A Bound for the Total Derivative of a Function from Rn to Rm

# A Bound for the Total Derivative of a Function from Rn to Rm

Recall from The Jacobian Matrix of Differentiable Functions from Rn To Rm page that if $S \subseteq \mathbb{R}^n$ is open, $\mathbf{c} \in S$, $\mathbf{f} : S \to \mathbb{R}^m$, and $\mathbf{f}$ is differentiable at $\mathbf{c}$ then the total derivative of $\mathbf{f}$ at $\mathbf{c}$ is identically the Jacobian matrix of $\mathbf{f}$ at $\mathbf{c}$ and the Jacobian matrix of $\mathbf{f}$ at $\mathbf{c}$ is:

(1)
\begin{align} \quad \mathbf{D} \mathbf{f} (\mathbf{c}) = \begin{bmatrix} D_1 f_1 (\mathbf{c}) & D_2 f_1 (\mathbf{c}) & \cdots & D_n f_1 (\mathbf{c}) \\ D_1 f_2 (\mathbf{c}) & D_2 f_2 (\mathbf{c}) & \cdots & D_n f_2 (\mathbf{c}) \\ \vdots & \vdots & \ddots & \vdots \\ D_1 f_m (\mathbf{c}) & D_2 f_m (\mathbf{c}) & \cdots & D_n f_m (\mathbf{c}) \\ \end{bmatrix} \end{align}

We also noted that for all $\mathbf{v} \in \mathbb{R}^n$ and if $\mathbf{e}_1', \mathbf{e}_2', ..., \mathbf{e}_m'$ are the standard basis vectors for $\mathbb{R}^m$ then:

(2)
\begin{align} \quad \mathbf{f}'(\mathbf{c})(\mathbf{v}) = \sum_{k=1}^{m} [\mathbf{v} \cdot \nabla f_k (\mathbf{c}] \mathbf{e}_k' \end{align}

With this formula we can state a nice theorem which gives us a bound for the total derivative of $\mathbf{f}$ at $\mathbf{c}$ evaluated at $\mathbf{v}$.

 Theorem 1: Let $S \subseteq \mathbb{R}^n$, $\mathbf{c} \in S$, $\mathbf{f} : S \to \mathbb{R}^m$, and $\mathbf{f}$ be differentiable at $\mathbf{c}$. Then for all $\mathbf{v} \in \mathbb{R}^n$ we have that $\| \mathbf{f}'(\mathbf{c})(\mathbf{v}) \| \leq M \| \mathbf{v} \|$ where $\displaystyle{M = \sum_{k=1}^{m} \| \nabla f_k(\mathbf{c}) \|}$.

In the proof below we use both the triangle inequality and the Cauchy-Schwarz inequality. For the triangle inequality, we note that $\| [\mathbf{v} \cdot \nabla f_k(\mathbf{c})] \mathbf{e}_k' \| = \mid \mathbf{v} \cdot \nabla f_k(\mathbf{c} \mid$ since $\| \mathbf{e}_k' \| = 1$. We then use the Cauchy-Schwarz inequality which says that for any vectors $\mathbf{u}, \mathbf{v}$ we have that $\mid \mathbf{u} \cdot \mathbf{v} \mid \leq \| \mathbf{u} \| \| \mathbf{v} \|$.

• Proof: We use the formula above and take the norm of both sides to get:
(3)
\begin{align} \quad \| \mathbf{f}'(\mathbf{c})(\mathbf{v}) \| = \biggr \| \sum_{k=1}^{m} [\mathbf{v} \cdot \nabla f_k (\mathbf{c})] \mathbf{e}_k' \biggr \| \end{align}
• Applying the triangle inequality yields:
(4)
\begin{align} \quad \| \mathbf{f}'(\mathbf{c})(\mathbf{v}) \| & \leq \sum_{k=1}^{m} \| [\mathbf{v} \cdot \nabla f_k (\mathbf{c}) ] \mathbf{e}_k' \| \\ & \leq \sum_{k=1}^{m} \mid \mathbf{v} \cdot \nabla f_k (\mathbf{c}) \mid \end{align}
• We now apply the Cauchy-Schwarz inequality:
(5)
\begin{align} \quad \| \mathbf{f}'(\mathbf{c})(\mathbf{v}) \| & \leq \sum_{k=1}^{m} \| \mathbf{v} \| \| \nabla f_k(\mathbf{c}) \| \\ & \leq \| \mathbf{v} \| \sum_{k=1}^{m} \| \nabla f_k(\mathbf{c}) \| \end{align}
• Set $\displaystyle{M = \sum_{k=1}^{m} \| \nabla f_k(\mathbf{c}) \|}$. Then from above we have that $\| \mathbf{f}'(\mathbf{c})(\mathbf{v}) \| \leq M \| v \|$. $\blacksquare$