Proof of Mean Value Theorem on a Closed Real Interval and on Any Interval

Proof of (i). Since $a<b$, $0<b-a$ (Difference and Quotient Identities in ℝ (i)), so $b-a\ne0$; let

$$m=\frac{f(b)-f(a)}{b-a}\in\mathbb{R},$$

so that $m\,(b-a)=f(b)-f(a)$ (Difference and Quotient Identities in ℝ (ii)). Let $h:[a,b]\to\mathbb{R}$ be the map $t\mapsto f(t)-f(a)-m\,(t-a)$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (v)). For $s,t\in[a,b]$,

$$h(t)-h(s)=\bigl(f(t)-f(s)\bigr)-m\,(t-s).$$

$h$ is continuous on $[a,b]$. Continuity is read in $(\mathbb{R},d_{\mathbb{R}})$ relative to $[a,b]$, as in The Real Line: Standing Notation and Background for Calculus on Intervals (vi). Let $\iota:[a,b]\to\mathbb{R}$ be the map $t\mapsto t$; it is continuous on $[a,b]$ by Continuity of Identity Maps, Compositions and Reciprocals (i), with $(X,d)=(\mathbb{R},d_{\mathbb{R}})$ and $A=[a,b]$. Put $c=m\,a-f(a)\in\mathbb{R}$, and let $k_{c}:[a,b]\to\mathbb{R}$ be the constant map with value $c$. For $t\in[a,b]$, $h(t)=f(t)+(-m)\,\iota(t)+k_{c}(t)$, so $h$ is the map $\bigl(f+(-m)\iota\bigr)+k_{c}$ formed as in Continuity of Sums and Products of Real-Valued Functions on a Metric Space (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv)). Let $s\in[a,b]$. Then $f$ and $\iota$ are continuous at $s$ relative to $[a,b]$ (Continuous Map Between Metric Spaces (ii)), $k_{c}$ is by Continuity of Sums and Products of Real-Valued Functions on a Metric Space (i), $(-m)\iota$ is by Continuity of Sums and Products of Real-Valued Functions on a Metric Space (iv), and $h$ is by Continuity of Sums and Products of Real-Valued Functions on a Metric Space (ii), applied twice. As $s$ was arbitrary, $h$ is continuous on $[a,b]$, by Continuous Map Between Metric Spaces (ii).

$h$ is differentiable at every $s\in(a,b)$, with $h'(s)=f'(s)-m$. Let $s\in(a,b)$, and let $L=f'(s)$, the derivative of $f$ at $s$ as in Derivative of a Real Function on an Interval (ii). For $t\in[a,b]$ with $t\ne s$, $t-s\neq0$, and the identity above gives $\frac{h(t)-h(s)}{t-s}=\frac{f(t)-f(s)}{t-s}-m$ (Difference and Quotient Identities in ℝ (ii), with $f(t)-f(s)$, $m$ and $t-s$ for its $a$, $m$ and $b$); hence

$$\frac{h(t)-h(s)}{t-s}-(L-m)=\frac{f(t)-f(s)}{t-s}-L .$$

Let $\varepsilon>0$. By Derivative of a Real Function on an Interval (i) there is $\delta>0$ such that every $t\in[a,b]$ with $0<|t-s|<\delta$ satisfies $\bigl|\frac{f(t)-f(s)}{t-s}-L\bigr|<\varepsilon$, hence $\bigl|\frac{h(t)-h(s)}{t-s}-(L-m)\bigr|<\varepsilon$. So $L-m$ is a derivative of $h$ at $s$ relative to $[a,b]$, by Derivative of a Real Function on an Interval (i); $h$ is differentiable at $s$ by Derivative of a Real Function on an Interval (ii), and $h'(s)=L-m=f'(s)-m$, the derivative being unique by Uniqueness of the Derivative of a Real Function on an Interval (i).

Conclusion. $h(a)=f(a)-f(a)-m\cdot0=0$ and $h(b)=f(b)-f(a)-m\,(b-a)=0$, so $h(a)=h(b)$. By Rolle's Theorem on a Closed Real Interval (i) applied to $h$ there is $\xi\in(a,b)$ with $h'(\xi)=0$, that is $f'(\xi)=m$. Then $f'(\xi)\,(b-a)=m\,(b-a)=f(b)-f(a)$.

Proof of (ii). Let $x,y\in I$ with $x<y$, and let $J=[x,y]$. By Basic Facts about Intervals of the Real Line (ii), $J\subseteq I$, and by Basic Facts about Intervals of the Real Line (v), $J$ is an interval containing two distinct points, with $(x,y)\subseteq J$. Let $g=f|_{J}:J\to\mathbb{R}$ be the restriction of $f$ to $J$, with $g(t)=f(t)$ for every $t\in J$ (Domains and Restrictions of Maps (iii)).

Since $f$ is continuous on $I$, it is continuous at every point of $I$ relative to $I$, in particular at every $t\in J$, as $J\subseteq I$. So for every $t\in J$, $g$ is continuous at $t$ relative to $J$ by Restriction Stability of Continuity and of the Derivative (i); hence $g$ is continuous on $J$, by The Real Line: Standing Notation and Background for Calculus on Intervals (vi).

Let $t\in(x,y)$. Then $t\in J$, as $(x,y)\subseteq J$, so $t\in I$, as $J\subseteq I$. Since moreover $x<t<y$ with $x,y\in I$, $f$ is differentiable at $t$ relative to $I$ by hypothesis. By Restriction Stability of Continuity and of the Derivative (ii), $g$ is differentiable at $t$ relative to $J$ and $g'(t)=f'(t)$.

So $g$ satisfies the hypotheses of (i) with $a=x$ and $b=y$, which gives $\xi\in(x,y)$ with $g(y)-g(x)=g'(\xi)\,(y-x)$. Here $x<\xi<y$, $\xi\in J\subseteq I$, $g(y)=f(y)$, $g(x)=f(x)$ and $g'(\xi)=f'(\xi)$, so $f(y)-f(x)=f'(\xi)\,(y-x)$.