What does $C^1([0,1])$ mean?

Question

I know this question has been asked a lot, but please bare with me as there is a nuance that has confused me.

I thought I understood the definition of $C^1([0,1])$. However, I recently asked a question here and the comments made me realise I may have a fundamental misunderstanding. I have always thought that $f \in C^1([0,1])$ meant that $f$ is continuous on $[0,1]$ and is continuously differentiable on $[0,1]$.

However, as noted to me in the comments on the referenced question, I've realised I don't understand what it means to say a function is continuously differentiable on a closed set. Which of the following does it mean:

1. $f$ is continuously differentiable on $(0,1)$ and the left and right derivatives of $f$ exist at $0$ and $1$, respectively, and the resulting derivative is continuous on $[0,1]$.
2. There is an open set $U$ containing $[0,1]$, over which $f$ is continuously differentiable.

Or, are the statements $(1)$ and $(2)$ equivalent? The reason I ask is that I always assumed the former definition was correct, and therefore I thought that for an arbitrary compact set, $T\subset \mathbb{R}^N$, you could say something similar like:

$$\textbf{(1*)} \quad f \in C^1(T) ~\text{ iff } ~f \text{ is continuously differentiable on Int}(T) \text{ and, for any}$$

$$\text{ sequence of points tending to }\partial T\text{ the limit of the}$$

$$\text{ derivative at those points is well defined and is a continuous$$ $$\text{ extension of the derivative on Int(}T) $$

with the statement being vacuously true for isolated points of $T$. However, I now realize this isn't standard, and the texts I have been referring to have been a little ambiguous on this point (for instance in page 22 of this text). An alternative notion for compact $T$ based on statement $2$ above could have been given by:

$$\textbf{(2*)} \quad f \in C^1(T) ~\text{ iff } ~\exists \text{ an open }T'\supset T, \text{ such that }f \in C^1(T')$$

Would $\textbf{(1*)}$ and $\textbf{(2*)}$ be equivalent (and if not, what is a counterexample)? It seems to me that $\textbf{(1*)}$ resembles the conditions of Whitney's Extension Theorem in some way, and $\textbf{(2*)}$ in some way resembles how differentiability is defined on a smooth manifold with boundary that is a subset of $\mathbb{R}^N$ (in the sense of John Lee's Introduction to Smooth Manifolds).

Answer 1

You are close, but not quite there (or perhaps you just took too many shortcuts in writing $1$ and $2$). Replace $1$ by this:

$1'$: $f$ is differentiable on $(0,1)$, the right derivative exists at $0$, the left derivative exists at $1$, and the resulting function $f' : [0,1] \to \mathbb R$ is continuous.

Also, replace $2$ by this:

$2'$: There is an open set $U \subset \mathbb R$ containing $[0,1]$ and an extension of $f : [0,1] \to \mathbb R$ to $f : U \to \mathbb R$, such that $f$ is differentiable on $U$ and $f' : [0,1] \to \mathbb R$ is continuous.

With those corrections, $1'$ and $2'$ are equivalent, and each of them is (by definition in the case of $1'$) equivalent to $f \in C^1[0,1]$.

Other equivalent variations are possible. For example in $2'$ the closing clause could be replaced by "$f' : U \to \mathbb R$ is continuous" in this position, and the statement would still be equivalent.

Answer 2

I think it is good learn from an example to understand such kind of concepts well.

Let $f(x) = x^2 \sin (1/x)$ if $x > 0$ and $f(0)=0$. Then $f$ is continuous, continuously differentiable on $x> 0$.

At $x = 0$, $$ \frac{f(h)-f(0)}{h} = h \sin (1/h) $$

implies the right derivative exists and is equal to $0$.

However, for $x>0$, $$ f^\prime(x) = 2x\sin (1/x) + x^2 \cos(1/x) \cdot (-1/x^2) = 2x \sin (1/x) - \cos (1/x) $$ implies that $f^\prime$ does not converge to $0$ as $x\to +0$.

I believe this example illustrates the reason why your definitions needs to be replaced as the other answer points out.