Expression for symmetric double derivative from first principles I want to start by stating that there have been multiple threads on SE all of which are possible duplicates. I am asking about the same issue because I still have something concrete to ask regarding this question. Duplicates are in 1, 2, 3, 4, 5, and 6. There may be more but these are the ones I found.
Question (From Stephen Abbott's Understanding Analysis): If $f$ is twice differentiable on an open interval containing $a$ and $f''$ is continuous at $a$, show
$$ \lim_{h \to 0} \frac{f(a+h)−2f(a)+f(a−h)}{h^2} = f''(a)$$
My attempt: While it is possible to use L'Hospital's theorem in this question to provide a short proof, I tried to find a proof starting from the basic principles. I had something like in here, where the expression for the double derivative in terms of limits turns out to be
$$f''(a) = \lim_{h\to0} \frac{  \lim_{h_1\to0}\frac{ f(a+h+h_1) - f(a+h)}{h_1} - \lim_{h_2\to0}\frac{ f(a+h_2) - f(a)}{h_2}  }{h}$$
which basically is a nested double limit on $f$. As we are given that the limits in the numerator exists, it's fine to change the expression to something like this
$$f''(a) = \lim_{h\to0} \frac{  \lim_{k\to0}\frac{ f(a+h+k) - f(a+h)}{k} - \lim_{k\to0}\frac{ f(a+k) - f(a)}{k}  }{h}$$
which further simplifies to
$$f''(a) = \lim_{h\to0} \lim_{k\to0} \frac{ f(a+h+k) - f(a+h) - f(a+k) + f(a)}{kh}$$
We now have an expression with two limits, each one being controlled by a different variable. This is not something that has been addressed in the text, atleast thus far. I wish to get a proof from this point onwards. Is that possible? Specifically, I wish to

*

*complete the proof starting from the above equation without using L'Hospital's theorem or Taylor series expansion. All we know at this point is the Mean Value Theorem, and it's variants like the generalized Mean Value Theorem, Rolle's Theorem etc. While L'Hospital's has been presented, I just do not wish to use it. I want to know if there is a proof starting from first principles, specifically starting from the above equation and concluding that it is equal to the double derivative if the double derivative is known to exist.


*know the reason as to why the added clause " $f''$ is continuous at $a$  " was given in the problem, as it seems that all we need is to know that the double derivative exists (based on the proof that uses L'Hospital's theorem).
 A: Your expansion of $f''(a)$ as a double limit is indeed correct based on the definition of the derivative. However, that definition requires $f'$ to be continuous at $a$. It turns out that the theorem you ask about actually holds under weaker conditions.
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$
Theorem 1. Take any open subset $D$ of $ℝ$ and any function $f : D→ℝ$ and any $x∈D$ such that $f(x+h) ∈ f(x) + c_1·h + c_2·h^2 + o(h^2)$ as $h → 0$. Then $\lfrac{f(x+h)-2·f(x)+f(x-h)}{h^2} → c_2$ as $h → 0$.
Proof. As given in this post, which is the second answer to one of the posts you cited.
Remark. This theorem is more general because it does not require $f$ to be differentiable at $x$, nor does it require $f$ to be continuous in some open interval around $x$. For example, let $f : ℝ→ℝ$ such that $f(0) = 0$ and $f(x) = \lfrac{x}{\lfloor 1/x \rfloor}$ for every $x∈ℝ_{≠0}$. Then $f(h) = \lfrac{h}{1/h+O(1)}$ $∈ h^2·(1+O(h))$ $⊆ h^2+o(h^2)$ as $h → 0$, so the theorem applies to $f$.
Theorem 2. Take any open subset $D$ of $ℝ$ and any function $f : D→ℝ$ and any $x∈D$ such that $f$ is differentiable on $D$ and $f'$ is differentiable at $x$. Then there are constants $c_1,c_2$ such that $f(x+h) ∈ f(x) + c_1·h + c_2·h^2 + o(h^2)$ as $h → 0$.
Proof. Essentially the same as in that same linked post.
Remark. Together with Theorem 1, this gives a much stronger theorem than the one you cited from Abbott, since we do not need $f$ to be twice differentiable on an open interval around $x$, much less that its second derivative is continuous at $x$!
