Let $f \colon [x_0,x_1] \to \mathbb{R}$ be smooth and $|f^{\prime \prime}(x)|\leq L$ for some $L \in \mathbb{R}^+$. I want to have an estimation of the form
$$ \left| f^\prime(x) - \frac{f(x_1)-f(x_0)}{x_1-x_0} \right| \leq K(x_0,x_1,L) \,. $$
Any idea for a suitable $K$? Or do I need more assumptions for that?
Let $f: [x_0, x_1] \to \mathbb{R}$ be smooth, and $|f^{\prime \prime}(x)|\leq L$ for some $L \in \mathbb{R}^+$. We want to estimate the difference:
$\bigg| f^\prime(x) - \frac{f(x_1)-f(x_0)}{x_1-x_0} \bigg|$.
First, let's apply the mean value theorem to f on the interval $[x_0, x_1]$. There exists a point $c_1 \in (x_0, x_1)$ such that:
$f^\prime(c_1) = \frac{f(x_1) - f(x_0)}{x_1 - x_0}$.
Now, let's apply the mean value theorem to $f'$ on the interval $[x, c_1]$, where $x \in [x_0, x_1]$. There exists a point $c_2 \in (x, c_1)$ such that:
$f^{\prime\prime}(c_2) = \frac{f^\prime(c_1) - f^\prime(x)}{c_1 - x}$.
Since $|f^{\prime\prime}(x)| \leq L$, we have:
$\bigg|\frac{f^\prime(c_1) - f^\prime(x)}{c_1 - x}\bigg| = |f^{\prime\prime}(c_2)| \leq L$.
Multiplying both sides by $|c_1 - x|$, we get:
$|f^\prime(c_1) - f^\prime(x)| \leq L|c_1 - x|$.
Substituting the expression for $f^\prime(c_1)$, we have:
$\bigg| \frac{f(x_1)-f(x_0)}{x_1-x_0} - f^\prime(x) \bigg| \leq L|c_1 - x|$.
In this case, the estimation for the difference is:
$\bigg| f^\prime(x) - \frac{f(x_1)-f(x_0)}{x_1-x_0} \bigg| \leq L(x_1-x_0)$.
You might have to check for some special cases (e.g. $c_1=x_0$). Does that make sense, I'm fairly new to Math.