5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ And then there is the problem that real differentiability is quite a different beast from complex differentiability. So I don't really see the connection between the first and second parts of your question. (With real differentiability you are allowed to use partitions of unity and what not.) $\endgroup$ – Willie Wong Jun 10 '14 at 15:11
  • $\begingroup$ Yeah I know but up to that point as merely being a definition this should not depend on the ground field... $\endgroup$ – C-Star-W-Star Jun 10 '14 at 15:47
  • $\begingroup$ The complex motivation is maybe not strong enough, but the question seems interesting to me, and I really do not see why down-voting... $\endgroup$ – Pietro Majer Jun 10 '14 at 16:45
  • $\begingroup$ Heeeyy I guess I got an example :) but I'll need some help to rigorously prove it - I will post it as a answer... $\endgroup$ – C-Star-W-Star Jun 10 '14 at 18:07
  • $\begingroup$ @Freeze_S: my point is that there may be stronger rigidity from holomorphy than would be implied by real differentiability. The simple example being that if two holomorphic functions on $U\subset \mathbb{C}$ agree on a 1 (real) dimension curve, then they agree on $U$. So if your real motivation is about holomorphic functional calculus, I am not really sure the question you actually asked is relevant. $\endgroup$ – Willie Wong Jun 11 '14 at 7:14
3
$\begingroup$

For infinite dimensions, even beyond Banach spaces, see
Section 22: Whitney's extension revisited
(in particular, theorem 22.17)
Section 24: Smooth mappings on non-open domains
Section 25: Real analytic mappings on non-open domains
Section 26: Holomorphic mappings on non-open domains
of this book.

$\endgroup$
2
$\begingroup$

1.(Riesz-Dunford Functional Calculus)

Consider the function $f(z):=|z|^2$ defined on the real and imaginary axis only. Then around every point it has an extension to a continuously differentiable function within some neighborhood. But that extension is confined to the Cauchy Riemann equations and therefore it must be $f(z)=+z^2$ and $f(z)=-z^2$ simultaneously in every neighborhood of zero which is impossible. So the answer to the first problem is: No, in general there won't be an extension continuously differentiable in a whole neighborhood.

locally

2.(Manifolds with Boundary)

Besides this example still doesn't resolve(!) the second problem as one can choose the following extension:

enter image description here

$\endgroup$
  • 1
    $\begingroup$ This isn't differentiable at $z=0$. If it were, you could extend it by setting $f_{\mathbb R^2}(x,y)=f(x)$. More generally, if $A$ is a (genuine) closed subspace and it has a direct summand, you can just declare your extension to be constant with respect to the new coordinates. $\endgroup$ – Christian Remling Jun 10 '14 at 20:00
  • $\begingroup$ @ChristianRemling: I corrected it. Can u prove the last statement? $\endgroup$ – C-Star-W-Star Jun 13 '14 at 2:36
  • $\begingroup$ Since $z$ can lie on axes or diagonals, what do you mean by $z > 0$? And I am not convinced your extension is complex differentiable. Outside of $z = 0$, the function $z\bar{z} = |z|^2$ is not differentiable. Maybe you are thinking that on the imaginary axis you want to extend by $-z^2$? $\endgroup$ – Willie Wong Jun 13 '14 at 7:40
  • $\begingroup$ Yes so I took as domain both the axes and the diagonals otherwise I could just extend the function to $z^2$ everywhere with an appropriate twist. $\endgroup$ – C-Star-W-Star Jun 13 '14 at 10:23
  • $\begingroup$ Then for a point either on one of the axes or on one of the diagonals I can just choose a neighborhood not interscting with the other axes and diagonals and not with the origin and extend it everywhere in the neighborhood to $z^2$ with an appropriate twist so to get an extension that is differentiable everywhere within the chosen neighborhood. $\endgroup$ – C-Star-W-Star Jun 13 '14 at 10:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.