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:

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