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My lecturer for my Analysis class provided the following definition for differentiability at a point:

Let $(V, \|\cdot\|_V)$ and $(W, \|\cdot \|_W)$ be two finite-dimensional vector spaces. Let $f : D \to W$ be a function and let $a \in D$. We say that $f$ is differentiable in $a$ if there exists a linear map $L_a: V \to W$ such that, if we define the error function $\mathsf{Err}_a : D \to W$ through $$ \mathsf{Err}_a(x) := f(x) - f(a) - L_a (x - a) $$ it holds that $$ \lim_{x \to a} \frac{\|\mathsf{Err}_a(x)\|_W}{\|x - a\|_V} = 0. $$ And $L_a$ is the derivative of $f$ in $a$, denoted alternatively as $(Df)_a$.

This surely differs from the definition of differentiability in a point that I have got to known: Checking whether the limit $$\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$$ exists.

So far I have not seen first mentioned definition for differentiability in literature, nor do I understand the intuitive idea behind it. Any help on literary sources or explanation of any other kind is greatly appreciated.

Kevo Sender
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  • The definition you knew works only for a function of one variable. If the domain of the function is a vector space, then that formula makes no sense, because you can't divide by a vector. The definition given by your instructor is a rigorous way to say that the function can be approximated by a linear function. For a function of one variable, it is equivalent to the formula you already know. – Deane Jan 18 '21 at 20:36
  • Take a look at my previous answers here and here for the motivation and heuristics of the definition, and how it relates to the single variable case. – peek-a-boo Jan 18 '21 at 22:39

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You can find this definition in Rudin's book Principles of Mathematical Analysis. The idea is that given a function $f:\mathbb{R}^n\to \mathbb{R}^m$ (dropping the abstract vector space notion) if it's differentiable at a point $p\in \mathbb{R}^n$ then the definitin you have given say's it can be well approximated by a linear map (ie a matrix). In the usual context we first learn where $f:\mathbb{R}\to \mathbb{R}$, let's see how we can obtain the new notion. Suppose $f$ is differentiable at $p$, then like you said the limit $$f'(p)=\lim_{t\to p}\frac{f(t)-f(p)}{t-p}$$ exists. This means, the difference quotient $$\varphi(t)=\frac{f(t)-f(p)}{t-p} \qquad (t\neq p)$$ is continuous at $t=p$. Which means for every $\epsilon>0$ there is a $\delta>0$ so that $$\Big | \frac{f(t)-f(p)}{t-p}-f'(p)\Big |< \epsilon \qquad \text{for} \quad |t-p|<\delta$$ We can interpret this as there existing a function say $E(p,t)$ so that $$f(t)-f(p)-f'(p)(t-p)=E(t,p)$$ which must satisfy $\lim_{t\to p}\frac{|E(p,t)|}{|t-p|}=0$.

Now, in the context of $\mathbb{R}$ what are the linear maps? They are in correspondence with $\mathbb{R}$ itself, ie $1\times 1$ matricies. So all along, the derivative that you knew and loved was always a linear map on a one dimensional vector space and to re-iterate it's the linear map that best approximates the given function about a base point.

Hope this helps!

Nick Castillo
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If $f : D \to \mathbb R$ for some open subset $D$ of $\mathbb R$, $a \in D$, $$\lim_{x \to a} \frac{f(x)-f(a)}{x-a} = f'(a)$$ and you consider the linear map $L_a : \mathbb R \to \mathbb R$ given by $L_a(t) = f'(a)t$, then of course $$\lim_{x \to a} \frac{|f(x)-f(a)-L_a(x-a)|}{|x-a|} = \lim_{x \to a} \bigg| \frac{f(x)-f(a)}{x-a}-f'(a) \bigg| = 0.$$ The first one definition of derivative is called the Fréchet derivative of $f$ at $a$, and it generalizes the classical notion of derivative.

azif00
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