Proof of Differentiability at an Interior Point is a Local Property

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named in the step where it is cited.

By Derivative of a Real Function on an Interval, the hypothesis on $f|_{I}$ says: for every real $\varepsilon>0$ there is a real $\delta>0$ such that every $h\in\mathbb{R}$ with $0<|h|<\delta$ and $x_{0}+h\in I$ satisfies

$$\left|\frac{f|_{I}(x_{0}+h)-f|_{I}(x_{0})}{h}-L\right|<\varepsilon .$$

What is to be proved is the corresponding statement for $f$ on $J$: that for every real $\varepsilon>0$ there is a real $\delta'>0$ such that every $h\in\mathbb{R}$ with $0<|h|<\delta'$ and $x_{0}+h\in J$ satisfies

$$\left|\frac{f(x_{0}+h)-f(x_{0})}{h}-L\right|<\varepsilon .$$

This is meaningful because $x_{0}$ is by hypothesis an interior point of $J$.

Let $\varepsilon$ be a real number with $0<\varepsilon$. The choices are made in this order: $\varepsilon$ is given, then $\delta$, then $\delta'$. Choose $\delta>0$ as in the hypothesis, and let $\delta'$ be the least of the two numbers $\delta$ and $r$, so that $\delta'\le\delta$, $\delta'\le r$ and $\delta'$ is one of $\delta$ and $r$, by claim 9 of Elementary Order Arithmetic in an Ordered Field; since both $\delta$ and $r$ are positive, so is $\delta'$.

Let $h\in\mathbb{R}$ satisfy $0<|h|<\delta'$ and $x_{0}+h\in J$, and put $y=x_{0}+h$. Then

$$|y-x_{0}|=|h|<\delta'\le r ,$$

so $y$ is a point of $J$ with $|y-x_{0}|<r$, and the hypothesis on $r$ places $y$ in $I$. Consequently $f|_{I}(y)=f(y)$, and $f|_{I}(x_{0})=f(x_{0})$ because $x_{0}\in I$. Moreover $0<|h|<\delta'\le\delta$ and $x_{0}+h=y\in I$, so the displayed bound of the hypothesis applies to this $h$ and yields

$$\left|\frac{f(x_{0}+h)-f(x_{0})}{h}-L\right| =\left|\frac{f|_{I}(x_{0}+h)-f|_{I}(x_{0})}{h}-L\right|<\varepsilon .$$

As $\varepsilon>0$ was arbitrary, the real number $L$ satisfies the condition of Derivative of a Real Function on an Interval for $f$ at the interior point $x_{0}$ of $J$. Hence $f$ is differentiable at $x_{0}$ with derivative $L$, which is the assertion.