Proof of Restriction Stability of Continuity and of the Derivative

Since $J\subseteq I$ and $J$ contains two distinct points, so does $I$; so derivatives relative to $I$ and relative to $J$ are as in Derivative of a Real Function on an Interval (i). Every $t\in J$ lies in $I$ and satisfies $f|_{J}(t)=f(t)$ (Domains and Restrictions of Maps (iii)); in particular $f|_{J}(x)=f(x)$. Hence for every $t\in J$ with $t\ne x$ the difference quotients agree:

$$\frac{f|_{J}(t)-f|_{J}(x)}{t-x}=\frac{f(t)-f(x)}{t-x}.$$

We use this agreement of quotients below without repeating it.

Proof of (i). By The Real Line: Standing Notation and Background for Calculus on Intervals (vi), continuity at $x$ relative to $I$ and relative to $J$ is that of Continuous Map Between Metric Spaces (i) for $(X,d_X)=(Y,d_Y)=(\mathbb{R},d_{\mathbb{R}})$. So, as $f$ is continuous at $x$ relative to $I$, Restricting a Continuous Map to a Smaller Subset Preserves Continuity (i), applied with these spaces, $A=I$, $B=J$, our $f$ and our $x\in J$, shows that $f|_{J}$ is continuous at $x$ relative to $J$.

Proof of (ii). Let $L=f'(x)$, the derivative of $f$ at $x$ as in Derivative of a Real Function on an Interval (ii). Let $\varepsilon>0$. By Derivative of a Real Function on an Interval (i) there is $\delta>0$ such that every $t\in I$ with $0<|t-x|<\delta$ satisfies $\bigl|\frac{f(t)-f(x)}{t-x}-L\bigr|<\varepsilon$. Let $t\in J$ with $0<|t-x|<\delta$. Then $t\in I$, and by the agreement of quotients

$$\Bigl|\frac{f|_{J}(t)-f|_{J}(x)}{t-x}-L\Bigr|=\Bigl|\frac{f(t)-f(x)}{t-x}-L\Bigr|<\varepsilon .$$

As $\varepsilon>0$ was arbitrary, $L$ is a derivative of $f|_{J}$ at $x$ relative to $J$, by Derivative of a Real Function on an Interval (i). So $f|_{J}$ is differentiable at $x$ by Derivative of a Real Function on an Interval (ii), and since the derivative of $f|_{J}$ at $x$ is unique by Uniqueness of the Derivative of a Real Function on an Interval (i), $(f|_{J})'(x)=L=f'(x)$.

Proof of (iii). Let $u,v\in J$ with $u<x<v$, suppose that $f|_{J}$ is differentiable at $x$, and let $L=(f|_{J})'(x)$, the derivative as in Derivative of a Real Function on an Interval (ii). Let $\varepsilon>0$. First, by Derivative of a Real Function on an Interval (i) applied to $f|_{J}$, choose $\delta_{1}>0$ such that every $t\in J$ with $0<|t-x|<\delta_{1}$ satisfies $\bigl|\frac{f|_{J}(t)-f|_{J}(x)}{t-x}-L\bigr|<\varepsilon$. Then let

$$\delta=\min\{\delta_{1},\,x-u,\,v-x\},$$

which is positive since each of $\delta_{1}$, $x-u$ and $v-x$ is, the last two as $u<x$ and $x<v$ (Difference and Quotient Identities in ℝ (i)). Let $t\in I$ with $0<|t-x|<\delta$. Since $|t-x|<x-u$ and $|t-x|<v-x$, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (ix) gives $-(x-u)<t-x$ and $t-x<v-x$. As $-(x-u)=u-x$ (Difference and Quotient Identities in ℝ (i)), adding $x$ to both sides of each inequality gives $u<t<v$, by Rules of Arithmetic and Order in an Ordered Field (iv). As $u,v\in J$ and $J$ is an interval, $t\in J$ by Basic Facts about Intervals of the Real Line (iii). Moreover $|t-x|<\delta\le\delta_{1}$, so by the agreement of quotients

$$\Bigl|\frac{f(t)-f(x)}{t-x}-L\Bigr|=\Bigl|\frac{f|_{J}(t)-f|_{J}(x)}{t-x}-L\Bigr|<\varepsilon .$$

As $\varepsilon>0$ was arbitrary, $L$ is a derivative of $f$ at $x$ relative to $I$, by Derivative of a Real Function on an Interval (i). So $f$ is differentiable at $x$ by Derivative of a Real Function on an Interval (ii), and since the derivative of $f$ at $x$ is unique by Uniqueness of the Derivative of a Real Function on an Interval (i), $f'(x)=L=(f|_{J})'(x)$.