Mapping continuously differentiable at a point

Question

I am reading Coleman's Calculus on Normed Vector Spaces. He defines what it means for a mapping $f: O \to F$ from an open subset $O$ of a normed vector space $E$ to a normed vector space $F$ to be differentiable at a point $x \in O$. Also, such a mapping is differentiable if it is differentiable at every point in its domain.

He than defines that such a mapping is continuously differentiable (or of class $C^1$) when it is differentiable and its differential mapping $f': O \to \mathcal{L}(E,F)$ is continuous.

He does not define what it means for a mapping to be continuously differentiable at a point (at least I could not find a definition), so I am wondering what such definition would be like. This is what I first thought of:

A mapping $f: O \to F$ is continuously differentiable at a point $x \in O$ if there is a neighborhood $N$ of $x$ such that $f$ is differentiable at $N$ and $f': N \to \mathcal{L}(E,F)$ is continuous at $x$.

My questions are:

1. Is it true that a mapping is differentiable at $x$ if, and only if, it has all partial differentials defined at a neigborhood of $x$ and continuous at $x$ (but not necessarily continuous at a neighborhood of $x$?
2. Is this definition useful in any other way?
3. Is it necessary to have $f$ differentiable at a neighborhood to talk about continuity at $x$? This seemed more intuitive to me but I am not sure this is needed.

Answer 1

Yes, the definition you give for continuous differentiability at a point is correct.

1. No, it's not true as written. Since you're dealing with general normed vector spaces, you must be careful with the notion of partial derivatives. Once you suitably modify the statement to account for the general normed vector space setting, the correct statements are the following:

• If all the partials are continuous at a point $x$ (this already requires existence in a neighborhood), then the function is differentiable at $x$ (proof using mean-value inequality). See this answer for some details sketched out.
• as a consequence, we can deduce that if all the partials exist and are continuous on an open set, then the function is itself differentiable on that open set. But in fact, based on the explicit formula for the derivative $Df_a=\sum_{i=1}^n(\partial_if)_a\circ \pi^i$, we can even deduce continuity of $Df$ on the open set.
• The converse of this second assertion is pretty trivial: if $Df$ exists and is continuous on an open set, then all the partials $\partial_if$ exist and are also continuous on that open set.

2. I'm not really sure how to answer this, except to say this is a useful definition, because it can sometimes make proving things easy (eg proving via a simple induction that if $f,g$ are $C^k$ then so is $f\circ g$... I'm sure there are more situations where this way of thinking of continuous differentiability makes things clearer, but unfortunately I can't think of examples off the top of my head right now).

3. Yes.