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Compare it to the definition of continuity of a function $f$, if $x \in B(a,\delta)$, then $f(x) \in B(f(a),\varepsilon)$. This definition doesn't need $f$ to be defined around a neighbourhood around $a$, in fact $a$ can be an isolated point in the domain of $f$.

(I just checked), when we talk about the (single-var case) derivative of $f$ at $x_0$, we do not need $f$ to be defined in a neighbourhood around $x_0$.

$$ \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0} = f'(x_0) \iff |x-x_0|< \delta \implies |\frac{f(x)-f(x_0)}{x-x_0} - f'(x_0)| < \varepsilon. $$

We just need that if $B(x_0,\delta)$ contains points, then the inequality above holds.

However, in the $f:\mathbb{R}^p \to \mathbb{R}$ case, $f$ differentiable at $a \iff \exists l(x)$ linear function st. $f(x) = f(a) + l(x-a) + \varepsilon(x)\cdot|x-a|$, where $\varepsilon(x) = 0$ if $x \to a$. Why do we need $f$ to be defined in some neighbourhood here? The equivalent condition for differentiability of $f$ at $a$ is:

$$ \lim_{x\to a} \frac{f(x)-f(a)-l(x-a)}{|x-a|} = 0 \iff \forall x \in B(a,\delta), |\frac{f(x)-f(a)-l(x-a)}{|x-a|}| < \kappa. $$

I don't understand why we need $f$ to be defined in some neighbourhood around $a$.

Jean Marie
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AyamGorengPedes
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  • In one dimension, if you allow an isolated point, then any value is "the" derivative there, since the set where $0<|x-x_0|<\delta$ can be empty. In multiple dimensions, you might have a domain narrowing to a line even if $a$ is a limit point, so the directional derivative is known in one direction but could be anything in another. – aschepler Mar 07 '22 at 17:50
  • @aschepler can you please expand on that? I didn't quite catch the meaning. I have an inkling, but I think it's better if you explain it further for me. – AyamGorengPedes Mar 07 '22 at 18:00
  • Honestly, I think the definition of continuity you describe is already the problem. It's the appropriate definition in advanced topology courses, but at the level of fundamental calculus it just causes confusions (like yours) when we permit limits to exist when the function isn't defined in a neighborhood of $a$. – Greg Martin Mar 07 '22 at 18:03
  • You are wrong in your definition of continuity, $f(x)$ need to exist, so you need $B(a, \delta)$ to be a subset of the domain of $f.$ Of course, you can change to $x \in B(a, \delta) \cap \mathsf{Dom}(f)...$ and you would be fine. – William M. Mar 07 '22 at 19:18
  • The definition of derivative also admits "relative to a domain" part. Namely, $f(x+h) = f(x) + L \cdot h + o(h)$ as $h \to 0$ for all $x, x+h \in \mathsf{Dom}(f).$ Of course, if $x$ is in the interior of $\mathsf{Dom}(f)$ you aren't adding anything new and when $x$ is frontier point, you can have something interesting. I am unsure if this is used, though. – William M. Mar 07 '22 at 19:21

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I think part of your confusion is that continuity depends both on the function and the domain. For instance, the function $$ f(x)= \begin{cases} 0, x\in \mathbb{Q}\\ 1, x\in \mathbb{R}\setminus\mathbb{Q} \end{cases} $$ isn't continuous on $\mathbb{R}$, but is continuous if you restrict to either of the rationals or irrationals - i.e. remove all of the discontinuities from your domain.

As for the derivative, a quick scan of my bookshelf (Baby Rudin, Bartle and Sherbert, Abbot) shows that it is common to require that $f$ is defined in an open ball containing $x$ for computing $f'(x)$ in the 1 dimensional case. Requiring that the function is defined on an open ball forces the object that you get out as the (total) derivative is a linear approximation in all directions from the point $x$. Allowing derivatives for functions not defined on an interval is basically just restricting the possible directions for your derivative to approximate in - with the derivative for an isolated point being 0, since there are no "directions".

Also note that if you don't require definition on an open ball, then you can take derivatives at discontinuities, which will break the nice formulas we learn in calculus for computing derivatives of combinations of functions in the larger space.

Jacob Smith
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