Proof of Mean Value Theorem on a Closed Real Interval

Throughout, $|\cdot|$ is the absolute value on $\mathbb{R}$. Since $a<b$, we have $b-a>0$, so $b-a\ne0$ and the quotient

is defined. Define $g:[a,b]\to\mathbb{R}$ by

Step 1: $g$ is continuous on $[a,b]$. The identity function $\mathrm{id}:[a,b]\to\mathbb{R}$, $\mathrm{id}(x)=x$, is continuous on $[a,b]$: for $x\in[a,b]$ and real $\varepsilon>0$ the choice $\delta=\varepsilon$ works, since every $y\in[a,b]$ with $d_{\mathbb{R}}(x,y)<\delta$ satisfies $d_{\mathbb{R}}(\mathrm{id}(y),\mathrm{id}(x))=d_{\mathbb{R}}(y,x)=d_{\mathbb{R}}(x,y)<\varepsilon$, using the symmetry required of a metric. Now apply Continuity of Sums and Products of Real-Valued Functions on a Metric Space with $X$ the real line and $A=[a,b]$: by its claim 5 the function $(-m)\,\mathrm{id}$ is continuous on $[a,b]$; by its claim 1, applied at each $x\in[a,b]$, the constant function $k:[a,b]\to\mathbb{R}$ with value $m\,a$ is continuous on $[a,b]$; and by its claim 5, applied twice, $g=f+\bigl((-m)\,\mathrm{id}+k\bigr)$ is continuous on $[a,b]$.

Step 2: $g$ is differentiable at every $x\in(a,b)$ with $g'(x)=f'(x)-m$. Fix $x\in(a,b)$, an interior point of $[a,b]$, and write $L=f'(x)$. Let $\varepsilon>0$ be real, and let $\delta>0$ be as provided for this $\varepsilon$ by the differentiability of $f$ at $x$. For every real $h$ with $0<|h|<\delta$ and $x+h\in[a,b]$,

so

As $\varepsilon>0$ was arbitrary, $g$ is differentiable at $x$ with $g'(x)=L-m=f'(x)-m$.

Step 3: conclusion. We compute $g(a)=f(a)+(-m)\,a+m\,a=f(a)$ and

so $g(a)=g(b)$. By Steps 1 and 2 the function $g$ satisfies the hypotheses of Rolle's Theorem on a Closed Real Interval, so there exists $c\in(a,b)$ with $g'(c)=0$, that is, $f'(c)-m=0$, that is,