Estimating probability mass given to an interval via First Order Taylor Approximation $F(x)$ is a cumulative distribution function (CDF) on a symmetric continuous interval, e.g., $[-\alpha, \alpha]$ for some $\alpha \in \mathbb{R}$. We can also assume the distribution is continuous & symmetric. However, the CDF is not necessarily twice differentiable (i.e., the PDF may not be differentiable). For example, the underlying distribution can be (and in my case it is) a Laplace distribution.
Let $\hat{x} \in (-\alpha, \alpha)$ be an arbitrary point and $x$ be such that $x \in (\hat{x}, \hat{x} + \beta)$ for some $\beta > 0$.
I have the following claim, that intuitively makes sense, but hard to formally prove:
The approximation
$$ \begin{align}
\frac{F(\hat{x} + \beta) - F(\hat{x})}{\beta}(x - \hat{x})&   \stackrel{(1)}{\approx} F'(\hat{x}) (x - \hat{x}) \\
& \stackrel{(2)}{\approx} F(x) - F(\hat{x})
\end{align}
  $$
can be as tight as desired by decreasing $\beta$. Formally, for any small $\epsilon > 0$ there exists a $\beta(\epsilon)$ such that for all $0 < \beta < \beta(\epsilon)$ we can achieve:
$$\left| \left( \frac{F(\hat{x} + \beta) - F(\hat{x})}{\beta}(x - \hat{x}) \right) - \left( F(x) - F(\hat{x}) \right) \right|  < \epsilon.$$
Notice that at step $(1)$ we use the definition of a derivative and at step $(2)$ we use a First Order Taylor Approximation.
Attempt: I tried using the Taylor's Inequality for the error in first order Taylor Approximation. However, that requires $F(x)$ to be twice differentiable. So I am not sure how this convergence can be proved.
Note: In another Math.SE discussion the claim above was mentioned, and I am suspected this may not be valid if the PDF is not differentiable. However, the author claims this is only due the compactness of $[-\alpha, \alpha]$ and differentiability of $F(x)$. So I wanted to have a new discussion about this more general result because i) this is a long discussion for a follow-up in another post, and ii) this seems to be very useful result if we can prove it here.
 A: Let $f := F'$ be the PDF. I'm also assuming that $\beta$ is small enough that $[\hat{x},\hat{x} + \beta] \subset [-\alpha,\alpha]$. By compactness of $[-\alpha,\alpha]$, $f$ is uniformly continuous on $[-\alpha,\alpha]$. That is, there exists a non-decreasing, continuous function $h:[0,\infty)\to[0,\infty)$ depending on $F$ such that $h(0) = 0$ and,
$$|f(a) - f(b)| \leq h(|b-a|) \text{ when } a,b \in [-\alpha,\alpha].$$
Thus,
$$\sup_{y \in [\hat{x},x]} f(y) - \inf_{z \in [\hat{x},x]} f(z) := \overline{f} - \underline{f} \leq h(\beta).$$
But then,
\begin{align}
\underline{f}(x - \hat{x}) &\leq F(x) - F(\hat{x}) &\leq \overline{f}(x - \hat{x})\\
\underline{f}(x - \hat{x}) &\leq f(\hat{x})(x - \hat{x}) &\leq \overline{f}(x - \hat{x}).
\end{align}
Thus,
$$\left|F(x) - F(\hat{x}) - f(\hat{x})(x - \hat{x})\right| \leq \left|(\overline{f} - \underline{f})(x - \hat{x})\right| \leq h(\beta)(x - \hat{x}) \leq \beta h(\beta).$$
In conclusion, $\overset{(2)}{\approx}$ in the statement of your problem does hold asymptotically as $\beta \to 0$, but the specific estimate given by Taylor's theorem may fail when $f$ is not differentiable. Instead you get an estimate involving $h$, which exists because we are working on a compact space.
A: Given $\varepsilon,$ one can find $\delta$ so small that for all $\widehat x + \beta\in(\widehat x - \delta,\widehat x+\delta)\cap (\widehat x -1,\widehat x + 1),$
$$
\left| \frac{F(\widehat x+\beta) - F(\widehat x\,)} \beta - F'(\widehat x\,) \right| < \varepsilon/2.
$$
If $x$ is between $\widehat x$ and $\widehat x+\beta,$ then because of the word "all" above, it follows that
$$
\left| \frac{F(x) - F(\widehat x\,)}{x-\widehat x} - F'(\widehat x\,) \right|< \varepsilon/2.
$$
From the triangle inequality it follows that
$$
\left| \frac{F(\widehat x + \beta) - F(\widehat x\,)} \beta - \frac{F(x) - F(\widehat x\,)}{x- \widehat x} \right| < \varepsilon.
$$
Since $|\widehat x - x|<1,$ multiplying both sides by $|\widehat x - x|$ leaves you with something${}<\varepsilon$ on the right side.
