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

Both claims follow from Derivatives Along a Segment for C^1 Functions on a Euclidean Open Set once the notation on the right-hand sides is unfolded.

Claim 1. Apply claim 2 of Derivatives Along a Segment for C^1 Functions on a Euclidean Open Set, taking for its open set $U$, for its base point the point $p$, for its direction the point $h$, for its interval $J$, and for its function $F$ the slice $g$. Its hypotheses hold: $f$ is of class $C^1$ on $U$, $J$ is an interval, and $p+t\,h\in U$ for every $t\in J$. Since $t_0$ is an interior point of $J$, that claim gives that $g$ is differentiable at $t_0$ with

the second equality being the definition $x_0=p+t_0\,h$. This is the first of the two asserted expressions.

Because $f$ is of class $C^1$ on $U$, clause 1 of C^k Maps on a Euclidean Open Set gives that the partial derivative of $f$ with respect to the $i$th variable exists at every point of $U$, and in particular at $x_0$; so the gradient $Df(x_0)$ is defined, and its $i$th coordinate is $\partial_i f(x_0)$. By the definition of the dot product,

which is the displayed sum, the two factors in each summand commuting. This proves claim 1.

Claim 2. Assume now that $f$ is of class $C^2$ on $U$. The function $g_1$ is exactly the function called $G$ in claim 4 of Derivatives Along a Segment for C^1 Functions on a Euclidean Open Set, for the same data as above, since both are given by $t\mapsto\sum_{i=1}^{n}\partial_i f(p+t\,h)\,h_i$. That claim therefore gives that $g_1$ is differentiable at the interior point $t_0$ of $J$, with

where $\partial_j\partial_i f$ is the iterated partial derivative of clause 4 of C^k Maps on a Euclidean Open Set, defined on all of $U$ by clause 2 there.

We rewrite this double sum. Put $a_{ij}=\partial_j\partial_i f(x_0)\,h_i h_j$ for $i,j\in\{1,\dots,n\}$. By Interchange of a Finite Double Sum,

Renaming the bound index $j$ to $i$ and the bound index $i$ to $j$ on the right-hand side — a change of the names of the summation variables only, the order of summation being left as it stands — turns it into

the last equality because the real factors in each summand commute. Hence

Finally, $f$ is of class $C^2$ on $U$ and $x_0\in U$, so the Hessian matrix $D^2f(x_0)$ is defined, and its entry in row $i$ and column $j$ is $\partial_i\partial_j f(x_0)$. By claim 4 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum, applied with $M=D^2f(x_0)$ and $w=z=h$, and with the matrix-vector product,

which is the sum just computed. This proves claim 2. $\blacksquare$