Continuity (Real Scalar Fields)#

In the case of real scalar fields, the definition of continuity reduces to the following.

Theorem: Continuity of Real Scalar Fields

A real scalar field \(f: \mathcal{D} \subseteq \mathbb{R}^n \to \mathbb{R}\) is continuous at \(\mathbf{p} \in \mathcal{D}\) if and only if its limit at \(\mathbf{p}\) is \(f(\mathbf{p})\) or \(\mathbf{p}\) is an isolated point of \(\mathcal{D}\):

\[\lim_{\boldsymbol{x} \to \mathbf{p}} f(\boldsymbol{x}) = f(\mathbf{p})\]

Proof

TODO

Example: \(f(\boldsymbol{x}) = \mathbf{a}^{\mathsf{T}} \boldsymbol{x}\)

Consider the real scalar field \(f: \mathbb{R}^n \to \mathbb{R}\) defined as

\[f(\boldsymbol{x}) = \mathbf{a}^{\mathsf{T}} \boldsymbol{x},\]

where \(\mathbf{a} \in \mathbb{R}^n\) is some fixed real vector.

It is continuous on \(\mathbb{R}^n\).

Example: \(f(\boldsymbol{x}) = \boldsymbol{x}^{\mathsf{T}} A\boldsymbol{x}\)

Consider the real scalar field \(f: \mathbb{R}^n \to \mathbb{R}\) defined as

\[f(\boldsymbol{x}) = \boldsymbol{x}^{\mathsf{T}} A\boldsymbol{x},\]

where \(A \in \mathbb{R}^{n \times n}\) is some fixed real matrix.

It is continuous on \(\mathbb{R}^n\).

Example

Consider the real scalar field \(f: \mathbb{R}^2 \to \mathbb{R}\) defined as follows:

\[f(x, y) = \begin{cases}\frac{2xy}{x^2 + y^2} & \text{if} & \begin{bmatrix}x & y\end{bmatrix}^{\mathsf{T}} \ne \mathbf{0} \\ 2 & \text{if} & \begin{bmatrix}x & y\end{bmatrix}^{\mathsf{T}} = \mathbf{0} \end{cases}\]

It is continuous on \(\mathbb{R}^2 \setminus \{\mathbf{0}\}\). To see this, we examine the following sequence \((\mathbf{z}_k)_{k \in \mathbb{N}}\):

\[\mathbf{z}_k = \begin{bmatrix}\frac{1}{k} \\ \frac{1}{k}\end{bmatrix}\]

For \(k \to \infty\), we see that \((\mathbf{z}_k)_{k \in \mathbb{N}}\) converges to \(\mathbf{0}\):

\[\lim_{k \to \infty} \mathbf{z}_k = \lim_{k \to \infty} \begin{bmatrix}\frac{1}{k} \\ \frac{1}{k}\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}\]

The sequence \((f(\mathbf{z}_k))_{k \in \mathbb{N}}\) converges to \(1\):

\[\lim_{k \to \infty} f(\mathbf{z}_k) = \lim_{k \to \infty} \frac{2 \times \frac{1}{k} \times \frac{1}{k}}{\left(\frac{1}{k}\right)^2 + \left(\frac{1}{k}\right)^2} = \lim_{k \to \infty} \frac{\frac{2}{k^2}}{\frac{2}{k^2}} = 1 \ne 2\]

This means that the limit of \(f\) at \(\mathbf{0}\) cannot be \(2\) and so \(f\) cannot be continuous there.

Theorem: Continuity of Sums, Products and Quotients

Let \(f: \mathcal{D}_f \subseteq \mathbb{R}^n \to \mathbb{R}\) and let \(g: \mathcal{D}_g \subseteq \mathbb{R}^n \to \mathbb{R}\) be real scalar fields.

If \(f\) and \(g\) are continuous on \(S \subseteq \mathcal{D}_f \cap \mathcal{D}_g\), then:

\(\lambda f + \mu g\) is also continuous on \(S\) for all \(\lambda, \mu \in \mathbb{R}\).
\(fg\) is also continuous on \(S\).
\(f / g\) is also continuous on \(S\) provided that \(g(\boldsymbol{x}) \ne 0\) for all \(\boldsymbol{x} \in S\).

Proof

TODO

The Extreme Value Theorem for Real Scalar Fields

Let \(f: \mathcal{D} \subseteq \mathbb{R}^n \to \mathbb{R}\) be a real scalar field.

If \(f\) is continuous on \(S \subseteq \mathcal{D}\) and \(S\) is compact, then there exist at least one \(\mathbf{x}_{\text{of min}} \in S\) and at least one \(\mathbf{x}_{\text{of max}} \in S\) such that

\[f(\mathbf{x}_{\text{of min}}) \le f(\boldsymbol{x}) \le f(\mathbf{x}_{\text{of max}})\]

for all \(\boldsymbol{x} \in S\).

Proof

TODO