Is there a sense in which a function converges to its total derivative?

Question

In short: Is there a way to rigorously define the total derivative (of $F$ at $x$) as a function $dF_x$ which is (1) linear and (2) a usual topological limit of some function $\mu:X\to Y$, where $X$ and $Y$ are topological spaces?

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Motivation: Let $F:\mathbb R^n\to\mathbb R^m$. The total derivative of $F$ at $x\in\mathbb R^n$ is defined as the linear map $dF_x:\mathbb R^n\to\mathbb R^m$ such that

$$\lim_{v\to0}\frac{\|F(x+v)-F(x)-(dF_x)v\|}{\|v\|}=0,$$

if such a linear map exists. This definition makes sense, but it is a bit indirect. The idea is that $F$ looks more and more like $dF_x$ near $x$, so it seems that the more natural definition should be something like $\lim_{v\to0} F(x+v)-F(x)=(dF_x)v$. The following points are suggestive:

• Normally when $\lim_{x\to c}\|f(x)-L\|=0$, this can be formulated as $\lim_{x\to c}f(x)=L$.
• The division by $\|v\|$ in the above definition is highly reminiscent of the operator norm.

I don't think there is any trouble extending the notion of operator norm to general continuous functions (although it may be infinite), via $\|F\|_\text{op}=\sup_{0\neq v\in\operatorname{dom}(F)}\frac{\|F(v)\|}{\|v\|}.$ Then, defining $\hat F_x(v)=F(x+v)-F(x)$ , we might be able to manipulate the first definition into something like

$$\lim_{v\to0}\hat F_x(v)=dF_x,$$

where it is understood that the limit is with respect to the operator norm. Or rather, as we look at $F$ restricted to smaller neighborhoods of $x$, the operator norm of $\hat F_x-dF_x$ restricted to this neighborhood approaches $0$. $F$ would approach many functions in this sense (including itself, for example), but the derivative would be the only linear function it approaches in this sense.

However, when I try making this rigorous, I run into some challenges. Topologically, the limit of a function $F$ near $x$ must be a value in the range of $F$, not a function with the same domain and range of $F$; so this formulation will require speaking of the limit of some function besides $F$, probably mapping into function spaces or spaces of germs of functions near $x$. This is where my knowledge gets more limited and I hit a wall.

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Update: Here's my current progress. Let $C^0_x=\{f\in C^0(U,\mathbb R^m)\,|\,U\text{ is a neighborhood of }x\}$. Define function addition/subtraction on $C^0_x$ as usual function addition/subtraction restricted to the intersection of both functions' domains. Topologize this space using the metric $d(f,g)=\sup_{t\in\operatorname{dom}(f-g)\setminus\{x\}}\frac{\|f(t)-g(t)\|}{\|t-x\|}$. This metric can take infinite values, but this seems ok according to this. Finally, let $\mu_{x,F}:(0,\infty)\to C^0_x$ be defined by $\mu_{x,F}(\delta)=F\big|_{B_\delta(x)}$. Then, perhaps, we can define $dF_x$ (if it exists) as the unique linear map defined on all of $\mathbb R^n$ such that

$$\lim_{\delta\to0}\mu_{x,F}(\delta)=F(x)+dF_x.$$

Does this work? Is there a more elegant or standard way to do this?

Answer 1

Let $C_x^0 (\mathbb R^m)$ be the set of equivalence classes of continuous functions $f:U\to \mathbb R^m$ defined on a neighborhood $U$ of $x$, where two functions are equivalent if they agree on a neighborhood of $x$ (possibly smaller than the intersection of the two domains, in order to have an equivalence relation).

The algebra structure under pointwise operations is compatible with this equivalence relation, making $C_x^0(\mathbb R^m)$ an algebra over $\mathbb R$.

Linear maps $L(\mathbb R^n,\mathbb R^m)$ are a subspace of $C_x^0(\mathbb R^m)$, with the embedding $A\mapsto[y\in\mathbb R^n \mapsto A(y-x)]$.

Let $[f]\in C_x^0(\mathbb R^n)$ and define : $$\| [f]\| = \limsup_{y\to x}\frac{\|f(y)\|}{\|y-x\|} = \lim_{\epsilon \to 0} \sup_{y\in B_\epsilon(x)\backslash\{0\}}\frac{\|f(y)\|}{\|y-x\|} $$

This is well defined on the equivalence classes, but may be infinite, so it does define a norm. To remedy this, we take : $$d([f],[g]) = \min( 1,\|[f-g]\|)$$

This is well defined on equivalence classes and finite, but is still not a distance function, because, if $f$ is $C^1$ and has $f(x) = 0$ and $\text df(x) = 0$, then $d(f,0) =0$. But this allows us to characterize differentiable functions and define there derivative !

Explicitely, a function $f$ is differentiable at $x$ if $\inf_{u \in L(\mathbb R^n,\mathbb R^m)} d([f],[f(x) + u]) =0$ and its derivative is the $u \in L(\mathbb R^n,\mathbb R^m)$ for which this minimum is reached.