Pointwise linear interpolation error from a second-derivative bound

Define $L(t):=f(a)+(f(b)-f(a))(t-a)\mathrm{/}(b-a)$. Then $L$ is affine and has the required endpoint values; any affine function agreeing with $f$ at the distinct endpoints has the same slope and intercept, so it equals $L$. By [F1] and [F2], $f^{\mathrm{\prime }\mathrm{\prime }}$ is bounded on $[a,b]$, so $M$ is a finite nonnegative real.

The desired inequality is equality at $x=a$ and $x=b$. Fix an interior point $x\in (a,b)$ and put $c_x:=(f(x)-L(x))\mathrm{/}((x-a)(b-x))$. The denominator is positive. Define $g(t):=f(t)-L(t)-c_x(t-a)(b-t)$; then $g(a)=g(x)=g(b)=0$.

Rolle's theorem [F3] on $[a,x]$ and $[x,b]$ gives $s_1\in (a,x)$ and $s_2\in (x,b)$ with $g^{\mathrm{\prime }}(s_1)=g^{\mathrm{\prime }}(s_2)=0$. Applying [F3] to $g^{\mathrm{\prime }}$ on $[s_1,s_2]$ gives $\xi \in (s_1,s_2)$ with $g^{\mathrm{\prime }\mathrm{\prime }}(\xi )=0$. All applications are valid because $f$ is $C^2$ on a neighborhood of $[a,b]$.

Since $L^{\mathrm{\prime }\mathrm{\prime }}=0$ and $((t-a)(b-t){)}^{\mathrm{\prime }\mathrm{\prime }}=-2$, step 3.1 yields $0=g^{\mathrm{\prime }\mathrm{\prime }}(\xi )=f^{\mathrm{\prime }\mathrm{\prime }}(\xi )+2c_x$. Thus $\mathrm{\mid }{c}_x\mathrm{\mid }=\mathrm{\mid }{f}^{\mathrm{\prime }\mathrm{\prime }}(\xi )\mathrm{\mid }\mathrm{/}2\le M\mathrm{/}2$. Multiplying by the positive denominator in the definition of $c_x$ proves the claimed bound at this $x$; together with the endpoint cases in step 2.1 this proves the result.