepsilon\)

Then \(||(p_{n,1}-p_{1}),...,(p_{n,d}-p_{d})||^2<\varepsilon^2\), then \(\sum_{i=1}^{d}(p_{n,i}-p_{i})^2<\varepsilon^{2}\)

Since \((p_{n,i}-p_{i})^{2}\geq 0\), then each \((p_{n,i}-p_{i})^2\) should be bounded by \(\varepsilon^{2}\)

Then \(|p_{n,i}-p_i|<\varepsilon\)

\(\Leftarrow\)) Let \(|p_{n,i}-p_{i}|<\frac{\varepsilon}{\sqrt{d}}\), then \(\sum_{i=1}^{d}(p_{n,i}-p_{i})^{2}<\varepsilon^{2}\)​

Limits of functions

Definition

Let \(D \subseteq R^d\) be a domain.

We say that a function \(f: D \to R\) converges to a number \(L\) at \(p \in R^d\), and write \(\lim_{x \to p} f(x) = L\) if for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that \(|f(x) - L| < \varepsilon \quad \forall x \in B(p, \delta)\)

Remarks

1. The known properties of limits of functions also hold here (sum of limits, products, sandwich lemma, etc.). The proofs are completely analogous. (Exercise)

2. Since our domain \(D\) is in \(R^d\), we do not have one-sided limits if \(d > 1\). we have infinitely many ways to get closed to a point than right or left.

But if we take two different ways for approaching a point and the limit is different, then the limit does not exists!

Let us see an example in \(R^d, d = 2\): Consider the function \(f : R^{d} - \{(0,0)\} \to R\) given by \(f(x, y) = \frac{xy}{x^2 + y^2} \quad \forall x, y \in R, \quad x \ne 0 \ne y\).

First we approach limit by \(y=x\), then \(f(x,x) = \frac{x^2}{2x^2} = \frac{1}{2}\) if \(x \neq 0\)

Then we approach limit by \(y=0\), then \(f(x,0) = \frac{x \cdot 0}{x^2+0^2} = 0\) if \(x \neq 0\)

The limit is not same, then \(\lim_{x\to(0,0)}f(x,y)\) doesn't exist

Definition

Let \(D \subseteq R^d\) and \(f: D \to R\) be a function on \(D\). Let \(p \in D\) be an interior point.

We say that \(f\) is continuous at \(p\) if for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that \(|f(x) − f(p)| < \varepsilon \quad \forall x \in B(p, \delta)\)

In other words, if \(\lim_{x \to p} f(x) = f(p)\)

We say that \(f\) is continuous in \(D\) if it is continuous in every point of \(D\).

The following theorems are a generalization of the results for functions over \(R\). Their proofs are identical.

Theorem

Let \(f, g: D \to R\) be two continuous functions at a point \(p \in D\).
Then the function \(f + g\) is continuous at \(p\).

Proof

Let \(\varepsilon > 0\). Since \(f\) is cont. at \(p\), \(g\) is cont at \(p\), then

\(\exists\delta_{1}>0\), \(|f(x)-f(p)| < \frac{\varepsilon}{2}\) if \(d(x,p) < \delta_1\), \(x \in B(p, \delta_1)\)
​\(\exists\delta_{2}>0\), \(|g(x)-g(p)| < \frac{\varepsilon}{2}\) if \(d(x,p) < \delta_2\), \(x \in B(p, \delta_{2})\)

So, \(|(f+g)(x) - (f+g)(p)| = |f(x)+g(x) - f(p) - g(p)|\)
\(= |f(x)-f(p) + g(x)-g(p)|\) \(\leq |f(x)-f(p)| + |g(x)-g(p)|\) \(< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} < \varepsilon\) if \(d(x,p) < \min\{\delta_1, \delta_2\} = \delta\)

Theorem

Let \(f: D \to R\) and \(g: R \to R\) be two continuous functions. Then the composition \(g \circ f: D \to R\) is also a continuous function.

Proof

Since \(f\) is continuous at \(t\), then \(\forall \epsilon_1 > 0\), \(\exists \delta_1 > 0\) such that \(d(f(t),f(s))<\epsilon_{1}\) if \(|t-s| < \delta_1\)

Since \(g\) is continuous at \(x\), then \(\forall \epsilon_2 > 0\), \(\exists \delta_2 > 0\) such that \(|g(x)-g(c)|<\epsilon_{2}\) if \(d(x,c) < \delta_2\)

Let \(x=f(t)\) and \(c=f(s)\), then \(|g(f(t))-g(f(s))|<\epsilon\) if \(d(f(t),f(s))<\delta_{2}\)

This happens if \(|t-s| < \delta_1\)

Theorem (Intermediate value)

Let \(D \subseteq R^d\) be a connected subset and \(f: D \to R\) be a continuous function on \(D\). Let \(p, q \in D\) be such that \(f(p) < a < f(q)\) for some \(a \in R\).

Then there exists \(t \in D\) such that \(f(t) = a\).

Proof

Since \(D\) is connected, it is path connected. Hence, there exists a continuous function \(g:[0,1]\to D\) such that \(g(0)=p\), \(g(1)=q\) and \(g(t) \in D\)

Take the composition \(f\circ g:[0,1]\rightarrow{\mathbb{R}}\) continuous \((f\circ g)(t)=f(g(t))\)

And \((f\circ g)(0)=f(g(0))=f(p)\), \((f \circ g)(1) = f(g(1)) = f(q)\)

And in hypothesis \(f(p)<a<f(q)\iff (f\circ g)(0)<a<(f\circ g)(1)\)

By intermediate value theorem in one dimension, \(\exists c \in [0,1]\) such that \(f(g(c)) = a\)​\(\implies f(g(c))=f(t)=a\) where \(t = g(c)\)

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