Exercise 5.3.12

Let $𝜖>0$ and choose ${\delta }_1>0$ so every $|h|<{\delta }_1$ has

Choose ${\delta }_2>0$ so every $0<|x-a|<{\delta }_2$ has $|f^{″}(x)-f^{″}(a)|<𝜖$ (this is where we use the continuity of $f^{″}$ at $a$.) and set $\delta =\min <!--\; FUNCTION\; APPLICATION\; -->\{{\delta }_1,{\delta }_2\}$.

Without loss of generality assume $h>0$, apply MVT on $(a,a+h)$ to get $c\in (a,a+h)$ with

Likewise MVT on $(a-h,a)$ gives $d\in (a-h,a)$ with

Meaning for this specific $h$ our estimate for $f^{″}(a)$ is

Note that $d<a<c$ and $|c-d|<h$, the right-hand side is essentially a central difference estimate for $f^{″}(a)$. We can prove this using the mean value theorem on $(d,c)$ to get a $c^{\prime }\in (d,c)$ with

Recall our choice of $\delta$ ensures that $|x-c|<\delta$ implies $|f^{″}(x)-f^{″}(a)|<𝜖$, setting $x=c^{\prime }$ (note $|c^{\prime }-a|<\delta$) gives (putting everything together)