Let $f : \mathbb{R}^{N \times 1} \to \mathbb{R}^{M \times 1}$ be a
D4363: Euclidean real function such that
| (i) |
\begin{equation}
\exists \, A \in \mathbb{R}^{M \times N} \text{ and } b \in \mathbb{R}^{M \times 1} :
\forall \, x \in \mathbb{R}^{N \times 1} :
f(x) = A x + b
\end{equation}
|
Let $L : \mathbb{R}^{N \times 1} \to \mathbb{R}^{M \times 1}$ be a
D4364: Real function such that
| (i) |
\begin{equation}
\forall \, x \in \mathbb{R}^{N \times 1} :
L(x) = A x
\end{equation}
|
Let $\varepsilon > 0$. If $x, x_0 \in \mathbb{R}^{N \times 1}$ such that $x \neq x_0$, then we have
\begin{equation}
\begin{split}
\frac{|f(x) - f(x_0) - L(x - x_0)|}{|x - x_0|}
& = \frac{|A x + b - (A x_0 + b) - A (x - x_0)|}{|x - x_0|} \\
& = \frac{|A x + b - A x_0 - b - A x + A x_0)|}{|x - x_0|} \\
& = \frac{|0|}{|x - x_0|} \\
& = 0 \\
& < \varepsilon
\end{split}
\end{equation}
Since $\varepsilon > 0$ was arbitrary, the claim follows. $\square$
