How to take "limits" of operators?

Question

Fix $h\ne0\in\mathbb{R}$. Define $\Delta_h\in\text{End}(\mathbb{R})$ by writing $\Delta_h(f):=f((-)+h)$ for all $f\in\text{Hom}_\text{Set}(\mathbb{R},\mathbb{R})$. Denote identity element of $\text{End}(\mathbb{R})$ by $I$. Then, for all functions $f$ from $\mathbb{R}$ to itself, we have $$(\Delta_h+I)(f)=f((-)+h).$$ Clearly $(\Delta_h+I)^k(f)=f((-)+kh)=\Delta_{kh}(f)$, for all $k\in\mathbb{N}$. Since $\text{End}(\mathbb{R})$ is a ring, the binomial formula holds and so $$\Delta_h^n=(\Delta_h+I-I)^n=\sum_{k=0}^n\binom{n}{k}(-I)^{n-k}(\Delta_h+I)^k.$$ It follows that, for all $n\in\mathbb{N}$ and for all $f\in\text{Hom}_\text{Set}(\mathbb{R},\mathbb{R})$, the identity $$\Delta^n_h(f)=\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}\Delta_{kh}(f),$$ holds. Now suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is $n$-times differentiable. I want to make sense of the limit $$\lim_{h\rightarrow 0}\frac{\Delta_h^n(f)}{h^n};$$ in fact, I want to somehow show that an appropriate notion of a limit exists where the above is equal to $\frac{d^n}{dx^n}f$. Is such a thing possible? Where might I find the required definitions, properties, constructions, etc.? I know nothing beyond first year analysis, so any help would be appreciated.

Answer 1

I'll assume $f \in C^{m}(\mathbb{R})$. The first kind of convergence to investigate is pointwise convergence. Fix $x \in \mathbb{R}$. Note that by the mean value theorem, for any $g \in C^1(\mathbb{R})$, $(\frac{\Delta_h}{h}g)(x) = g'(x + \theta(h)h)$ for some $\theta(h) \in (0, 1)$. Also note that by linearity of differentiation, $\frac{d}{dx}$ and $\frac{\Delta_h}{h}$ commute, i.e. $\frac{d}{dx}\frac{\Delta_h}{h} = \frac{\Delta_h}{h}\frac{d}{dx}$. By the mean value theorem, \begin{align} g_h(x) &:= (\frac{\Delta_h^m}{h^m}f)(x) \\ &= (\frac{d}{dx}\frac{\Delta_h^{m - 1}}{h^{m - 1}}f)(x + \theta_1(h)h) \\ &= (\frac{\Delta_h^{m - 1}}{h^{m - 1}}\frac{d}{dx}f)(x + \theta_1(h)h) \\ &= (\frac{\Delta_h^{m - 1}}{h^{m - 1}}f')(x + \theta_1(h)h), \end{align} where $\theta_1(h) \in (0, 1)$. Iterating this gives \begin{align} g_h(x) &= (\frac{\Delta_h^{m - 1}}{h^{m - 1}}f')(x + \theta_1(h)h) \\ &= (\frac{\Delta_h^{m - 2}}{h^{m - 2}}f'')(x + \theta_1(h)h + \theta_2(h)h) \\ &= \dots \\ &= f^{(m)}(x + \theta_1(h)h + \dots + \theta_m(h)h). \end{align}

Note that $|\theta_1(h)h + \dots + \theta_m(h)h| \leq m|h| \to 0$ as $h \to 0$. Since $f^{(m)}$ is continuous at $x$, it follows that $g_h(x) \to f^{(m)}(x)$ as $h \to 0$. This kind of analysis is a basis for a proof of commutativity of mixed partial derivatives.

A stronger kind of convergence to investigate is whether $g_h \to f^{(m)}$ uniformly on compact subsets of $\mathbb{R}$ as $h \to 0$. This is true. To prove this, it suffices to prove uniform convergence on every closed ball $\overline{B_R(0)}$ centered at $0$. To prove it, adapt the analysis for the pointwise case, and use the fact that a continuous function on a compact set it uniformly continuous, i.e. has a modulus of continuity. By induction, it follows that if $f \in C^{\infty}(\mathbb{R})$, then for every $k \geq 0$, $g^{(k)}_h \to f^{(m + k)}$ uniformly on compact subsets of $\mathbb{R}$ as $h \to 0$.