Proof of A $C^1$ Map is Differentiable at Every Point

Let $|\cdot|$ be the absolute value, write $\partial_{m}f$ for $\partial f/\partial x_{m}$, and for $z\in\mathbb{R}^{n}$ write $\lVert z\rVert$ for the Euclidean distance from $z$ to the origin, the nonnegative real with $\lVert z\rVert^{2}=\sum_{i=1}^{n}z_{i}^{2}$. For $z\in\mathbb{R}^{n}$, an index $m$ and $u\in\mathbb{R}$, write $z[m{:}u]$ for the point whose $m$th coordinate is $u$ and whose other coordinates are those of $z$. Claim numbers refer to Elementary Order Arithmetic in an Ordered Field and to Properties of the Absolute Value in an Ordered Field as indicated.

As in earlier arguments we use that squares are strictly monotone on nonnegative reals: if $0\le\alpha$, $0\le\beta$ and $\alpha<\beta$, then $0<\beta$ by claim 2, so $\beta\alpha<\beta\beta$ by claim 10 and $\alpha\alpha\le\beta\alpha$, whence $\alpha^{2}<\beta^{2}$ by claim 2; consequently $\alpha\le\beta$ whenever $\alpha^{2}\le\beta^{2}$. In particular $|z_{i}|\le\lVert z\rVert$ for each $i$, since $z_{i}^{2}\le\lVert z\rVert^{2}$ and $|z_{i}|^{2}=z_{i}^{2}$ by claim 4 of Properties of the Absolute Value in an Ordered Field.

Step 1 (a ball inside $U$). Since $a\in U$ and $U$ is open, there is $r$ with $0<r$ such that every point of $\mathbb{R}^{n}$ at Euclidean distance less than $r$ from $a$ lies in $U$.

Step 2 (telescoping). Let $h=(h_{1},\dots,h_{n})\in\mathbb{R}^{n}$ with $\lVert h\rVert<r$. Define points $p_{0},\dots,p_{n}$ by $p_{0}=a$ and $p_{m}=p_{m-1}[m{:}a_{m}+h_{m}]$ for $m\in\{1,\dots,n\}$, so that $p_{m}$ has $l$th coordinate $a_{l}+h_{l}$ for $l\le m$ and $a_{l}$ for $l>m$; in particular $p_{n}=a+h$.

More generally, for $m\in\{1,\dots,n\}$ and $u$ between $a_{m}$ and $a_{m}+h_{m}$ inclusive, the point $p_{m-1}[m{:}u]$ differs from $a$ only in coordinates $l\le m$, by $h_{l}$ for $l<m$ and by $u-a_{m}$ for $l=m$, and $|u-a_{m}|\le|h_{m}|$; so the square of its distance to $a$ is at most $\sum_{l}h_{l}^{2}=\lVert h\rVert^{2}$, whence that distance is at most $\lVert h\rVert<r$ and the point lies in $U$. In particular every $p_{m}$ lies in $U$, and

Step 3 (a mean value point in each coordinate). Fix $m$. If $h_{m}=0$ then $p_{m}=p_{m-1}$ and $f(p_{m})-f(p_{m-1})=0=\partial_{m}f(q_{m})h_{m}$ with $q_{m}=p_{m-1}$.

Suppose $h_{m}\ne0$. Since $\lVert h\rVert<r$, claim 1 gives $0<r-\lVert h\rVert$, so by claim 8 the element $\rho=\bigl(r-\lVert h\rVert\bigr)\cdot2^{-1}$ satisfies $0<\rho$ and $\rho+\rho=r-\lVert h\rVert$, that is $\lVert h\rVert+\rho+\rho=r$; adding $\lVert h\rVert+\rho$ to $0<\rho$ gives $\lVert h\rVert+\rho<r$, again by claim 1. Let $J$ be the open interval consisting of those $u$ with $a_{m}-\bigl(|h_{m}|+\rho\bigr)<u$ and $u<a_{m}+\bigl(|h_{m}|+\rho\bigr)$.

Let $u\in J$. Adding $-a_{m}$ to both inequalities, claim 1 gives $-\bigl(|h_{m}|+\rho\bigr)<u-a_{m}$ and $u-a_{m}<|h_{m}|+\rho$, so $|u-a_{m}|<|h_{m}|+\rho$ by claim 9 of Properties of the Absolute Value in an Ordered Field. The point $p_{m-1}[m{:}u]$ agrees with $a$ in the coordinates $l>m$ and differs from it by $h_{l}$ in the coordinates $l<m$ and by $u-a_{m}$ in the coordinate $m$; hence the square of its Euclidean distance to $a$ equals $\sum_{l<m}h_{l}^{2}+(u-a_{m})^{2}$. Here $\sum_{l<m}h_{l}^{2}\le\lVert h\rVert^{2}-h_{m}^{2}$, since the omitted terms $h_{l}^{2}$ with $l>m$ are nonnegative, and $(u-a_{m})^{2}=|u-a_{m}|^{2}\le\bigl(|h_{m}|+\rho\bigr)^{2}$ by claim 4 of that lemma and the monotonicity of squares. Moreover $|h_{m}|\rho\le\lVert h\rVert\rho$: this is claim 10 when $|h_{m}|<\lVert h\rVert$, and an equality when $|h_{m}|=\lVert h\rVert$. Using $|h_{m}|^{2}=h_{m}^{2}$, again by claim 4, we get

By the monotonicity of squares the distance from $p_{m-1}[m{:}u]$ to $a$ is therefore at most $\lVert h\rVert+\rho$, hence less than $r$; so $p_{m-1}[m{:}u]\in U$ for every $u\in J$.

Let $G:J\to\mathbb{R}$ be given by $G(u)=f\bigl(p_{m-1}[m{:}u]\bigr)$. For each $u_{0}\in J$, apply Slice Function and the Partial Derivative to $f$ at the point $p_{m-1}[m{:}u_{0}]$ of $U$ in the $m$th variable: it gives a positive radius $\rho_{0}$ and identifies the slice function on $(u_{0}-\rho_{0},u_{0}+\rho_{0})$, which agrees with $G$ there, as differentiable at $u_{0}$ with derivative $\partial_{m}f\bigl(p_{m-1}[m{:}u_{0}]\bigr)$, the partial derivative existing because $f$ is of class $C^{1}$. By An Open Interval is an Interval All of Whose Points Are Interior both intervals have all points interior, so shrinking the $\delta$ in the defining condition confines the increments to the overlap, where the two functions agree; hence $G$ is differentiable at $u_{0}$ with $G'(u_{0})=\partial_{m}f\bigl(p_{m-1}[m{:}u_{0}]\bigr)$.

Both $a_{m}$ and $a_{m}+h_{m}$ lie in $J$: indeed $0\le|h_{m}|$ and $0<\rho$ give $0<|h_{m}|+\rho$ and $|h_{m}|<|h_{m}|+\rho$ by claim 1, so claim 9 of Properties of the Absolute Value in an Ordered Field, applied to $0$ and to $h_{m}$ with $c=|h_{m}|+\rho$, yields the two pairs of strict inequalities required. Applying Mean Value Theorem on an Open Interval to $G$ on $J$ with these two points in increasing order yields $\xi_{m}$ strictly between them with $G(a_{m}+h_{m})-G(a_{m})=G'(\xi_{m})h_{m}$, that is, with $q_{m}=p_{m-1}[m{:}\xi_{m}]$,

Since $|\xi_{m}-a_{m}|<|h_{m}|$, Step 2 shows $q_{m}$ is at distance at most $\lVert h\rVert$ from $a$.

Step 4 (the estimate). Combining Steps 2 and 3,

Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. Since $n\ge1$, the element $n$, a sum of copies of $1$, is positive by claims 6 and 3, so $\varepsilon'=\varepsilon\,n^{-1}$ is positive by claims 7 and 5. Each $\partial_{m}f$ is continuous at $a$, by the definition of class $C^{1}$; taking the least of the finitely many radii by repeated use of claim 9, there is $\theta$ with $0<\theta$ such that every $z\in U$ at distance less than $\theta$ from $a$ satisfies $|\partial_{m}f(z)-\partial_{m}f(a)|<\varepsilon'$ for every $m$.

Let $\delta$ be the least, by claim 9, of $r$ and $\theta$, so $0<\delta$. Suppose $h\in\mathbb{R}^{n}$ satisfies $0<\sum_{i}h_{i}^{2}<\delta^{2}$ and $a+h\in U$. Then $\lVert h\rVert<\delta$ by the monotonicity of squares, so $\lVert h\rVert<r$ and Steps 2 and 3 apply, and each $q_{m}$ is at distance at most $\lVert h\rVert<\theta$ from $a$, so $|\partial_{m}f(q_{m})-\partial_{m}f(a)|<\varepsilon'$.

Writing $c_{m}=\partial_{m}f(q_{m})-\partial_{m}f(a)$ and using claim 5 of Properties of the Absolute Value in an Ordered Field repeatedly, then claim 4 of that lemma and claim 10 together with $|h_{m}|\le\lVert h\rVert$,

Both sides are nonnegative, so squaring and using claim 4 of Properties of the Absolute Value in an Ordered Field and the monotonicity of squares,

This is exactly the defining condition of differentiability of $f$ at $a$, with $m=1$ there, the required partial derivatives existing because $f$ is of class $C^{1}$.