Bounding the difference between a function and a line connecting its endpoints by Taylor's Theorem

Question

I am unsure how the acquire the following result of the Lemma from using Taylor's Theorem. How exactly would I go about proving this?

Thank you in advance.

Answer 1

Without loss of generality one can assume that $s=0$ and $t=1$. Otherwise apply the following to the function $f: [0, 1] \to \Bbb R^d$, defined by $$ f(x) = F(s + x(t-s)) \quad \Longrightarrow \quad f''(u) = (t-s)^2 F''(s + x(t-s)) \, . $$

Let's first consider the case $d=1$, i.e. a function $F: [0, 1] \to \Bbb R$. The difference $g(x) = F(x) - L(x)$ satisfies

• $g(0) = g(1) = 0$, because $L$ interpolates $F$ at $x=0, 1$.
• $g''(x) = F''(x)$, because $L$ is linear.

So we have to show that $$ \tag{*} |g(x)| \le \frac 18 \max_{0 \le \xi \le 1} |g''(\xi)| $$ for $0 \le x \le 1$.

(The "trick" is now to apply Taylor's theorem at a point where $g$ attains a maximum, and the first derivative vanishes.)

Let $a \in [0, 1]$ be the point where $|g|$ attains its maximum. If $a = 0$ or $a=1$ then $g$ is identically zero and we are done.

Otherwise $g'(a)=0$, and we can apply Taylor's theorem (with $x_1 = a, x_2 = 0$): $$ 0 = g(0) = g(a) + (0 - a)g'(a) + \frac 12 (0-a)^2 g''(\xi) = g(a) + \frac 12 a^2 g''(\xi) $$ for some $\xi$ in the interval $(0, a)$. If $0 < a \le \frac 12$ then it follows that $$ |g(a)| \le \frac 12 |a|^2 |g''(\xi)| \le \frac 18 \max_{0 \le \xi \le 1} |g''(\xi)| \, . $$ If $\frac 12 \le a < 1$ then we apply Taylor's theorem with $x_1 = a, x_2 = 1$: $$ 1 = g(1) = g(a) + (1 - a)g'(a) + \frac 12 (1-a)^2 g''(\xi) = g(a) + \frac 12 (1-a)^2 g''(\xi) $$ for some $\xi$ in the interval $(a, 1)$ and the conclusion is the same. This proves $(*)$ in all cases.

For $d > 1$ the above estimate can be applied to each component of $F - L$.