Is the derivative map a surjection of differentiability classes?

Question

It is clear to me from basic calculus that differentiability is a stronger condition than continuity, hence $C^{1}\subset C^{0}$.

But, there is something that has been bothering me recently. I stumbled across the same quandary a few years back, but I can’t seem to remember how I reconciled it…

For simplicity, assume $X$ is an interval and consider the differential map $D:C^{1}(X)\rightarrow C^{0}(X)$.

Clearly D is a function since by definition, given $f\in C^{1}(X)$, it’s derivative exists and is continuous, and also any pair of continuous functions that are the derivative of the same continuously differentiable function, must be equal.

Here’s the thing that is tripping me up…

By the fundamental theorem of calculus, given any continuous function $g\in C^{0}(X)$, shouldn’t we be able to construct a continuously differentiable function in the form of its antiderivative $f(x)=\int_{a}^{x}g(u)\,du$ on any subinterval $[a,x]\subset X$ since a continuous function is Riemann integrable?

But, this would seem to imply that $D$ Is surjective, which is certainly not the case in general. Otherwise, we would have $C^{1}(X)=C^{0}(X)$, which is not true.

So, where is this line of reasoning incorrect?

José Carlos Santos · Accepted Answer · 2021-08-22T17:29:44.733

1

The fact that there is a surjective map $f$ from a subset $A$ of a set $B$ into $B$ does not imply that $B=A$. For instance, take $f\colon\Bbb N\longrightarrow\Bbb Z$ such that$$f(n)=\begin{cases}\frac{n-1}2&\text{ if $n$ is odd}\\-\frac n2&\text{ if $n$ is even.}\end{cases}$$Then $f$ is surjective, but $\Bbb N\ne\Bbb Z$.

edited Aug 22 '21 at 17:29

answered Aug 22 '21 at 17:07

José Carlos Santos

427,504

It does if we already know that B is a subset of A. – The Emerging Son Aug 22 '21 at 17:16
Then what's wrong with my example? Isn't $\Bbb Z$ a subset of $\Bbb R$? – José Carlos Santos Aug 22 '21 at 17:18
I apologize, I meant to say, that it does if we already know that A is a subset of B.If a map is a surjection, then the cardinality of the codomain is less than or equal to the cardinality of the domain. $A \subset B$ and $card(A)\geq card(B)$ implies $B \subset A$ and hence $A=B$ – The Emerging Son Aug 22 '21 at 17:21
I've edited my answer. What do you think now? – José Carlos Santos Aug 22 '21 at 17:27
Note that it is false that $\operatorname{card}(A)\geqslant\operatorname{card}(B)\implies B\subset A$. – José Carlos Santos Aug 22 '21 at 17:28
Ok I see what you’re saying. You’re correct. – The Emerging Son Aug 22 '21 at 17:29
If my answer was useful, perhaps that you could mark it as the accepted one. – José Carlos Santos Aug 22 '21 at 22:11

Is the derivative map a surjection of differentiability classes?

1 Answers1