Differentiable Functions from Rn to Rm and Their Components

# Differentiable Functions from Rn to Rm and Their Components

Recall from the Differentiable Functions from Rn to Rm and Their Total Derivatives page that if $S \subseteq \mathbb{R}^n$ is open, $\mathbf{c} \in S$, and $\mathbf{f} : S \to \mathbb{R}^m$ then $\mathbf{f}$ is said to be differentiable at $\mathbf{c}$ if there exists a linear function $\mathbf{T}_{\mathbf{c}} : \mathbb{R}^n \to \mathbb{R}^m$ such that for all $\mathbf{v} \in \mathbb{R}^n$ we have that:

(1)
\begin{align} \quad \mathbf{f}(\mathbf{c} + \mathbf{v}) = \mathbf{f}(\mathbf{c}) + \mathbf{T}_{\mathbf{c}}(\mathbf{v}) + \| \mathbf{v} \| \mathbf{E}_{\mathbf{c}} (\mathbf{v}) \end{align}

Where $\mathbf{E}_{\mathbf{c}} (\mathbf{v}) \to \mathbf{0}$ as $\mathbf{v} \to \mathbf{0}$.

In the following theorem we will see that if $\mathbf{f}$ is differentiable at $\mathbf{c}$ with $\mathbf{f} = (f_1, f_2, ..., f_m)$ (here $f_1, f_2, ..., f_m : S \to \mathbb{R}$ are the components of $\mathbf{f}$) then $f_1, f_2, ..., f_m$ are also differentiable at $\mathbf{c}$, and furthermore, the converse is also true in that if the components $f_1, f_2, ..., f_m$ are all differentiable at $\mathbf{c}$ then $\mathbf{f}$ is differentiable at $\mathbf{c}$.

 Theorem 1: Let $S \subseteq \mathbb{R}^n$ be open, $\mathbf{c} \in S$, and $\mathbf{f} : S \to \mathbb{R}^m$ where $\mathbf{f} = (f_1, f_2, ..., f_m)$. Then $\mathbf{f}$ is differentiable at $\mathbf{c}$ if and only if $f_1, f_2, ..., f_m$ are all differentiable at $\mathbf{c}$.
• Proof: $\Rightarrow$ Suppose that $\mathbf{f}$ is differentiable at $\mathbf{c}$. Then there exists a linear function $\mathbf{T}_{\mathbf{c}} : \mathbb{R}^n \to \mathbb{R}^m$ such that for all $\mathbf{v} \in \mathbb{R}^n$ we have that:
(2)
\begin{align} \quad \mathbf{f}(\mathbf{c} + \mathbf{v}) = \mathbf{f}(\mathbf{c}) + \mathbf{T}_{\mathbf{c}}(\mathbf{v}) + \| \mathbf{v} \| \mathbf{E}_{\mathbf{c}} (\mathbf{v}) \end{align}
• Where $\mathbf{E}_{\mathbf{c}} (\mathbf{v}) \to \mathbf{0}$ as $\mathbf{v} \to \mathbf{0}$.
• Then $\mathbf{T}_{\mathbf{c}}(\mathbf{v}) = (T_{1, \mathbf{c}}(\mathbf{v}), T_{2, \mathbf{c}}(\mathbf{v}), ..., T_{m, \mathbf{c}}(\mathbf{v}))$ where for each $k \in \{ 1, 2, ..., m \}$ we have that $T_{k, \mathbf{c}} : \mathbb{R}^n \to \mathbb{R}$ are linear functions.
• Furthermore, $\mathbf{E}_{\mathbf{c}} (\mathbf{v}) = (E_{1, \mathbf{c}}(\mathbf{v}), E_{2, \mathbf{c}}(\mathbf{v}), ..., E_{m, \mathbf{c}}(\mathbf{v}))$ where for each $k \in \{ 1, 2, ..., m \}$ we have that $E_{k, \mathbf{c}} : \mathbb{R}^n \to \mathbb{R}$. Since $\mathbf{E}_{\mathbf{c}}(\mathbf{v}) \to \mathbf{0}$ as $\mathbf{v} \to \mathbf{0}$ this means component-wise that for each $k$, $E_{k, \mathbf{c}}(\mathbf{v}) \to 0$ as $\mathbf{v} \to \mathbf{0}$, and:
(3)
\begin{align} \quad (f_1(\mathbf{c} + \mathbf{v}), f_2(\mathbf{c} + \mathbf{v}), ..., f_k(\mathbf{c} + \mathbf{v})) = (f_1(\mathbf{c}), f_2(\mathbf{c}), ..., f_m(\mathbf{c})) + (T_{1, \mathbf{c}}(\mathbf{v}), T_{2, \mathbf{c}}(\mathbf{v}), ..., T_{m, \mathbf{c}}(\mathbf{v})) + \| \mathbf{v} \| (E_{1, \mathbf{c}}(\mathbf{v}), E_{2, \mathbf{c}}(\mathbf{v}), ..., E_{m, \mathbf{c}}(\mathbf{v})) \\ \end{align}
• Hence for each $k \in \{ 1, 2, ..., m \}$, from the equality above we must have the equality of the components on the lefthand side with the righthand side and so $T_{k, \mathbf{c}} : \mathbb{R}^n \to \mathbb{R}$ is a linear function such that for all $\mathbf{v} \in \mathbb{R}^n$:
(4)
\begin{align} \quad f_k(\mathbf{c} + \mathbf{v}) = f_k(\mathbf{c}) + T_{k, \mathbf{c}}(\mathbf{v}) + \| \mathbf{v} \| E_{k, \mathbf{c}} (\mathbf{v}) \end{align}
• Where $E_{k, \mathbf{c}} (\mathbf{v}) \to 0$ as $\mathbf{v} \to \mathbf{0}$ so $f_k$ is differentiable at $\mathbf{c}$.
• $\Leftarrow$ Conversely, if $f_1, f_2, ..., f_m$ are all differentiable at $\mathbf{c}$ then set $\mathbf{T}_{\mathbf{c}} : \mathbb{R}^n \to \mathbb{R}^m$ to be $\mathbf{T}_{\mathbf{c}} = (T_{1, \mathbf{c}}, T_{2, \mathbf{c}}, ..., T_{m, \mathbf{c}})$ and $\mathbf{E}_{\mathbf{c}} = (E_{1, \mathbf{c}}, E_{2, \mathbf{c}}, ..., E_{m, \mathbf{c}})$. Then $\mathbf{T}_{\mathbf{c}} : \mathbb{R}^n \to \mathbb{R}^m$ is a linear map such that for all $\mathbf{v} \in \mathbb{R}^n$ we have that:
(5)
\begin{align} \quad \mathbf{f}(\mathbf{c} + \mathbf{v}) &= (f_1(\mathbf{c} + \mathbf{v}), f_2(\mathbf{c} + \mathbf{v}), ..., f_m(\mathbf{c} + \mathbf{v})) \\ &= (f_1(\mathbf{c}) + T_{1, \mathbf{c}}(\mathbf{v}) + \| \mathbf{v} \| E_{1, \mathbf{c}}(\mathbf{v}), f_2(\mathbf{c}) + T_{2, \mathbf{c}}(\mathbf{v}) + \| \mathbf{v} \| E_{2, \mathbf{c}}(\mathbf{v}), ..., f_m(\mathbf{c}) + T_{m, \mathbf{c}}(\mathbf{v}) + \| \mathbf{v} \| E_{m, \mathbf{c}}(\mathbf{v})) \\ &= (f_1(\mathbf{c}), f_2(\mathbf{c}), ..., f_m(\mathbf{c})) + (T_{1, \mathbf{c}}(\mathbf{v}), T_{2, \mathbf{c}}(\mathbf{v}), ..., T_{m, \mathbf{c}}(\mathbf{v})) + \| \mathbf{v} \| (E_{1, \mathbf{c}}(\mathbf{v}), E_{2, \mathbf{c}}(\mathbf{v}), ..., E_{m, \mathbf{c}}(\mathbf{v})) \\ &= \mathbf{f}(\mathbf{c})(\mathbf{v}) + \mathbf{T}_{\mathbf{c}}(\mathbf{v}) + \| \mathbf{v} \| \mathbf{E}_{\mathbf{c}}(\mathbf{v}) \end{align}
• Where $\mathbf{E}_{\mathbf{c}}(\mathbf{v}) \to \mathbf{0}$ as $\mathbf{v} \to \mathbf{0}$ (since the conponents of $\mathbf{E}_{\mathbf{c}}$ individually go to $0$ as $\mathbf{v} \to \mathbf{0}$. So $\mathbf{f}$ is differentiable at $\mathbf{c}$. $\blacksquare$