[proofplan] We use the openness of $U$ to restrict to small increments $h$ for which the whole segment from $a$ to $a+h$ lies in $U$. Along that segment, the one-variable [fundamental theorem of calculus](/theorems/632) expresses $f(a+h)-f(a)$ as the integral of the derivative of $f$ applied to $h$. After subtracting the linear term $Df_a(h)$, the remainder is controlled by the uniform oscillation of $Df$ on the segment. Since $Df$ is continuous at $a$, that oscillation tends to zero as $h \to 0$. [/proofplan] [step:Choose a ball on which all small line segments stay inside $U$] For each $\rho>0$, let \begin{align*} B(a,\rho):=\{x \in \mathbb{R}^n: |x-a|<\rho\} \end{align*} denote the open Euclidean ball of radius $\rho$ centred at $a$. Since $U$ is open and $a \in U$, there exists $\rho>0$ such that \begin{align*} B(a,\rho) \subset U. \end{align*} Define \begin{align*} V_\rho := \{h \in \mathbb{R}^n : |h|<\rho\}. \end{align*} For every $h \in V_\rho$ and every $t \in [0,1]$, the point $a+th$ belongs to $U$, because \begin{align*} |a+th-a| = t|h| < \rho. \end{align*} Thus the line segment from $a$ to $a+h$ is contained in $U$ whenever $h \in V_\rho$. [/step] [step:Represent the increment of $f$ by integrating the derivative along the segment] Fix $h \in V_\rho$. Define the curve \begin{align*} \gamma_h: [0,1] \to U, \quad t \mapsto a+th. \end{align*} Define also the one-variable map \begin{align*} \varphi_h: [0,1] \to \mathbb{R}^m, \quad t \mapsto f(\gamma_h(t)). \end{align*} Since $f \in C^1(U;\mathbb{R}^m)$ and $\gamma_h$ is smooth, the chain rule gives that $\varphi_h$ is $C^1$ and \begin{align*} \varphi_h'(t)=Df_{a+th}(h) \end{align*} for every $t \in [0,1]$. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $[0,1]$. By the one-variable fundamental theorem of calculus for $C^1$ curves in $\mathbb{R}^m$, \begin{align*} f(a+h)-f(a)=\int_0^1 Df_{a+th}(h)\,d\mathcal{L}^1(t). \end{align*} [guided] Fix $h \in V_\rho$. The point of introducing a one-variable curve is that Taylor's formula at first order is obtained by moving from $a$ to $a+h$ along the straight line segment and integrating the instantaneous rate of change. Define \begin{align*} \gamma_h: [0,1] \to U, \quad t \mapsto a+th. \end{align*} The previous step verifies that $\gamma_h(t) \in U$ for every $t \in [0,1]$, so the composition with $f$ is well-defined. Define \begin{align*} \varphi_h: [0,1] \to \mathbb{R}^m, \quad t \mapsto f(\gamma_h(t)). \end{align*} Because $f \in C^1(U;\mathbb{R}^m)$, the derivative $Df_x:\mathbb{R}^n \to \mathbb{R}^m$ exists for every $x \in U$ and depends continuously on $x$ as a map into $\mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$, the [vector space](/page/Vector%20Space) of linear maps from $\mathbb{R}^n$ to $\mathbb{R}^m$ equipped with the operator norm induced by the Euclidean norms. Since $\gamma_h$ is a smooth curve with derivative $\gamma_h'(t)=h$, the chain rule gives \begin{align*} \varphi_h'(t)=Df_{\gamma_h(t)}(\gamma_h'(t))=Df_{a+th}(h). \end{align*} The map $\varphi_h$ is therefore a $C^1$ map from the compact interval $[0,1]$ into $\mathbb{R}^m$. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $[0,1]$. Applying the one-variable fundamental theorem of calculus for $C^1$ curves in $\mathbb{R}^m$, we obtain \begin{align*} \varphi_h(1)-\varphi_h(0)=\int_0^1 \varphi_h'(t)\,d\mathcal{L}^1(t). \end{align*} Substituting the definitions of $\varphi_h(1)$, $\varphi_h(0)$, and $\varphi_h'(t)$ gives \begin{align*} f(a+h)-f(a)=\int_0^1 Df_{a+th}(h)\,d\mathcal{L}^1(t). \end{align*} This is the exact place where the linear approximation appears: the integral averages the derivative along the segment, and continuity of $Df$ will show that this average differs from $Df_a$ by a small operator. [/guided] [/step] [step:Define the remainder on all admissible increments] For every $h \in V$, define \begin{align*} r(h):=f(a+h)-f(a)-Df_a(h). \end{align*} Then \begin{align*} f(a+h)=f(a)+Df_a(h)+r(h) \end{align*} for every $h \in V$. If $h \in V_\rho$, the integral representation from the previous step and linearity of $Df_a:\mathbb{R}^n \to \mathbb{R}^m$ give \begin{align*} r(h)=\int_0^1 \bigl(Df_{a+th}-Df_a\bigr)(h)\,d\mathcal{L}^1(t). \end{align*} Indeed, \begin{align*} Df_a(h)=\int_0^1 Df_a(h)\,d\mathcal{L}^1(t), \end{align*} because $\mathcal{L}^1([0,1])=1$. [/step] [step:Estimate the remainder by the oscillation of $Df$ on the segment] Let $h \in V_\rho \setminus \{0\}$. By the triangle inequality for the Euclidean norm in $\mathbb{R}^m$ and the definition of the operator norm on $\mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$, \begin{align*} |r(h)| \leq \int_0^1 \|(Df_{a+th}-Df_a)\|_{\mathrm{op}}\,|h|\,d\mathcal{L}^1(t). \end{align*} Therefore \begin{align*} \frac{|r(h)|}{|h|}\leq \sup_{t \in [0,1]}\|Df_{a+th}-Df_a\|_{\mathrm{op}}. \end{align*} [/step] [step:Use continuity of $Df$ at $a$ to prove the small-o estimate] Let $\varepsilon>0$. Let $\mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$ denote the vector space of linear maps from $\mathbb{R}^n$ to $\mathbb{R}^m$, equipped with the operator norm induced by the Euclidean norms. Since $f \in C^1(U;\mathbb{R}^m)$, the derivative map \begin{align*} Df: U \to \mathcal{L}(\mathbb{R}^n,\mathbb{R}^m), \quad x \mapsto Df_x \end{align*} is continuous at $a$. Hence there exists $\delta \in (0,\rho)$ such that, whenever $x \in U$ and $|x-a|<\delta$, \begin{align*} \|Df_x-Df_a\|_{\mathrm{op}}<\varepsilon. \end{align*} If $h \in V \setminus \{0\}$ and $|h|<\delta$, then $h \in V_\rho$ because $\delta<\rho$. Hence, for every $t \in [0,1]$, \begin{align*} |a+th-a|=t|h|<\delta. \end{align*} Thus \begin{align*} \sup_{t \in [0,1]}\|Df_{a+th}-Df_a\|_{\mathrm{op}}\leq \varepsilon. \end{align*} Combining this with the previous step gives \begin{align*} \frac{|r(h)|}{|h|}\leq \varepsilon \end{align*} whenever $0<|h|<\delta$. Therefore \begin{align*} \frac{|r(h)|}{|h|}\to 0 \end{align*} as $h \to 0$ with $h \in V \setminus \{0\}$. This proves the first-order Taylor approximation at $a$. [/step]