Differentiable Functions from Rn to Rm are Continuous

# Differentiable Functions from Rn to Rm are Continuous

Recall from the Differentiability and the Total Derivative of Functions from Rn to Rm 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 $T_c(h) : \mathbb{R}^m \to \mathbb{R}^n$ called the total derivative of $\mathbf{f}$ at $\mathbf{c}$ such 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 we have that $\mathbf{E}_{\mathbf{c}}(\mathbf{v}) \to \mathbf{0}$ as $\mathbf{v} \to \mathbf{0}$.

We will now look at a nice theorem which is an analogue to one we are familiar with - that is, a function $\mathbf{f}$ being differentiable at a point $\mathbf{c}$ implies that $\mathbf{f}$ is continuous at $\mathbf{c}$.

 Theorem 1: Let $S \subseteq \mathbb{R}^m$ be open, $\mathbf{c} \in \mathbb{R}^n$, and $\mathbf{f} : S \to \mathbb{R}^m$. If $\mathbf{f}$ is differentiable at $\mathbf{c}$ then $\mathbf{f}$ is continuous at $\mathbf{c}$.
• Proof: Suppose that $\mathbf{f}$ is differentiable at $\mathbf{c}$. Then:
(2)
\begin{align} \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 we have that $\mathbf{E}_{\mathbf{c}}(\mathbf{v}) \to \mathbf{0}$ as $\mathbf{v} \to \mathbf{0}$. To show that $\mathbf{f}$ is continuous at $\mathbf{c}$ we must show that $\displaystyle{\lim_{\mathbf{x} \to \mathbf{c}} \mathbf{f}(\mathbf{x}) = \mathbf{f}(\mathbf{c})}$ or equivalently show that $\displaystyle{\lim_{\mathbf{v} \to \mathbf{0}} \mathbf{f} (\mathbf{c} + \mathbf{v}) = \mathbf{f}(\mathbf{c})}$. We have by $(*)$ that:
(3)
\begin{align} \quad \lim_{\mathbf{v} \to \mathbf{0}} \mathbf{f}(\mathbf{c} + \mathbf{v}) & = \lim_{\mathbf{v} \to \mathbf{0}} \left [ \mathbf{f}(\mathbf{c}) + \mathbf{T}_{\mathbf{c}}(\mathbf{v}) + \| \mathbf{v} \| \mathbf{E}_{\mathbf{c}}(\mathbf{v}) \right ] \\ & = \mathbf{f} (\mathbf{c}) + \lim_{\mathbf{v} \to \mathbf{0}} \mathbf{T}_{\mathbf{c}} (\mathbf{v}) + \lim_{\mathbf{v} \to \mathbf{0}} \| \mathbf{v} \| \mathbf{E}_{\mathbf{c}} (\mathbf{v}) \end{align}
• We know that $\displaystyle{\lim_{\mathbf{v} \to \mathbf{0}} \| \mathbf{v} \| = \mathbf{0}}$ and by definition $\displaystyle{\lim_{\mathbf{v} \to \mathbf{0}} \mathbf{E}_{\mathbf{c}}(\mathbf{v}) = \mathbf{0}}$ so $\displaystyle{\lim_{\mathbf{v} \to \mathbf{0}} \| \mathbf{v} \| \mathbf{E}_{\mathbf{c}} (\mathbf{v}) = \mathbf{0}}$. All that remains to show is that $\displaystyle{\lim_{\mathbf{v} \to \mathbf{0}} \mathbf{T}_{\mathbf{c}} (\mathbf{v}) = \mathbf{0}}$. Let $\mathbf{u}_1, \mathbf{u}_2, ..., \mathbf{u}_m$ be the standard basis vectors for $\mathbb{R}^n$. Then for each $\mathbf{v} \in \mathbb{R}^n$ we have that:
(4)
\begin{align} \quad \mathbf{v} = v_1\mathbf{u}_1 + v_2\mathbf{u}_2 + ... + v_n\mathbf{u}_n \end{align}
• Since $\mathbf{T}_{\mathbf{c}}$ is a linear function we have that then:
(5)
\begin{align} \quad \mathbf{T}_{\mathbf{c}} ( \mathbf{v}) = v_1 \mathbf{T}_{\mathbf{c}} (\mathbf{u}_1) + v_2 \mathbf{T}_{\mathbf{c}} (\mathbf{u}_2) + ... + v_n\mathbf{T}_{\mathbf{c}} (\mathbf{u}_n) \end{align}
• Note that the values of $\mathbf{T}_{\mathbf{c}} (\mathbf{u}_1), \mathbf{T}_{\mathbf{c}} (\mathbf{u}_2), ..., \mathbf{T}_{\mathbf{c}} (\mathbf{u}_n)$ are fixed, and as $\mathbf{v} \to \mathbf{0}$ we have that $v_1, v_2, ..., v_n \to 0$, so indeed $\displaystyle{\lim_{\mathbf{v} \to \mathbf{0}} \mathbf{T}_{\mathbf{c}} (\mathbf{v}) = \mathbf{0}}$. Hence:
(6)
\begin{align} \quad \lim_{\mathbf{v} \to \mathbf{0}} \mathbf{f} (\mathbf{c} + \mathbf{v}) = \mathbf{f}(\mathbf{c}) \end{align}
• So $\mathbf{f}$ is continuous at $\mathbf{c}$. $\blacksquare$