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 $\mathbf{c}$. $\blacksquare$