Continuity of Real Vector Functions

Continuity of a Real Vector Function

Let $f:\mathcal{D}\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^n$ be a real vector function.

We say that $f$ is continuous at $\mathbf{a}\in \mathcal{D}$ iff its limit for $\mathbf{x}\to \mathbf{a}$ is $f(\mathbf{a})$

$$\underset{\mathbf{x}\to \mathbf{a}}{\lim }f(\mathbf{x})=f(\mathbf{a})$$

or $\mathbf{a}$ is an isolated point of $D$.

We say that $f$ is continuous on $\mathcal{D}$ if it is continuous at every $\mathbf{a}\in \mathcal{D}$.

Definition: Piecewise Continuity

Let $f:D\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^n$ be a real vector function.

We say that $f$ is piecewise continuous if $D$ can be expressed as the union of a finite collection of disjoint sets $D=D_1\cup \cdots \cup D_p$ such that the restrictions $f{|}_{D_1},\dots ,f{|}_{D_p}$ are continuous.

Theorem: Continuity via Component Functions

A real vector function $f:\mathcal{D}\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^n$ is continuous at ${\mathbf{x}}_0\in \mathcal{D}$ if and only if its component functions $f_1,\dots ,f_n$ are continuous at ${\mathbf{x}}_0$.

PROOF

TODO

Theorem: Continuity via Limits of Scalar Fields

A real vector function $f:\mathcal{D}\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^n$ is continuous at $\mathbf{a}\in \mathcal{D}$ if and only if $\mathbf{a}\in \mathcal{D}$ is an isolated point of $\mathcal{D}$ or the following limit is zero.

$$\underset{\mathbf{x}\to \mathbf{a}}{\lim }||f(\mathbf{x})-f(\mathbf{a})||=0$$

PROOF

We need to prove two things:

• (I) If $f$ is continuous at $\mathbf{a}\in \mathcal{D}$, then ${\lim }_{\mathbf{x}\to \mathbf{a}}||f(\mathbf{x})-f(\mathbf{a})||=0$.
• (II) If ${\lim }_{\mathbf{x}\to \mathbf{a}}||f(\mathbf{x})-f(\mathbf{a})||=0$, then $f$ is continuous at $\mathbf{a}$.

Proof of (I):

To prove that

$$\underset{\mathbf{x}\to \mathbf{a}}{\lim }||f(\mathbf{x})-f(\mathbf{a})||=0,$$

we need to show that for each $\epsilon >0$, there exists some $\delta >0$ such that if $\mathbf{x}\in \mathcal{D}$ and $0<||\mathbf{x}-\mathbf{a}||<\delta$, then $|||f(\mathbf{x})-f(\mathbf{a})||-0|<\epsilon$, i.e. $||f(\mathbf{x})-f(\mathbf{a})||<\epsilon$.

Now, since $f$ is continuous, we have

$$\underset{\mathbf{x}\to \mathbf{a}}{\lim }f(\mathbf{x})=f(\mathbf{a})$$

In other words, for each open ball $B_{\epsilon }(f(\mathbf{a}))$ around $\mathbf{a}$, there exists some open ball $B_{\delta }(\mathbf{a})$ around $\mathbf{a}$ such that for all $\mathbf{x}\in \mathcal{D}$ different from $\mathbf{a}$, if $\mathbf{x}\in B_{\delta }(\mathbf{a})$, then $f(\mathbf{x})\in B_{\epsilon }(f(\mathbf{a}))$. By recalling the definition of an open ball, we can restate the previous sentence as the following - for each $\epsilon >0$, there exists some $\delta >0$ such that for all $\mathbf{x}\in \mathcal{D}$, if $0<||\mathbf{x}-\mathbf{a}||<\delta$, then $||f(\mathbf{x})-f(\mathbf{a})||<\epsilon$, which is precisely what we set out to prove.

Proof of (II):

To prove that $f$ is continuous at $\mathbf{a}$, we need to prove that

$$\underset{\mathbf{x}\to \mathbf{a}}{\lim }f(\mathbf{x})=f(\mathbf{a})$$

Restating this using the definition of a limit, we need to prove that for each open ball $B_{\epsilon }(f(\mathbf{a}))$ around $f(\mathbf{a})$, there exists some open ball $B_{\delta }(\mathbf{a})$ around $\mathbf{a}$ such that for all $\mathbf{x}\in \mathcal{D}$ different from $\mathbf{x}$, if $\mathbf{x}\in B_{\delta }(\mathbf{a})$, then $f(\mathbf{x})\in B_{\epsilon }(f(\mathbf{a}))$. Further paraphrasing this using the definition of an open ball, we need to prove that for each $\epsilon >0$, there exists some $\delta >0$ such that for all $\mathbf{x}\in \mathcal{D}$, if $0<||\mathbf{x}-\mathbf{a}||<\delta$, then $||f(\mathbf{x})-f(\mathbf{a})||<\epsilon$.

We are given that

$$\underset{\mathbf{x}\to \mathbf{a}}{\lim }||f(\mathbf{x})-f(\mathbf{a})||=0$$

Using the definition of the limit for scalar fields, we know that for each $\epsilon >0$, there exists some $\delta >0$ such that for all $\mathbf{x}\in \mathcal{D}$, if $||\mathbf{x}-\mathbf{a}||$, then $|||f(\mathbf{x})-f(\mathbf{a})||-0|<\epsilon$. Since $|||f(\mathbf{x})-f(\mathbf{a})||-0|=||f(\mathbf{x})-f(\mathbf{a})||$, this just means that for each $\epsilon >0$, there exists some $\delta >0$ such that for all $\mathbf{x}\in \mathcal{D}$, if $||\mathbf{x}-\mathbf{a}||$, then $||f(\mathbf{x})-f(\mathbf{a})||<\epsilon$, which is what we wanted to prove.

Theorem: Properties of Continuous Real Vector Functions

If $f,g:\mathcal{D}\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^n$ are continuous at ${\mathbf{x}}_0\in \mathcal{D}$, then so are $f+g$ and $\lambda f$ for all $\lambda \in \mathbb{R}$.

PROOF

TODO

Theorem: Continuity of Composition

If $g:\mathcal{D}\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^n$ and $f:g(\mathcal{D})\to {\mathbb{R}}^p$ are continuous, then so is their composition $f\circ g$.

PROOF

TODO