The function given is \(F: \mathbb{R}^2 \rightarrow \mathbb{R}^2\), defined by \(F(x, y) = (ax + by + c, dx + ey + f)\) for real numbers \(a, b, c, d, e, f\). This function can be written in terms of its component functions as:

• \(F_1(x, y) = ax + by + c\)
• \(F_2(x, y) = dx + ey + f\)

Both \(F_1(x, y)\) and \(F_2(x, y)\) are polynomial functions of \(x\) and \(y\).

Function Continuity

A function \(F: \mathbb{R}^n \rightarrow \mathbb{R}^m\) is continuous if and only if each of its component functions is continuous. In this case, \(F_1\) and \(F_2\) are component functions of \(F\). Polynomials are known to be continuous everywhere on their domain (\(\mathbb{R}^2\)). Since \(F_1\) and \(F_2\) are polynomial functions, they are continuous on \(\mathbb{R}^2\). Therefore, the function \(F\) is continuous.

Function Uniform Continuity

A function \(F: \mathbb{R}^n \rightarrow \mathbb{R}^m\) is uniformly continuous if for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all points \(\mathbf{v}_1, \mathbf{v}_2\) in the domain, if \(||\mathbf{v}_1 - \mathbf{v}_2|| < \delta\), then \(||F(\mathbf{v}_1) - F(\mathbf{v}_2)|| < \epsilon\). The given function \(F(x,y)\) is a linear transformation plus a constant vector. A general linear transformation \(L(\mathbf{v}) = M\mathbf{v}\) defined on a finite-dimensional vector space is always uniformly continuous. The function \(F(x,y)\) can be written in matrix form:

\(F\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a & b \\ d & e \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} c \\ f \end{pmatrix}\)

Let \(L\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a & b \\ d & e \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\). \(L\) is a linear transformation. The function \(F(x,y) = L(x,y) + (c, f)\). For any two points \((x_1, y_1)\) and \((x_2, y_2)\):

\(F(x_1, y_1) - F(x_2, y_2) = (L(x_1, y_1) + (c, f)) - (L(x_2, y_2) + (c, f))\)

\(F(x_1, y_1) - F(x_2, y_2) = L(x_1, y_1) - L(x_2, y_2)\)

Since \(L\) is linear, \(L(x_1, y_1) - L(x_2, y_2) = L((x_1, y_1) - (x_2, y_2))\). For a linear transformation \(L\) on a finite-dimensional space, there exists a constant \(M\) such that \(||L(\mathbf{v})|| \le M ||\mathbf{v}||\) for all \(\mathbf{v}\). Thus:

\(||F(x_1, y_1) - F(x_2, y_2)|| = ||L((x_1, y_1) - (x_2, y_2))|| \le M ||(x_1, y_1) - (x_2, y_2)||\)

If we choose \(\delta = \epsilon / M\) (assuming \(M > 0\)), then \(||(x_1, y_1) - (x_2, y_2)|| < \delta\) implies \(||F(x_1, y_1) - F(x_2, y_2)|| < M\delta = M(\epsilon/M) = \epsilon\). If \(M=0\) (i.e., \(a=b=d=e=0\)), \(F\) is a constant function, which is trivially uniformly continuous. Therefore, \(F\) is uniformly continuous.

Function Differentiability

A function \(F: \mathbb{R}^2 \rightarrow \mathbb{R}^2\) with component functions \(F_1\) and \(F_2\) is differentiable at a point if the partial derivatives of \(F_1\) and \(F_2\) exist and are continuous at that point. Let's compute the first-order partial derivatives of \(F_1\) and \(F_2\):

\(\frac{\partial F_1}{\partial x} = \frac{\partial}{\partial x}(ax + by + c) = a\)

\(\frac{\partial F_1}{\partial y} = \frac{\partial}{\partial y}(ax + by + c) = b\)

\(\frac{\partial F_2}{\partial x} = \frac{\partial}{\partial x}(dx + ey + f) = d\)

\(\frac{\partial F_2}{\partial y} = \frac{\partial}{\partial y}(dx + ey + f) = e\)

The partial derivatives \(a, b, d, e\) are constants. Constants are continuous functions everywhere on \(\mathbb{R}^2\). Since the first-order partial derivatives of the component functions exist and are continuous, the function \(F\) is differentiable everywhere on \(\mathbb{R}^2\).

Higher-Order Partial Derivatives

We have found the first-order partial derivatives are constants \(a, b, d, e\). Let's compute the second-order partial derivatives:

\(\frac{\partial^2 F_1}{\partial x^2} = \frac{\partial}{\partial x}(a) = 0\)

\(\frac{\partial^2 F_1}{\partial y \partial x} = \frac{\partial}{\partial y}(a) = 0\)

\(\frac{\partial^2 F_1}{\partial x \partial y} = \frac{\partial}{\partial x}(b) = 0\)

\(\frac{\partial^2 F_1}{\partial y^2} = \frac{\partial}{\partial y}(b) = 0\)

Similarly, all second-order partial derivatives of \(F_2\) will also be zero: \(\frac{\partial^2 F_2}{\partial x^2} = 0\), \(\frac{\partial^2 F_2}{\partial y \partial x} = 0\), \(\frac{\partial^2 F_2}{\partial x \partial y} = 0\), \(\frac{\partial^2 F_2}{\partial y^2} = 0\). Since all first-order partial derivatives are constants, all partial derivatives of order two and higher are zero. Since these derivatives are all constants (specifically, 0), they exist everywhere. Thus, the function \(F\) has partial derivatives of all orders.

Based on the analysis, the function \(F(x, y) = (ax + by + c, dx + ey + f)\) possesses all the listed properties: it is continuous, uniformly continuous, differentiable, and has partial derivatives of all orders.