Do progressively more differentiable functions (once differentiable, then $C^1$, then twice differentiable, etc.) form an infinite nested hierarchy?

Question

I know that the Baire category theorem can be used to show that

The space of functions $\mathbb{R} \to \mathbb{R}$ differentiable at at least one point is a meager subset of the space of all continuous function $\mathbb{R} \to \mathbb{R}$.
The space of functions $\mathbb{R} \to \mathbb{R}$ that are analytic is a meager subset of the space of all infinitely continuously differentiable (smooth) functions $\mathbb{R} \to \mathbb{R}$.

This suggests a nested hierarchy of meager subsets, with each class of functions with a strictly stronger differentiability property a meager subset of the previous class.

Question: Is this intuition correct? Do "progressively more strongly differentiable" functions form a hierarchy of meager subsets?

I believe it suffices (because of induction) to show the following, so I will rephrase the question to be more precise:

Question: Inside of the space of all differentiable functions $\mathbb{R} \to \mathbb{R}$, is the space of all functions whose derivative is continuous at at least one point (i.e. the space of all functions that are differentiable and continuously differentiable at at least one point) a meager subset?

It's not important to me either way whether "space of all differentiable functions $\mathbb{R} \to \mathbb{R}$" is interpreted to mean either:

"space of all functions $\mathbb{R} \to \mathbb{R}$ which are differentiable at every point in $\mathbb{R}$" or
"space of all functions $\mathbb{R} \to \mathbb{R}$ which are differentiable at at least one point of $\mathbb{R}$".

Obviously the result being true for the former implies the result also being true for the latter but not necessarily vice versa.

This has been asked elsewhere on the internet before but did not seem to receive a satisfactory answer there. I also was unable to find the answer to this question while searching this site, but I acknowledge it might be a duplicate.

The map $$C^{k+1}(\Bbb R) \mapsto \Bbb C \oplus C^{k}(\Bbb R), \quad f \mapsto (f(0), f' )$$ is a topological isomorphism if you give $C^k$ the topology of locally uniform convergence of all derivatives of order $≤k$. Note that the image of $C^{k+2}\subseteq C^{k+1}$ is $\Bbb C\oplus C^{k+1}\subseteq \Bbb C\oplus C^{k}$. So $C^{k+2}$ is meagre in $C^{k+1}$ iff $C^{k+1}$ is meagre in $C^{k}$. — s.harp, May 22 '21 at 11:23
My answer to How discontinuous can a derivative be? should clarify the type of question to ask and give some results of interest. Unfortunately, I have to leave right this moment to pick up something (groceries due to be picked up right now at the grocery store), so I can't say more now. This answer might also be of interest. — Dave L. Renfro, May 22 '21 at 16:08
@s.harp Thank you! This does seem to clarify the issue of whether an induction argument would be valid/sufficient. — hasManyStupidQuestions, May 24 '21 at 13:09
@DaveL.Renfro This also looks very helpful! I will definitely read through and think about those answers in depth the next time I think about this question/problem. — hasManyStupidQuestions, May 24 '21 at 13:10
For results involving spaces between continuous functions and continuously differentiable functions, see Über die Höldersche bedingung by Auerbach/Banach (1931; see also Exercise 17.24 on pp. 135-136 of Eduard Čech's 1969 book Point Sets) AND Über die differenzierbarkeit stetiger funktionen by Vojtěch Jarník (1933; see Satz II on p. 50) (continued) — Dave L. Renfro, May 24 '21 at 16:37
AND Baire category and the existence of very continuous functions (Mathematics Stack Exchange question) AND The Sobolev embeddings are usually sharp by Fraysse/Jaffard (2005; BTW, I don't know much about Sobolev stuff) AND The prevalence of continuous nowhere differentiable functions by Brian R. Hunt (see Proposition 5 on p. 717). — Dave L. Renfro, May 24 '21 at 16:38
@DaveL.Renfro Thank you so much! Obviously I need to fill in some more dots, but this is already more than 90% of the way there! I will post/update for everyone on the site if/when I get around to doing this. — hasManyStupidQuestions, May 25 '21 at 16:44

Do progressively more differentiable functions (once differentiable, then $C^1$, then twice differentiable, etc.) form an infinite nested hierarchy?

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