Limits and Continuity

Notes based on section 1.5 of Vector Calculus, Linear Algebra, and Differential Forms, 5th Edition by John H. Hubbard and Barbara Burke Hubbard.

There’s always a piece of me in the proofs. Many are the result of my own direct attempts; others are reconstructions of proofs from the book, rewritten in my own words after learning them.

I’ve also added some notes and expanded some of the examples to make them clearer.

Open and Closed Sets

Open ball. For any $\mathbf{x} \in \mathbb{R}^n$ and $r > 0$, the open ball of radius $r$ centered at $\mathbf{x}$ is the subset

$$ B_r(\mathbf{x}) := \{\mathbf{y} \in \mathbb{R}^n : \lVert\mathbf{x} - \mathbf{y}\rVert < r\}. $$

Open set of $\mathbb{R}^n$. A subset $U \subset \mathbb{R}^n$ is open if for every $x \in U$, there exists $r > 0$ such that $B_r(\mathbf{x}) \subset U$.

Closed set of $\mathbb{R}^n$. A subset $C \subset \mathbb{R}^n$ is closed if its complement $C^c = \mathbb{R}^n - C$ is open.

Note

A set can be both open and closed. For example, $\emptyset$ and $\mathbb{R}^n$ are both open and closed.
A set can be neither open nor closed. For example, $[0, 1)$ is neither open nor closed.

Neighborhood. A neighborhood of a point $\mathbf{x} \in \mathbb{R}^n$ is a subset $X \subset \mathbb{R}^n$ such that there exists $\epsilon > 0$ with $B_\epsilon(\mathbf{\mathbf{x}}) \subset X$.

Often, we deal with sets that are neither open or closed. But every set is contained in a smallest closed set, called its closure, and largest open set, called its interior.

Closure. The closure of $A \subset \mathbb{R}^n$, denoted $\overline{A}$, is the set of $\mathbf{x} \in \mathbb{R}^n$ such that for every $r > 0$, $B_r(\mathbf{x}) \cap A \neq \emptyset$.

Interior. The interior of $A \subset \mathbb{R}^n$, denoted $\overset{\circ}{A}$, is the set of $\mathbf{x} \in \mathbb{R}^n$ such that there exists $r > 0$ with $B_r(\mathbf{x}) \subset A$.

Boundary. The boundary of $A \subset \mathbb{R}^n$, denoted $\partial A$, is the set of $\mathbf{x} \in \mathbb{R}^n$ such that every neighborhood of $\mathbf{x}$ overlaps both $A$ and $A^c$.

Thus,

The closure of $A$ is the union of $A$ and its boundary: $\overline{A} = A \cup \partial A$.
The interior of $A$ is the intersection of $A$ minus its boundary: $\overset{\circ}{A} = A - \partial A$.
The boundary of $A$ is closure minus interior: $\partial A = \overline{A} - \overset{\circ}{A}$.

Convergence and Limits

Convergence. A sequence $i \mapsto \mathbf{a}_i \in \mathbb{R}^n$ converges to $\mathbf{a} \in \mathbb{R}^n$ if

$$ \forall \epsilon > 0, \exists M | m > M \implies \lVert\mathbf{a}_m - \mathbf{a}\rVert < \epsilon. $$

We call $\mathbf{a}$ the limit of the sequence.

Proposition 1.5.13 (Convergence in terms of coordinates). A sequence $m \mapsto \mathbf{a}_m \in \mathbb{R}^n$ converges to $\mathbf{a} \in \mathbb{R}^n$ if and only if each coordinate converges.

Proof. Click to Expand

($\implies$) Assume $m \mapsto \mathbf{a}_m$ converges to $\mathbf{a}$. This means for any $\epsilon > 0$, there exists $M$ such that for all $m > M$ we have:

$$ \begin{align*} &\lVert\mathbf{a}_m - \mathbf{a}\rVert < \epsilon \\ \implies & \sqrt{ \left((a_m)_1-a_1\right)^2 + \cdots + \left((a_m)_n-a_n\right)^2 } < \epsilon \\ \implies & \left((a_m)_i-a_i\right)^2 < \epsilon^2 \\ \implies & \lVert(a_m)_i-a_i\rVert < \epsilon \end{align*} $$

Which means $m \mapsto (a_m)_i$ converges to $a_i$.

($\impliedby$) Fix $\epsilon$. For each $a_i$ we find corresponding $M_i$ for $\epsilon_i = \epsilon / \sqrt{n}$. Then we choose $M = \max\\{M_1,...,M_n\\}$. For $m > M$, $\lVert\mathbf{a} - \mathbf{a}_m\rVert < \sqrt{n \cdot \left(\dfrac{\epsilon}{\sqrt{n}}\right)^2} = \epsilon$. Which means $m \mapsto \mathbf{a}_m$ converges to $\mathbf{a}$.

Proposition 1.5.14 (Elegance is not required). Let $u$ be such that $u(\epsilon) \to 0$ as $\epsilon \to 0$. Then a sequence $i \mapsto \mathbf{a}_i$ converges to $\mathbf{a}$ if either of the following equivalent statements hold:

For every $\epsilon > 0$, there exists $M$ such that for all $m > M$, $\lVert\mathbf{a}_m - \mathbf{a}\rVert < u(\epsilon)$.
For every $\epsilon > 0$, there exists $M$ such that for all $m > M$, $\lVert\mathbf{a}_m - \mathbf{a}\rVert < \epsilon$.

Proof. Click to Expand

$lim_{\epsilon \to 0} u(\epsilon) = 0$ means that for every $\delta > 0$ there exists $\epsilon_0$ such that $u(\epsilon) < \delta$ for all $\epsilon < \epsilon_0$.

[TODO]

Proposition 1.5.15 (Limit of sequence is unique). If the sequence $i \mapsto \mathbf{a}_i$ of points in $\mathbb{R}^n$ converges to $\mathbf{a}$ and $\mathbf{b}$, then $\mathbf{a} = \mathbf{b}$.

Proof. Click to Expand

Assume $\mathbf{a} \ne \mathbf{b}$. Then $\epsilon_0 = \dfrac{\lVert\mathbf{a}-\mathbf{b}\rVert}{2} > 0$.

By definition of convergence, there exists $M_1$ such that for all $m > M_1$, $\lVert\mathbf{a} - \mathbf{a}_m\rVert < \epsilon_0$. Similar $M_2$ exists for $\mathbf{b}$.

Then for all $m > \max \\{M_1, M_2\\}$, we have

$$ \begin{align*} \lVert\mathbf{a} - \mathbf{b}\rVert &\le \lVert\mathbf{a} - \mathbf{a}_m\rVert + \lVert\mathbf{b} - \mathbf{a}_m\rVert \quad \text{(triangle inequality)} \\ &< \epsilon_0 + \epsilon_0 \\ &= \lVert\mathbf{a} - \mathbf{b}\rVert \end{align*} $$

Hence $\lVert\mathbf{a} - \mathbf{b}\rVert < \lVert\mathbf{a} - \mathbf{b}\rVert$, a contradiction. So, $\mathbf{a} = \mathbf{b}$.

Theorem 1.5.16 (The arithmetic of limits of sequence). Let $i \mapsto \mathbf{a}_i$ and $i \mapsto \mathbf{b}_i$ be two sequences of points in $\mathbb{R}^n$, and let $i \mapsto c_i$ be a sequence of numbers. Then

If $i \mapsto \mathbf{a}_i$ and $i \mapsto \mathbf{b}_i$ both converge, then so does $i \mapsto \mathbf{a}_i + \mathbf{b}_i$, and

$$ \lim_{i\to\infty} \left(\mathbf{a}_i + \mathbf{b}_i\right) \= \lim_{i\to\infty} \mathbf{a}_i \+ \lim_{i\to\infty} \mathbf{b}_i \tag{1.5.20} $$

If $i \mapsto \mathbf{a}_i$ and $i \mapsto c_i$ both converge, then so does $i \mapsto c_i \mathbf{a}_i$, and

$$ \lim_{i\to\infty} c_i \mathbf{a}_i \= \left(\lim_{i\to\infty} c_i \right) \left( \lim_{i\to\infty} \mathbf{a}_i \right) \tag{1.5.21} $$

If $i \mapsto \mathbf{a}_i$ and $i \mapsto \mathbf{b}_i$ both converge, then so does their dot product $i \mapsto \vec{\mathbf{a}_i} \cdot \vec{\mathbf{b}_i}$, and

$$ \lim_{i\to\infty} \left(\vec{\mathbf{a}_i} \cdot \vec{\mathbf{b}_i} \right) \= \left(\lim_{i\to\infty} \vec{\mathbf{a}_i} \right) \cdot \left( \lim_{i\to\infty} \vec{\mathbf{b}_i} \right) \tag{1.5.22} $$

If $i \mapsto \mathbf{a}_i$ is bounded and $i \mapsto c_i$ converges to $0$, then

$$ \lim_{i\to\infty} c_i \mathbf{a}_i \= 0 \tag{1.5.23} $$

Proposition 1.5.17 (Sequence in closed set).

Let $i \mapsto \mathbf{x}_i$ be a sequence in a closed set $C \subset \mathbb{R}^n$ converging to $\mathbf{x} \in \mathbb{R}^n$. Then $\mathbf{x} \in C$.
Conversely, if every convergent sequence in $C \subset \mathbb{R}^n$ converges to a point in $C$, then $C$ is closed.

Proof. Click to Expand

Assume $\mathbf{x} \notin C$. Then $\mathbf{x} \in C^c$ which is open. So, there exists $r > 0$ such that $B_r(\mathbf{x}) \subset C^c$. Since all $\mathbf{x}_i \in C$, then $\mathbf{x}_i \notin B_r(\mathbf{x})$, which means for all $i$ we have $\lVert\mathbf{x}_i - \mathbf{x}\rVert \ge r$ for every $i$. Then if we choose $\epsilon = r/2$, we can’t find an $M$ such that for $m > M$ we have $\lVert\mathbf{x}_m - \mathbf{x}\rVert < \epsilon$. So, $i \mapsto \mathbf{x}_i$ doesn’t converge to $\mathbf{x}$. A contradiction.
Assume $C$ is not closed. Choose $\mathbf{x} \in \partial C \notin C$. Since $\mathbf{x} \in \overline{C}$, then for all $r > 0$, $B_r(\mathbf{x}) \cap C \ne \emptyset$. We choose $i \mapsto \mathbf{x}_i$ such that $\mathbf{x}_i \in B_{1/i}(\mathbf{x}) \cap C$.
This sequence converges to $\mathbf{x}$: For a given $\epsilon > 0$, for all $m > 1/\epsilon$ we have $\lVert\mathbf{x} - \mathbf{x}_m\rVert < 1/m < \epsilon$.
Since $\mathbf{x} \not\in C$, this sequence doesn’t converge to a point in $C$. A contradiction. Hence, $C$ is closed.

Subsequences

Subsequence. $j \mapsto a_{i(j)}$ where $i(k) > i(j)$ when $k > j$.

Proposition 1.5.19 (Subsequence of convergent sequence converges). If a sequence $k \mapsto \mathbf{a}_k$ converges to $\mathbf{a}$, then any subsequence of the sequence converges to the same limit $\mathbf{a}$.

Limits of Functions

Limit of a function. Let $X$ be a subset of $\mathbb{R}^n$ and $\mathbf{x}_0$ a point in $\overline{X}$. A function $\mathbf{f}: X \to \mathbb{R}^p$ has the limit $\mathbf{a}$ at $\mathbf{x}_0$:

$$ \lim_{\mathbf{x}\to\mathbf{x}_0} \mathbf{f}(\mathbf{x}) = \mathbf{a} $$

if for all $\epsilon > 0$ there exists $\delta > 0$ such that for all $\mathbf{x} \in X$,

$$ \lVert\mathbf{x} - \mathbf{x}_0\rVert < \delta \implies \lVert\mathbf{f}(\mathbf{x}) - \mathbf{a}\rVert < \epsilon $$

Note

With this definition, $\mathbf{x}_0$ need not be in $X$. It only needs to be in the closure of $X$. But if $\mathbf{x}_0$ is in $X$, then $\mathbf{f}(\mathbf{x}_0)$ must be equal to $\mathbf{a}$.

For example, limit of $f: \mathbb{R} \to \mathbb{R}$ doesn’t exist at $0$: $$ > f(x) = \begin{cases} > 1 & \text{if } x \ne 0 \\ > 0 & \text{if } x = 0 > \end{cases} > $$

Contrast this with the usual 1-dimensional definition of limit:

Definition (From Zorich, Mathematical Analysis I). We shall say (following Cauchy) that the function $f: E \to \mathbb{R}$ tends to $A$ as $x$ tends to $a$, or that $A$ is the limit of $f$ as $x$ tends to $a$, if for every $\epsilon > 0$ there exists $\delta > 0$ such that $\lvert f(x) - A \rvert < \epsilon$ for every $x \in E$ such that $0 < \lvert x - a \rvert < \delta$.

The difference is the “greater than 0”: $0 < \lvert x - a \rvert < \delta$ instead of $\lVert\mathbf{x} - \mathbf{x}_0\rVert < \delta$.

Proposition 1.5.21 (Limit of function is unique). Let $f: X \to \mathbb{R}^n$ be a function. If $f$ has a limit at $\mathbf{x}_0 \in \overline{X}$, the limit is unique.

Proof. Click to Expand

Assume it’s not unique and $\lim_{\mathbf{x}\to\mathbf{x}_0}\mathbf{f}(\mathbf{x})$ takes two values of $\mathbf{a} \ne \mathbf{b}$. Let $\epsilon = \dfrac{\lVert\mathbf{a} - \mathbf{b}\rVert}{2} > 0$.

Since $\mathbf{a}$ is a limit, then there exists $\delta_a$ such that for all $\mathbf{x} \in X$, $\lVert\mathbf{x} - \mathbf{x}_0\rVert < \delta_a$ implies $\lVert\mathbf{f}(\mathbf{x}) - \mathbf{a}\rVert < \epsilon$. Similar $\delta_b$ exists for $\mathbf{b}$.

Then for all $\mathbf{x} \in X$ such that $\lVert\mathbf{x} - \mathbf{x}_0\rVert < \min\\{\delta_a, \delta_b\\}$, we have:

$$ \begin{align*} \lVert\mathbf{a}-\mathbf{b}\rVert &\le \lVert\mathbf{f}(\mathbf{x}) - \mathbf{a}\rVert+\lVert\mathbf{f}(\mathbf{x}) - \mathbf{b}\rVert \quad\text{(triangle inequality)} \\ &< \epsilon + \epsilon \\ &= \lVert\mathbf{a}-\mathbf{b}\rVert \end{align*} $$

Hence $\lVert\mathbf{a}-\mathbf{b}\rVert < \lVert\mathbf{a}-\mathbf{b}\rVert$, a contradiction. So, $\mathbf{a} = \mathbf{b}$

Theorem 1.5.22 (Limit of a composition). Let $U \subset \mathbb{R}^n$, $V \subset \mathbb{R}^m$, and $\mathbf{f}: U \to V$ and $\mathbf{g}: V \to \mathbb{R}^k$ be mappings, so that $\mathbf{g} \circ \mathbf{f}$ is defined on $U$. If $\mathbf{x}_0 \in \overline{U}$ and

$$ \mathbf{y}_0 = \lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{f}(\mathbf{x}) \quad \text{and} \quad \mathbf{z}_0 \= \lim_{\mathbf{y} \to \mathbf{y}_0} \mathbf{g}(\mathbf{y}) $$

both exist, then $\lim_{\mathbf{x} \to \mathbf{x}_0} \left(\mathbf{g} \circ \mathbf{f}\right)(\mathbf{x})$ exists and is equal to $\mathbf{z}_0$.

Proof. Click to Expand

Since $\lim_{\mathbf{y}\to\mathbf{y}_0} \mathbf{g}(\mathbf{y}) \= \mathbf{z}_0$,

$$ \forall \epsilon > 0, \exists \eta > 0 \quad\text{s.t.}\quad\lVert \mathbf{y} - \mathbf{y}_0 \rVert < \eta \implies \lVert \mathbf{g}(\mathbf{y}) - \mathbf{z}_0 \rVert < \epsilon $$

Since $\lim_{\mathbf{x}\to\mathbf{x}_0} \mathbf{f}(\mathbf{x}) \= \mathbf{y}_0$,

$$ \forall \eta > 0, \exists \delta > 0 \quad\text{s.t.}\quad\lVert \mathbf{x} - \mathbf{x}_0 \rVert < \delta \implies \lVert \mathbf{f}(\mathbf{x}) - \mathbf{y}_0 \rVert < \eta $$

Combining these, we get

$$ \forall \epsilon > 0, \exists \delta > 0 \quad\text{s.t.}\quad\lVert \mathbf{x} - \mathbf{x}_0 \rVert < \delta \implies \lVert \mathbf{g}(\mathbf{f}(\mathbf{x})) - \mathbf{z}_0 \rVert < \epsilon $$

which is equivalent to saying $\lim_{\mathbf{x} \to \mathbf{x}_0} \left(\mathbf{g} \circ \mathbf{f}\right)(\mathbf{x}) = \mathbf{z}_0$.

Note

With the standard definition of limit of real functions (see the previous note), this theorem is not true. For example, consider $f, g: \mathbb{R} \to \mathbb{R}$

$$ > f(x) = \begin{cases} > x \sin\left(\dfrac{1}{x}\right) & \text{if } x \ne 0 \\ > 0 & \text{if } x = 0 > \end{cases} \quad\text{and}\quad > g(x) = \begin{cases} > 1 & \text{if } x \ne 0 \\ > 0 & \text{if } x = 0 > \end{cases} > $$

Then using the standard definition of limit, we have $\lim_{x \to 0} f(x) = 0$ and $\lim_{x \to 0} g(x) = 1$. But $\lim_{x \to 0} (g \circ f)(x)$ does not exist: $\sin(1/x) = 0$ for $x = 1/(k \pi)$, $k \in \mathbb{Z}$. So, in any radius $r > 0$ of $0$, we have infinitely many points where $f(x) = 0$ and infinitely many points where $f(x) \ne 0$. Hence, the limit of $(g \circ f)(x)$ does not exist.

With our definition, $\lim_{y \to 0} g(y)$ does not exist.

Proposition 1.5.25 (Convergence by coordinates). Suppose

$$ U \subset \mathbb{R}^n, \quad \mathbf{f} = \begin{pmatrix} f_1 \\ \vdots \\ f_m \end{pmatrix} : U \to \mathbb{R}^m, \quad \text{and} \quad \mathbf{x}_0 \in \overline{U}. $$

Then $\lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{f}(\mathbf{x}) = \begin{pmatrix}a_1\\\\ \vdots\\\\ a_m \end{pmatrix}$ if and only if $\lim_{\mathbf{x} \to \mathbf{x}_0} f_i(\mathbf{x}) = a_i$ for all $i = 1, \ldots, m$.

Proof. Click to Expand

($\implies$) Since $\forall i \in \\{1,\ldots,m\\}$ we have $\lVert\mathbf{f}(\mathbf{x})-\mathbf{f}(\mathbf{y})\rVert\ge\lvert f_i(\mathbf{x})-f_i(\mathbf{y}) \rvert$, then for each $\epsilon$ the same $\delta$ that works for the limit of vector function works also for the limit of each coordinate function.
($\impliedby$) Fix $\epsilon$, and for each coordinate function $f_i$ find the $\delta_i$ corresponding to $\epsilon/\sqrt{m}$. Then use $\delta=\min\\{\delta_1,\ldots,\delta_m\\}$ for the vector function.

Theorem 1.5.26 (Limits of functions). Let $U \subset \mathbb{R}^n$, and let $\mathbf{f}, \mathbf{g}: U \to \mathbb{R}^m$, $h: U \to \mathbb{R}$.

If $\lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{f}(\mathbf{x})$ and $\lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{g}(\mathbf{x})$ exist, then

$$ \lim_{\mathbf{x} \to \mathbf{x}_0} (\mathbf{f} + \mathbf{g})(\mathbf{x}) = \lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{f}(\mathbf{x}) + \lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{g}(\mathbf{x}). $$

If $\lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{f}(\mathbf{x})$ and $\lim_{\mathbf{x} \to \mathbf{x}_0} h(\mathbf{x})$ exist, then

$$ \lim_{\mathbf{x} \to \mathbf{x}_0} (h \mathbf{f})(\mathbf{x}) = \lim_{\mathbf{x} \to \mathbf{x}_0} h(\mathbf{x}) \cdot \lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{f}(\mathbf{x}). $$

If $\lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{f}(\mathbf{x})$ and $\lim_{\mathbf{x} \to \mathbf{x}_0} h(\mathbf{x}) \ne 0$ exist, then

$$ \lim_{\mathbf{x} \to \mathbf{x}_0} \left( \frac{\mathbf{f}}{h} \right)(\mathbf{x}) = \frac{\lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{f}(\mathbf{x})}{\lim_{\mathbf{x} \to \mathbf{x}_0} h(\mathbf{x})}. $$

Define $(\mathbf{f} \cdot \mathbf{g})(\mathbf{x}) := \mathbf{f}(\mathbf{x}) \cdot \mathbf{g}(\mathbf{x})$. If $\lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{f}(\mathbf{x})$ and $\lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{g}(\mathbf{x})$ exist, then

$$ \lim_{\mathbf{x} \to \mathbf{x}_0} (\mathbf{f} \cdot \mathbf{g})(\mathbf{x}) = \lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{f}(\mathbf{x}) \cdot \lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{g}(\mathbf{x}). $$

If $\mathbf{f}$ is bounded and $\lim_{\mathbf{x} \to \mathbf{x}_0} h(\mathbf{x}) = 0$, then

$$ \lim_{\mathbf{x} \to \mathbf{x}_0} (h \mathbf{f})(\mathbf{x}) = 0. $$

If $\lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{f}(\mathbf{x}) = 0$ and $h$ is bounded, then

$$ \lim_{\mathbf{x} \to \mathbf{x}_0} (h \mathbf{f})(\mathbf{x}) = 0. $$

Proof of 4. Click to Expand

Using 1.5.25, since limit of $\mathbf{f}$ exists, then limit of each coordinate function $f_i$ exits. Part 1 and 2 apply also when the target space is one-dimensional. Induction on part 1 implies similar result for finite-sums.

So, we have:

$$ \begin{align*} \lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{f}(\mathbf{x}) \cdot \lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{g}(\mathbf{x}) &= \sum_{i=1}^{m} \left(\lim_{\mathbf{x} \to \mathbf{x}_0}{f_i(\mathbf{x})}\right) \left(\lim_{\mathbf{x} \to \mathbf{x}_0} {g_i(\mathbf{x})}\right) \\ &= \sum_{i=1}^{m} \lim_{\mathbf{x} \to \mathbf{x}_0}{f_i(\mathbf{x}) g_i(\mathbf{x})} \tag{using part 2} \\ &=\lim_{\mathbf{x} \to \mathbf{x}_0} \sum_{i=1}^{m}{f_i(\mathbf{x}) g_i(\mathbf{x})} \tag{using part 1} \\ &= \lim_{\mathbf{x} \to \mathbf{x}_0} (\mathbf{f} \cdot \mathbf{g})(\mathbf{x}) \end{align*} $$

Continuous Functions

Continuous function. Let $X \subset \mathbb{R}^n$. A function $\mathbf{f}: X \to \mathbb{R}^m$ is continuous at $\mathbf{x}_0 \in X$ if $\lim_{\mathbf{x} \to \mathbf{x}_0} \mathbf{f}(\mathbf{x}) = \mathbf{f}(\mathbf{x}_0)$.

$\mathbf{f}$ is continuous on $X$ if it is continuous at every point $\mathbf{x}_0 \in X$.

Note

An example of a two-variable function that is continuous only at the origin:

$$ > f(x, y) = \begin{cases} > \dfrac{xy}{x^2 + y^2} & \text{if } (x, y) \ne (0, 0) \\ > 0 & \text{if } (x, y) = (0, 0) > \end{cases} > $$

Another example:

$$ > f(x, y) = \begin{cases} > \lVert(x,y)\rVert & \text{if } (x, y) \in \mathbb{Q}^2 \\ > 0 & \text{otherwise} > \end{cases} > $$

Proposition 1.5.28 (Criterion for continuity). Let $X \subset \mathbb{R}^n$. A function $\mathbf{f}: X \to \mathbb{R}^m$ is continuous at $\mathbf{x}_0 \in X$ if and only if for every sequence $i \mapsto \mathbf{x}_i$ in $X$ converging to $\mathbf{x}_0$, we have

$$ \lim_{i\to\infty} \mathbf{f}(\mathbf{x}_i) = \mathbf{f}(\mathbf{x}_0). $$

Proof. Click to Expand

[TODO: Add proof]

Theorem 1.5.29 (Combining continuous mappings). Let $U$ be a subset of $\mathbb{R}^n$, $\mathbf{f}$ and $\mathbf{g}$ mappings $U \to \mathbb{R}^m$, and $h$ a function $U \to \mathbb{R}$.

If $\mathbf{f}$ and $\mathbf{g}$ are continuous at $\mathbf{x}_0 \in U$, so is $\mathbf{f} + \mathbf{g}$.
If $\mathbf{f}$ and $h$ are continuous at $\mathbf{x}_0 \in U$, so is $h \mathbf{f}$.
If $\mathbf{f}$ and $h$ are continuous at $\mathbf{x}_0 \in U$, and $h(\mathbf{x}_0) \ne 0$, so is $\dfrac{\mathbf{f}}{h}$.
If $\mathbf{f}$ and $\mathbf{g}$ are continuous at $\mathbf{x}_0 \in U$, so is $\mathbf{f} \cdot \mathbf{g}$.
If $h$ is defined and continuous at $\mathbf{x}_0 \in \overline{U}$ with $h(\mathbf{x}_0) = 0$, and there exist $C, \delta > 0$ such that

$$ |\mathbf{\mathbf{f}}(\mathbf{x})| \le C \quad \text{for } \mathbf{x} \in U, \quad |\mathbf{x} - \mathbf{x}_0| < \delta, $$ (i.e. $\mathbf{f}$ is bounded near $\mathbf{x}_0$), then the map $$ \mathbf{x} \mapsto \begin{cases} h(\mathbf{x}) \mathbf{f}(\mathbf{x}) & \text{for } \mathbf{x} \in U \\ 0 & \text{if } \mathbf{x} = \mathbf{x}_0 \end{cases} $$

is continuous at $\mathbf{x}_0$.

Note

An example for part 5:

$$ > \mathbf{f}(\mathbf{x}) = \begin{cases} > x \sin\left(\dfrac{1}{x}\right) & \text{if } x \ne 0 \\ > 0 & \text{if } x = 0 > \end{cases} > $$

Here $\lvert\sin(1/x)\rvert \le 1$ for all $x \ne 0$ and value of $h(x)=x$ at $0$ is $0$. So, $\mathbf{f}$ is continuous at $0$.

A discontinuous example in which the boundedness condition is not satisfied:

$$ > \mathbf{f}(\mathbf{x}) = \begin{cases} > x \cdot \dfrac{1}{x} & \text{if } x \ne 0 \\ > 0 & \text{if } x = 0 > \end{cases} > $$

Then $g(x) = 1$ for all $x \ne 0$, but $g(0) = 0$, so $g$ is discontinuous at $x = 0$.

Theorem 1.5.30 (Composition of continuous functions). Let $U \subset \mathbb{R}^n$, $V \subset \mathbb{R}^m$, and $\mathbf{f}: U \to V$ and $\mathbf{g}: V \to \mathbb{R}^p$ be mappings, so that $\mathbf{g} \circ \mathbf{f}$ is defined on $U$. If $\mathbf{f}$ is continuous at $\mathbf{x}_0 \in U$ and $\mathbf{g}$ is continuous at $\mathbf{f}(\mathbf{x}_0)$, then $\mathbf{g} \circ \mathbf{f}$ is continuous at $\mathbf{x}_0$.

Corollary 1.5.31 (Continuity of polynomials and rational functions).

Any polynomial function $\mathbf{f}: \mathbb{R}^n \to \mathbb{R}$ is continuous on all of $\mathbb{R}^n$.
Any rational function $\mathbf{f}: \mathbb{R}^n \to \mathbb{R}$ is continuous on all of $\mathbb{R}^n$ except at points where the denominator is $0$.

(A rational function is a ratio of two polynomials.)

Uniform Continuity

Uniformly continuous function. Let $X \subset \mathbb{R}^n$. A function $\mathbf{f}: X \to \mathbb{R}^m$ is uniformly continuous on $X$ if for every $\epsilon > 0$ there exists $\delta > 0$ such that for all $\mathbf{x}, \mathbf{y} \in X$, if $\lVert\mathbf{x} - \mathbf{y}\rVert < \delta$, then $\lVert\mathbf{f}(\mathbf{x}) - \mathbf{f}(\mathbf{y})\rVert < \epsilon$.

Note

The difference between continuity and uniform continuity is that in the latter the $\delta$ is independent of the point $\mathbf{x}_0 \in X$. For example, $f(x) = x^2$ is continuous on $\mathbb{R}$, but not uniformly continuous.

Theorem 1.5.33 (Linear functions are uniformly continuous). Any linear function $\mathbf{f}: \mathbb{R}^n \to \mathbb{R}^m$ is uniformly continuous.

Proof. Click to Expand

Any linear function has a corresponding matrix $A$ such that $\mathbf{f}(\mathbf{x}) = A\mathbf{x}$. Then

$$ \begin{align*} \lVert\mathbf{f}(\mathbf{x})-\mathbf{f}(\mathbf{y})\rVert &= \lVert A \cdot \mathbf{x} - A \cdot \mathbf{y} \rVert \\ &= \lVert A \cdot (\mathbf{x} - \mathbf{y}) \rVert \\ &= \lVert A \rVert \cdot \lVert \mathbf{x} - \mathbf{y} \rVert \end{align*} $$

Where $\lVert A \rVert = \sup_{\lVert v\rVert=1} \lVert Av \rVert$ is the operator norm of $A$.

Then, for all $\epsilon$ if we take $\delta = \dfrac{\epsilon}{\lVert A \rVert + 1 }$ , we’ll have:

$$ \begin{align*} \lVert \mathbf{x} - \mathbf{y} \rVert < \delta &\implies \lVert \mathbf{x} - \mathbf{y} \rVert < \dfrac{\epsilon}{\lVert A \rVert + 1 } \\ &\implies \left(\lVert A \rVert + 1\right) \lVert \mathbf{x} - \mathbf{y} \rVert < \epsilon \\ &\implies \lVert A \rVert \cdot \lVert \mathbf{x} - \mathbf{y} \rVert < \epsilon \\ &\implies \lVert\mathbf{f}(\mathbf{x})-\mathbf{f}(\mathbf{y})\rVert < \epsilon \end{align*} $$

which means $\mathbf{f}$ is uniformly continuous.

Series of Vectors

Convergent series of vectors. A series $\sum_{i=1}^{\infty} \vec{\mathbf{a}_i}$ is convergent if the sequence of partial sums $n \mapsto \vec{\mathbf{s}_n} = \sum_{i=1}^{n} \vec{\mathbf{a}_i}$ is convergent. The limit of the series is defined as

$$ \sum_{i=1}^{\infty} \vec{\mathbf{a}_i} = \lim_{n \to \infty} \vec{\mathbf{s}_n} $$

Proposition 1.5.35 (Absolute convergence). If $\sum_{i=1}^{\infty} \lVert \vec{\mathbf{a}_i} \rVert$ converges, then $\sum_{i=1}^{\infty} \vec{\mathbf{a}_i}$ converges.

Note

This is a very important result. This can be used to prove:

Convergence of Newton’s method
Euler’s identity
That the geometric series of matrices can be treated like the geometric series of numbers

Complex exponentials and trigonometric functions

Proposition 1.5.36 (Complex exponentials). For any complex number $z$, the series $e^z = \sum_{n=0}^{\infty} \dfrac{z^n}{n!}$ converges.

Proposition 1.5.37. For any real number $t$ we have $e^{it} = \cos(t) + i \sin(t)$.

Geometric series of matrices

Proposition 1.5.38. Let $A$ be a square matrix. If $\lVert A \rVert < 1$, then the series $S = I + A + A^2 + \ldots$ converges to $(I - A)^{-1}$.

Corollary 1.5.39. If $\lVert A \rVert < 1$, then $(I - A)$ is invertible.

Corollary 1.5.40. The set of invertible $n \times n$ matrices is open.