ThmDex – An index of mathematical definitions, results, and conjectures.
Strong derivative for constant real function
Formulation 0
Let $f : \mathbb{R}^N \to \mathbb{R}$ be a D4364: Real function such that
(i) \begin{equation} \exists \, C \in \mathbb{R} : \forall \, x \in \mathbb{R}^N : f(x) = C \end{equation}
Let $L : \mathbb{R}^N \to \mathbb{R}$ be a D4364: Real function such that
(i) \begin{equation} \forall \, x \in \mathbb{R}^N : L(x) = 0 \end{equation}
Then $L$ is a D5681: Real function derivative for $f$ on $\mathbb{R}^N$.
Proofs
Proof 0
Let $f : \mathbb{R}^N \to \mathbb{R}$ be a D4364: Real function such that
(i) \begin{equation} \exists \, C \in \mathbb{R} : \forall \, x \in \mathbb{R}^N : f(x) = C \end{equation}
Let $L : \mathbb{R}^N \to \mathbb{R}$ be a D4364: Real function such that
(i) \begin{equation} \forall \, x \in \mathbb{R}^N : L(x) = 0 \end{equation}
Let $\varepsilon > 0$. If $x, x_0 \in \mathbb{R}^N$ such that $x \neq x_0$, then we have \begin{equation} \frac{|f(x) - f(x_0) - L(x - x_0)|}{|x - x_0|} = \frac{|C - C - 0|}{|x - x_0|} = 0 < \varepsilon \end{equation} Since $\varepsilon > 0$ was arbitrary, the claim follows. $\square$