This problem arose while reading Lee's Introduction to Smooth Manifolds.

Given Banach space: $E,F$
with a plain subset: $A\subseteq E$
and a function: $f:A\to F$

Call a function differentiable at some point if it admits an extension to the Banach space that is differentiable there: $$f_E:E\to F:\quad f_E\restriction_A=f$$ This definition is equivalent to the existence of an extension to some open neighborhood of the point that is differentiable there too: $$f_U:U\to F:\quad f_U\restriction_A=f$$

Now, I'm wondering wether it might happen that the extensions have to be chosen differently for different points. Can that happen???

I'm thinking there of some function on half space: $A=\mathbb{H}^n$
Obviously, locally at specific points it must look like: $$F_E(a_0+v):=2F(a_0)-F(a_0-v),v\notin \mathbb{H}^n$$ and globally at all points it must look like: $$F_E(a-n):=2F(a)-F(a+n),n\bot\partial\mathbb{H}^n$$

The reason for considering this is that when defining the holomorphic functional calculus we require the functions on the spectrum to extend to holomorphic functions on some neighborhood of the spectrum. If the above can not happen then one might simply think of holomorphic functions on the spectrum itself.

These ideas came to my mind while reading Lee's Introduction to Smooth Manifolds.
(Cf. discussion on p. 45.)

Definition

Let $E$ and $F$ be two Banach spaces together with a plain subset $A\subseteq E$.

Here, a partially defined function $f:A\to F$ is called differentiable at $a\in A$ if it admits an extension $\bar{f}_a:E\to F$ differentiable at $a$.

Remarks

Note the dependence of the extension on the point under consideration: $\bar{f}_a$

Also a function $f:U\to F$ with open domain $U$ is differentiable at $u\in U$ in the definition given above iff it is differentiable there in the ordinary sense.

The leading principle of this approach to differentiability is that a linear approximation foots on linear spaces. Plain subsets or opens in general aren't!

Problems

1. (Riesz-Dunford Functional Calculus)
   Let a function $f:A\to F$ be (continuously) differentiable in $A$ in the definition given above.
   Does it necessarily admit an extension $\bar{f}:E\to F$ that happens to be (continuously) differentiable on some whole neighborhood $U_A$ of $A$ rather than merely on $A$?
2. (Manifolds with Boundary)
   Let a function $f:A\to F$ be (continuously) differentiable in $A$ in the definition given above.
   Does it necessarily admit an extension $\bar{f}:A\to F$ that happens to be (continuously) differentiable at every point $a\in A$ simultaneously rather than for every point a separate extension $\bar{f}_a:E\to F$?

Explanation

1. (Riesz-Dunford Functional Calculus)
   The Riesz-Dunford Calculus applies only to functions that happen to be holomorphic on some neighborhood of the spectrum of an operator. A positive result here would pin the problem to holomorphic functions on the spectrum precisely.
2. (Manifolds with Boundary)
   On manifolds a map is differentiable on the boundary iff its coordinate expression has one-sided directional derivatives within half space. A negative result here would complicate the situation alot.
   Moreover, the definition given in Lee's book for differentiability of partially defined functions slightly varies from the one given above to the extend that it requires the existence of a common extension. The lack, however, here is that though differentiability is a local property it is defined pointwise. So from a structural point the definition given above shows consistency while for practical purposes the definition given in Lee's book is favourable. A positive result here would unveil them as equivalent and therefore justify the approach.

Attempts

1. (Riesz-Dunford Functional Calculus)

2. (Manifolds with Boundary)
   For some function on half space $f:\mathbb{H}^m\to\mathbb{R}^n$ to be differentiable in the sense given above it must hold that locally at specific points it extends infinitesimally as: $$F_E(a_0+v):=2F(a_0)-F(a_0-v),v\notin \mathbb{H}^n$$ while globally at all points it extends infinitesimally as: $$F_E(a-n):=2F(a)-F(a+n),n\bot\partial\mathbb{H}^n$$ These guiding constructions seem to clash. But this still requires a rigorous counterexample.

This problem arose while reading Lee's Introduction to Smooth Manifolds.

Given Banach space: $E,F$
with a subspace: $A\subseteq E$
and a function: $f:A\to F$

Call a function differentiable at some point if it admits an extension to the Banach space that is differentiable there: $$f_E:E\to F:\quad f_E\restriction_A=f$$ This definition is equivalent to the existence of an extension to some open neighborhood of the point that is differentiable there too: $$f_U:U\to F:\quad f_U\restriction_A=f$$

Now, I'm wondering wether it might happen that the extensions have to be chosen differently for different points. Can that happen???

I'm thinking there of some function on half space: $A=\mathbb{H}^n$
Obviously, locally at specific points it must look like: $$F_E(a_0+v):=2F(a_0)-F(a_0-v),v\notin \mathbb{H}^n$$ and globally at all points it must look like: $$F_E(a-n):=2F(a)-F(a+n),n\bot\partial\mathbb{H}^n$$

The reason for considering this is that when defining the holomorphic functional calculus we require the functions on the spectrum to extend to holomorphic functions on some neighborhood of the spectrum. If the above can not happen then one might simply think of holomorphic functions on the spectrum itself.

This problem arose while reading Lee's Introduction to Smooth Manifolds.

Given Banach space: $E,F$
with a plain subset: $A\subseteq E$
and a function: $f:A\to F$

Call a function differentiable at some point if it admits an extension to the Banach space that is differentiable there: $$f_E:E\to F:\quad f_E\restriction_A=f$$ This definition is equivalent to the existence of an extension to some open neighborhood of the point that is differentiable there too: $$f_U:U\to F:\quad f_U\restriction_A=f$$

Now, I'm wondering wether it might happen that the extensions have to be chosen differently for different points. Can that happen???

I'm thinking there of some function on half space: $A=\mathbb{H}^n$
Obviously, locally at specific points it must look like: $$F_E(a_0+v):=2F(a_0)-F(a_0-v),v\notin \mathbb{H}^n$$ and globally at all points it must look like: $$F_E(a-n):=2F(a)-F(a+n),n\bot\partial\mathbb{H}^n$$

The reason for considering this is that when defining the holomorphic functional calculus we require the functions on the spectrum to extend to holomorphic functions on some neighborhood of the spectrum. If the above can not happen then one might simply think of holomorphic functions on the spectrum itself.