# Is $L^p(\mathbb{R})$ minus the zero function contractible?

Is $L^p(\mathbb{R}) \setminus 0$ contractible? My intuition says that the answer is yes, but I'm afraid that this is based on thinking of this as somehow similar to a limit of $\mathbb{R}^n \setminus 0$ as n approaches $\infty$, which is of course nonsense. In any case, every contraction I've tried ends up making some function pass through $0$.

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Technically, this is a duplicate of: mathoverflow.net/questions/198/… , my answer there works for any space that is "stable" in the sense that $X \oplus \mathbb{R} \cong X$. – Loop Space Sep 15 '10 at 7:18
(Looking again at my answer to the contractiblity of the sphere, I realised one could weaken the conditions so the "stable" in the above is stronger than needed.) – Loop Space Sep 15 '10 at 7:50
Meta discussion on whether to close as duplicate or not: tea.mathoverflow.net/discussion/672/are-these-duplicates – Loop Space Sep 15 '10 at 8:09
All separable, infinite-dimensional Banach spaces are homeomorphic to each other (and to $\mathbb R^\infty$), so if you like you can use one of those other spaces. – Gerald Edgar Feb 24 '14 at 14:22

Here is something really cheap and dirty. Let $p<+\infty$. Take $g=\frac{1}{1+x^2}$. Then $f(x,t)=e^{-(1+|x|)t/(1-t)}f(x)$ ($0\le t\le 1$) is a continuous contraction of $L^p\setminus\{g\}$ to $0$. (the reason is that your only chance to hit $g$ is to start with it because $g(x)e^{s(1+|x|)}$ is not in $L^p$ for $s>0$).

Let's make it more interesting without making it more abstract. Can we find a uniformly continuous (both in space and time, as usual) contraction of the unit ball in $L^p$ without the center to a point?

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That's nice! I have no idea about your more general question. Maybe it's worth asking separately? – Nikita Sep 15 '10 at 4:07
I updated my answer to mathoverflow.net/questions/198/… to note that the homotopy described there is jointly continuous. – Loop Space Sep 15 '10 at 7:49
Fedja, Benyamini and Sternfeld proved in 1983 that the unit sphere of any infinite dimensional Banach space is Lipschitz contractible. A lot of related things are mentioned in a survey talk he gave in 2007: nim.nankai.edu.cn/activites/conferences/hy200707/invite-talks/… As Borcherds mentioned, the topological structure of infinite dimensional spaces is very simple (not that the proofs are easy!). The uniform and Lipschitz structures are complicated and there is a lot of current research on these topics. The book of Benyamini-Lindenstrauss is the standard reference. – Bill Johnson Sep 15 '10 at 12:42
Bill, I read that as a paper by "Fedja, Benyamini and Sternfeld" so was about to search for those three names! Then I wondered why Fedja didn't know about that result ... English is a funny language. – Loop Space Sep 16 '10 at 7:44
Oops! Sorry about that, Andrew. – Bill Johnson Sep 17 '10 at 2:53

An infinite dimensional Banach space is homeomorphic to itself minus a point. Maybe R.D. Anderson or V. Klee proved this.

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Do you know a reference for this? – Nikita Sep 15 '10 at 2:32
Start with projecteuclid.org/DPubS/Repository/1.0/… Also look for Bessaga's papers, maybe joint with Klee, in the 1960s in MathSciNet (which I cannot access right now). – Bill Johnson Sep 15 '10 at 2:54
Wow, that's a pretty serious body of work! I'm going to hold off marking this as the accepted answer for a while in the hope that someone will suggest an easier proof of the fact that I asked for (which is much weaker than these papers prove!). – Nikita Sep 15 '10 at 3:00

According to mathscinet, the results of Anderson given in "Topological properties of the Hilbert cube and the infinite product of open intervals" Trans. Amer. Math. Soc. 126 1967 200--216, and "Hilbert space is homeomorphic to the countable infinite product of lines" Bull. Amer. Math. Soc. 72 1966 515--519 show that all infinite dimensional separable Frechet spaces are homeomorphic, and that removing any countable union of compact sets from such a space leaves the homeomorphism type unchanged.

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Here is a simple proof for case of a Hilbert space $V$. Since $V$ minus the origin deformation retracts onto the unit sphere $S^\infty$, it suffices to show that $S^\infty$ is contractible, and that will follow if we can show that $S^\infty$ is a deformation retract of the unit disk. Below is a simple proof of that fact taken from my book "Critical Point Theory and Submanifold Geometry". (I have an old paper called "On the Homotopy Theory of Infinite Dimensional Manifolds" that proves much more general results of this nature. It appeared in vol.3 of Topology (1966).)

Proposition. If $D^\infty$ is the closed unit disk in an infinite dimensional Hilbert space $V$, and $S^\infty=\partial D^\infty$ is the unit sphere in $V$, then there is a deformation retraction of $D^\infty$ onto $S^\infty$.

Proof. Since $D^\infty$ is convex, it will suffice to show that there is a retraction of $D^\infty$ onto $S^\infty$. Now recall the standard proof of the Brouwer Fixed Point Theorem. If there were a fixed point free map $h:D^n\to D^n$ it would imply the existence of a deformation retraction $r$ of $D^n$ onto $S^{n-1}$; namely $r(x)$ is the point where the ray from $h(x)$ to $x$ meets $S^{n-1}$. If $n<\infty$ this would contradict the fact that $H_n(D^n,S^{n-1})=Z$, so there can be no such retraction and hence no such fix point free map. But when $n=\infty$ we will see that such a fixed point free map does exist, and hence so does the retraction $r$. This will be a consequence of two simple lemmas.

Lemma 1. $D^\infty$ has a closed subspace homeomorphic to $R$.

Proof Let $\{e_n\}$ be an orthonormal basis for $V$ indexed by $Z$, and define $F:R \to D^\infty$ by $F(t)=\cos({1\over2}(t-n)\pi)e_n +\sin({1\over2}(t-n)\pi)e_{n+1}$ for $n\le t \le n+1$. It is easily checked that $F$ is a homeomorphism of $R$ into $D^\infty$ with closed image. QED

Lemma 2. If a normal space $X$ has a closed subspace $A$ homeomorphic to $R$ then it admits a fixed point free map $H:X\to X$.

Proof. Since $A$ is homeomorphic to $R$ it admits a fixed point free map $h:A\to A$, corresponding to say translation by $1$ in $R$. Since $A$ is closed in $X$ and $X$ is normal, by the Tietze Extension Theorem $h$ can be extended to a continuous map $H:X\to A$, and we may regard $H$ as a map $H:X\to X$. If $x\in A$ then $x\not=h(x)=H(x)$, while if $x\in X\setminus A$ then, since $H(x)\in A$, again $H(x)\not=x$. QED

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I cannot help saying once again my enthusiasm for those papers of yours about Banach manifolds on Topology... they are simply great! – Pietro Majer Sep 15 '10 at 7:51
I first learnt of this sort of fun-and-games in infinite dimensions from those papers and I've been hooked ever since. Finite dimensions is just so dull! – Loop Space Sep 15 '10 at 8:04

I got here via this mathoverflow question. I find I like Dick Palais' simple proof so much that I want to make it simpler. $\newcommand{\from}{\colon}\newcommand{\One}{\mathbb{1}}$

Palais reduces the problem to finding a map $f\from D^\infty \to D^\infty$ without fixed points. Here is an easier proof that such a map exists. Let $T \from D^\infty \to D^\infty$ be the shift map: $T((x_0, x_1, x_2, \ldots)) = (0, x_0, x_1, x_2, \ldots)$. Note that $T$ is continuous and fixes the origin.

Set $\One = (1,0,0,0,\ldots)$. Working with the $L^2$ norm, define $$f(x) = \sqrt{1 - |x|^2} \cdot \One + T(x).$$ Since $f$ and $T$ agree on $S^\infty$, the map $f$ has no fixed points on the sphere. On the other hand, for all $x \in D^\infty$, the point $f(x)$ lies in $S^\infty$. So $f$ has no fixed points inside the ball. We are done.

Said another way: the Brouwer fixed-point theorem is "obviously" false for $D^\infty$.

EDIT: Problem 36 of the Scottish book, posed by Ulam, asks if $D^\infty$ deformation retracts to its boundary. The book goes on to say that Tychonoff found the required retraction. See page 178 of "Spaces and fixed point theory" by Khamsi and Kirk for a brief discussion.

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