Proof of The Increment Form of the Derivative, and Differentiability Implies Continuity

Preliminaries. By The Real Line: Standing Notation and Background for Calculus on Intervals (vi), a map $\chi:I\to\mathbb{R}$ is continuous at $x$ relative to $I$ if and only if for every real $\varepsilon>0$ there is a real $\delta>0$ such that every $t\in I$ with $|t-x|<\delta$ satisfies $|\chi(t)-\chi(x)|<\varepsilon$. For $t\in I$, $|t-x|=0$ holds if and only if $t=x$, by Rules of Arithmetic and Order in an Ordered Field (x); so for $t\in I$ with $t\neq x$ the quotient $q(t)=(f(t)-f(x))/(t-x)$ is defined. Finally, for $a,b\in\mathbb{R}$ and $t\in I$ with $t\neq x$:

$$(\ast)\qquad a=b\,(t-x)\quad\text{if and only if}\quad a/(t-x)=b,$$

because $(t-x)\,(t-x)^{-1}=1$ by Negatives, Differences, Reciprocals and Quotients (ii): multiply the left equation by $(t-x)^{-1}$, or the right one by $t-x$.

Proof of (i), only if. Let $L$ be a derivative of $f$ at $x$. By Maps Defined by Cases (i), applied to the property $t\neq x$ of elements $t\in I$ and the expressions $q(t)$ and $L$, there is a map $\varphi:I\to\mathbb{R}$ with $\varphi(t)=q(t)$ for $t\in I$ with $t\neq x$ and $\varphi(x)=L$. In particular $\varphi(x)=L$.

For $t=x$ both sides of $f(t)-f(x)=\varphi(t)\,(t-x)$ are $0$. For $t\in I$ with $t\neq x$, the equation $q(t)=\varphi(t)$ gives $f(t)-f(x)=\varphi(t)\,(t-x)$ by $(\ast)$.

For continuity, let $\varepsilon>0$ be real. By Derivative of a Real Function on an Interval (i) there is a real $\delta>0$ such that every $t\in I$ with $0<|t-x|<\delta$ satisfies $|q(t)-L|<\varepsilon$. Let $t\in I$ with $|t-x|<\delta$. If $t=x$, then $|\varphi(t)-\varphi(x)|=|0|=0<\varepsilon$, by Rules of Arithmetic and Order in an Ordered Field (x). If $t\neq x$, then $0<|t-x|$, by the preliminaries and Rules of Arithmetic and Order in an Ordered Field (x), so $|\varphi(t)-\varphi(x)|=|q(t)-L|<\varepsilon$. Hence $\varphi$ is continuous at $x$ relative to $I$.

Proof of (i), if. Let $\varphi$ be as stated, and let $\varepsilon>0$ be real. Continuity of $\varphi$ at $x$ relative to $I$ gives a real $\delta>0$ such that every $t\in I$ with $|t-x|<\delta$ satisfies $|\varphi(t)-\varphi(x)|<\varepsilon$. Let $t\in I$ with $0<|t-x|<\delta$. Then $t\neq x$, and $f(t)-f(x)=\varphi(t)\,(t-x)$ gives $q(t)=\varphi(t)$ by $(\ast)$. Hence $|q(t)-L|=|\varphi(t)-\varphi(x)|<\varepsilon$. So $L$ is a derivative of $f$ at $x$, by Derivative of a Real Function on an Interval (i).

Proof of (ii). Let $f$ be differentiable at $x$. By Derivative of a Real Function on an Interval (ii), $f'(x)$ is a derivative of $f$ at $x$, so (i) with $L=f'(x)$ gives a map $\varphi:I\to\mathbb{R}$, continuous at $x$ relative to $I$, with $f(t)-f(x)=\varphi(t)\,(t-x)$ for every $t\in I$. Let $c:I\to\mathbb{R}$ be the constant map with value $f(x)$, and $h:I\to\mathbb{R}$ the map $t\mapsto t-x$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (v)).

The map $h$ is continuous at $x$ relative to $I$: given a real $\varepsilon>0$, take $\delta=\varepsilon$; every $t\in I$ with $|t-x|<\delta$ satisfies $|h(t)-h(x)|=|(t-x)-0|=|t-x|<\varepsilon$, since $h(x)=x-x=0$.

By Continuity of Sums and Products of Real-Valued Functions on a Metric Space (i), applied in the metric space $(\mathbb{R},d_{\mathbb{R}})$ with the subset $I$, the map $c$ is continuous at $x$ relative to $I$; by Continuity of Sums and Products of Real-Valued Functions on a Metric Space (iii), applied to $\varphi$ and $h$, so is the map $\varphi h:t\mapsto\varphi(t)\,h(t)$; and by Continuity of Sums and Products of Real-Valued Functions on a Metric Space (ii), applied to $c$ and $\varphi h$, so is the map $c+\varphi h:t\mapsto c(t)+\varphi(t)\,h(t)$. For every $t\in I$,

$$c(t)+\varphi(t)\,h(t)=f(x)+\varphi(t)\,(t-x)=f(x)+\bigl(f(t)-f(x)\bigr)=f(t),$$

so $c+\varphi h=f$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv)). Hence $f$ is continuous at $x$ relative to $I$.