How is the Taylor formula applied here?

Question

Let $d\in\mathbb N$, $\Lambda\subseteq\mathbb R^d$ be connected and open and $A:\Lambda\times\Lambda\to\mathbb R$ be twice continuously differentiable. Let $x,x'\in\Lambda$ and $h,h'>0$. How can we show that $$\frac1{hh'}\left(A(x+he_i,x'+h'e_i)-A(x+he_i,x')-A(x,x'+h'e_i)+A(x,x')\right)\\=\int_0^1\int_0^1\frac{\partial^2A}{\partial x_i\partial x'_i}(x+\theta he_i,x'+\theta'h'e_i)\:{\rm d}\theta\:{\rm d}\theta',$$ where $e_i$ denotes the $i$th standard basis vector? Clearly, this is an application of Taylor's formula, but I don't see how it is appield.

Answer 1

I would actually not show this using Taylor expansion. It follows from (and generalizes) the mean value theorem.

First, let's define a new function $f : [0,h] \times [0,h'] \to \mathbb{R}$ by $$f(t,s) = A(x+te_i,x'+se_i)$$ Then, our problem is to prove $$\frac{1}{hh'}\left(f(h,h') - f(h,0) - f(0,h) + f(0,0)\right) = \int_0^1 \int_0^1 \frac{\partial^2}{\partial t \partial s} f(h\tau,h'\sigma)\;d\tau d\sigma$$ By changing variables in the integral, this becomes $$f(h,h') - f(h,0) - f(0,h') + f(0,0) = \int_0^{h'} \int_0^h \frac{\partial^2}{\partial t \partial s} f(\tau,\sigma)\;d\tau d\sigma$$ Now, define a new function $$g(t) = f(t,h') - f(t,0)$$ Then, by the Fundamental Theorem of Calculus, $$g(t) = \int_0^{h'} \frac{\partial}{\partial_s}f(t,\sigma)\;d\sigma$$ and $$f(h,h') - f(h,0) - f(0,h') + f(0,0) = g(h) - g(0) = \int_0^h \frac{\partial}{\partial t} g(\tau)\;d\tau$$

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From the previous work, it would suffice to show that $$\frac{\partial}{\partial t} \int_0^{h'} \frac{\partial}{\partial s} f(t,\sigma)\;d\sigma = \int_0^{h'} \frac{\partial^2}{\partial t\partial s} f(t,\sigma)\;d\sigma.$$

This follows from the fact that (since $f$ is $C^2$), the difference quotients $$\frac{\frac{\partial f}{\partial s}(t+k,\sigma) - \frac{\partial f}{\partial s}(t,\sigma)}{k}$$ converge to $\frac{\partial^2}{\partial t \partial s} f(t,\sigma)$ uniformly in $\sigma$ for all $\sigma \in [0,h']$. This is a particular case of the Leibniz integral rule.