Finding an upper limit on a continuous function

Question

I'm tasked with finding a smallest constant $c$ so that for a function $f\in C^4([0,1])$ and $x,y\in[0,1]$: $$\left\vert\frac{f(x)-f(y)}{x-y}-\frac{f'(x)+f'(y)}{2}\right\vert\le c\vert x-y\vert ^2\max_\limits{\xi\in[0,1]}\vert f'''(\xi)\vert$$ Since this is an excercise I was given on the Topic of Taylor series, my first thought was making use of Taylor's theorem and somehow set the upper bound using the Lagrange remainder but I'm not quite sure that's the way to go. Alternatively I thought about using the fact, that $f(x)$ is lipschitz continuous. Any nudge in the right direction would be greatly appreciated. Thanks in advance.

Answer 1

For $x \ne y$, there exists $\xi_1, \xi_2$ between $x,y$ such that:

\begin{align} f(x)&=f(y)+f'(y)(x-y)+\frac {f''(y)}2(x-y)^2+\frac{f'''(\xi_1)}6(x-y)^3\\ f'(x)&=f'(y)+f''(y)(x-y)+\frac {f'''(\xi_2)}2(x-y)^2\end{align}

Substituting:

\begin{align}&\quad\left\vert\frac{f(x)-f(y)}{x-y}-\frac{f'(x)+f'(y)}{2}\right\vert \\&=\left\vert\frac{f'(y)(x-y)+\frac {f''(y)}2(x-y)^2+\frac{f'''(\xi_1)}6(x-y)^3}{x-y}-\frac{2f'(y)+f''(y)(x-y)+\frac {f'''(\xi_2)}2(x-y)^2}{2}\right\vert\\ \\&=\left\vert f'(y)+\frac {f''(y)}2(x-y)+\frac{f'''(\xi_1)}6(x-y)^2-f'(y)-\frac {f''(y)}2(x-y)-\frac {f'''(\xi_2)}4(x-y)^2\right\vert \\&=\left\vert \frac{f'''(\xi_1)}6(x-y)^2-\frac {f'''(\xi_2)}4(x-y)^2\right\vert \\&\le\left(\left\vert \frac{f'''(\xi_1)}6\right\vert+\left\vert\frac {f'''(\xi_2)}4\right\vert\right)(x-y)^2 \end{align}

Answer 2

In fact the smallest constant for which the inequality holds is $c = \frac{1}{12}$.

For the proof, consider the map

$$\varphi(t) =f(x) - f(t) +\frac{t-x}{2}\left(f^\prime(t) + f^\prime(x)\right)-\frac{(t-x)^3}{12}A$$ where $A$ is such that $\varphi(y)=0$.

As $\varphi(x)=0$ Rolle's theorem ensures the existence of $c_1 \in (x,y)$ with $\varphi^\prime(c_1)=0$, i.e. $$0 = - f^\prime(c_1) +\frac{f^\prime(c_1) + f^\prime(x)}{2}+\frac{c_1-x}{2}f^{\prime\prime}(c_1)+\frac{(c_1-x)^2}{4}A.$$

Applying Taylor's theorem to $f^\prime$ it exists $c \in (x,c_1)$ with

$$f^\prime(x) - f^\prime(c_1)=(x-c_1)f^{\prime\prime}(c_1) + \frac{(x-c_1)^2}{2} f^{\prime \prime \prime}(c).$$

The two equalities above imply $A=f^{\prime \prime \prime}(c)$. Using $\varphi(y)=0$ in the definition of $\varphi$ we get for $x \neq y$

$$\frac{f(x)-f(y)}{x-y} - \frac{f^\prime(x) + f^\prime(x)}{2}= -\frac{(x-y)^2}{12}f^{\prime \prime \prime}(c)$$ and therefore the desired inequality

$$\left\vert\frac{f(x)-f(y)}{x-y}-\frac{f^{\prime}(x)+f^{\prime}(y)}{2}\right\vert \le \frac{\vert x-y\vert ^2}{12}\max_\limits{\xi\in[0,1]}\vert f^{\prime \prime \prime}(\xi)\vert.$$

For $f(x) = x^3$, the inequality becomes an equality for all $x \neq y$, proving that $c=\frac{1}{12}$ is the smallest constant for which the inequality is valid.