Proof of Derivatives of the Slice of a Function Along a Line

Write $\lVert\,\cdot\,\rVert$ for the Euclidean norm and $|\,\cdot\,|$ for the absolute value. Claims 1 and 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ give $\lVert z\rVert^{2}=z\cdot z=\sum_{i=1}^{n}z_iz_i$ and $\lVert z\rVert=d_E(z,0_{\mathbb{R}^n})$ for the Euclidean distance and the origin $0_{\mathbb{R}^n}$, and claim 5 gives $\lVert\mu z\rVert=|\mu|\,\lVert z\rVert$. Claims 2 and 4 of Properties of the Absolute Value in an Ordered Field are used as symmetry and multiplicativity, claims 2 and 3 of Properties of Finite Sums as additivity and homogeneity of finite sums, and claims 1 and 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field as the strict and weak comparison of squares of nonnegative elements. Order arithmetic is taken from Elementary Order Arithmetic in an Ordered Field (claim 5 product of positive elements, claim 7 inverse of a positive element, claim 8 halving, claim 10 strict compatibility with multiplication by a positive element). The field axioms of the field $\mathbb{R}$ and the vector space axioms of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space are used for rearrangement; in particular

Claim 1. Put $L=\sum_{i=1}^{n}\frac{\partial f}{\partial x_i}(x_0)\,h_i$. By Gradient of a Real-Valued Function on a Euclidean Open Set the $i$th coordinate of $Df(x_0)$ is $\frac{\partial f}{\partial x_i}(x_0)$, so $L=h\cdot Df(x_0)$ by Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$ and commutativity of multiplication. We show that $L$ has the property required of $g'(t_0)$.

Suppose first that $h$ is the origin. Then $p+t\,h=p$ for every $t$, so $g$ is constant and every difference quotient of $g$ vanishes; moreover $h_i=0$ for every $i$, so homogeneity of finite sums with the factor $0$ gives $L=0$. Hence for any $\varepsilon$ with $0<\varepsilon$ the choice $\delta=1$ works.

Assume now that $h$ is not the origin, so that $0<\lVert h\rVert$ by the axioms of Metric Space, the Euclidean distance being a metric by Euclidean Distance is a Metric on $\mathbb{R}^n$. By C^1 Maps on Euclidean Open Sets are Differentiable the function $f$, regarded as a map into $\mathbb{R}^m$ with $m=1$, is differentiable at $x_0$. For $k\in\mathbb{R}^n$ with $x_0+k\in U$ write

The defining condition of differentiability reads: for every $\varepsilon'$ with $0<\varepsilon'$ there is $\delta_0$ with $0<\delta_0$ such that $0<\sum_{i=1}^{n}k_i^{2}<\delta_0^{2}$ and $x_0+k\in U$ imply $R(k)^{2}\le\varepsilon'^{2}\sum_{i=1}^{n}k_i^{2}$. Since $\sum_{i=1}^{n}k_i^{2}=\lVert k\rVert^{2}$, since $R(k)^{2}=|R(k)|^{2}$ by multiplicativity of the absolute value, and since $\varepsilon'^{2}\lVert k\rVert^{2}=(\varepsilon'\lVert k\rVert)^{2}$, the weak comparison of squares of nonnegative elements turns that conclusion into

Let $\varepsilon$ with $0<\varepsilon$ be given. Put $\varepsilon'=\tfrac12\,\varepsilon\,\lVert h\rVert^{-1}$, which is positive, take $\delta_0$ as above for this $\varepsilon'$, and put $\delta=\delta_0\,\lVert h\rVert^{-1}$, which is positive. Let $s\in\mathbb{R}$ satisfy $0<|s|<\delta$ and $t_0+s\in J$, and put $k=s\,h$. Then $\lVert k\rVert=|s|\,\lVert h\rVert$ is positive, and multiplying $|s|<\delta$ by the positive element $\lVert h\rVert$ gives $\lVert k\rVert<\delta_0$; the strict comparison of squares then gives $0<\sum_{i=1}^{n}k_i^{2}<\delta_0^{2}$. Moreover $x_0+k=p+(t_0+s)h\in U$ because $t_0+s\in J$. Homogeneity of finite sums gives

so $g(t_0+s)-g(t_0)=f(x_0+k)-f(x_0)=R(k)+s\,L$ and therefore

Using multiplicativity of the absolute value, $|R(k)/s|=|R(k)|\,|s|^{-1}\le\varepsilon'\,\lVert k\rVert\,|s|^{-1}=\varepsilon'\,\lVert h\rVert=\tfrac12\,\varepsilon<\varepsilon$, the last step by halving. Hence $g$ is differentiable at $t_0$ with $g'(t_0)=L$.

Claim 2. Let $j\in\{1,\dots,n\}$. By clause 2 of C^2 Real-Valued Map on an Open Subset of Euclidean Space the function $\partial f/\partial x_j$ is of class $C^1$ on $U$, so claim 1 applied to it in place of $f$ shows that the function $\psi_j:J\to\mathbb{R}$ with $\psi_j(t)=\frac{\partial f}{\partial x_j}(p+t\,h)$ is differentiable at $t_0$ with

the second equality by Hessian Matrix of a C^2 Function.

By commutativity of multiplication, $g_1(t)=\sum_{j=1}^{n}h_j\,\psi_j(t)$ for every $t\in J$, so Derivative of a Finite Linear Combination of Real Functions, applied with the constants $h_j$, shows that $g_1$ is differentiable at $t_0$ with

the inner factor $h_j$ being moved inside the inner sum by homogeneity of finite sums. Interchanging the two summations by Interchange of a Finite Double Sum and applying claim 4 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum with $M=D^2f(x_0)$ and $w=z=h$ gives