Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $C^0$ be the set of all of all (edit: bounded) continuous functions from $\mathbb R \to \mathbb R$ with the sup norm. Then a "path" in $C^0$ is a continuous function $f \colon [0,1] \to C^0$. The question boils down to if there is path (or continuous deformation) which transforms any one bounded continuous function into another.

share|improve this question
2  
The sup norm isn't defined on all of $C^0$ since, for example, $f(x) =x $ is unbounded. –  Jason DeVito Aug 22 '10 at 20:26
    
ok sry, how about just the bounded continuous functions –  Matt Calhoun Aug 22 '10 at 20:29
5  
This is true because $\mathbb{R}$ is contractible. More generally, "paths of continuous functions" are also called homotopies, cf. en.wikipedia.org/wiki/Homotopy –  Akhil Mathew Aug 22 '10 at 23:57
    
The sup norm is a semimetric, so it's fine by me. –  scineram Mar 21 '11 at 17:42

3 Answers 3

up vote 11 down vote accepted

It suffices to find a path to the 0 function from any other continuous function.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be any continuous function and define $F(x,t) = tf(x)$. Assuming we can prove this is, in fact, continuous, note that $F(x,1) = f(x)$ and $F(x,0)$ is the $0$ function.

So, why is it continuous? Let $\epsilon > 0$. Let $K = \sup|f(x)|$ Choose $\delta > 0 $ such that $\delta K < \epsilon$.

Then, we have $|t_1 - t_2| < \delta$ implies $|t_1f(x) - t_2 f(x)| = |(t_1 - t_2)f(x)| \leq \delta K = \epsilon$, so that this map is continuous (in fact, uniformly continuous).

share|improve this answer
    
Are you assumming $K< \infty $? –  gary Aug 12 '11 at 5:40
    
@gary: In the original question, $f$ is assumed to be bounded. You're right that I should have explicitly written it in my proof, though. –  Jason DeVito Aug 12 '11 at 13:30
    
This proves that set of bounded continuous functions is convex. –  lhf Nov 3 '11 at 9:51

I think it's worth to develop a little bit more Akhil's interesting remark.

1. If instead of ${\cal C}^0(\mathbb{R}, \mathbb{R})$, you take ${\cal C}^0([a,b], \mathbb{R})$, where $[a,b] \subset \mathbb{R}$ is a compact interval, then something interesting happens: you can give ${\cal C}^0([a,b], \mathbb{R})$ the compact-open topology, and this topology coincides with the one induced by the sup norm (also called the topology of uniform convergence). More generally, the same is true for ${\cal C}^0 (X , Y)$, where $X$ is any compact topological space and $Y$ any metric space.

2. Now, for psychological reasons, let's use the notation $Y^X$ instead of ${\cal C}^0 (X , Y) $, and let $X$, $Y$ be any pair of topological spaces. Put the compact-open topology on $Y^X$ and let $Z$ be a third topological space. To every map $F: X\times Z \longrightarrow Y$ you can associate the map $\Phi (F) : Z \longrightarrow Y^X$ defined by

$$ (\Phi (F) (z)) (x)= F(x,z) \ . $$

Reciprocally, to every map $\gamma : Z \longrightarrow Y^X$ you can associate the map $\Psi (\gamma ): X \times Z \longrightarrow Y$ defined by

$$ \Psi (\gamma) (x,z) = (\gamma (z)) (x) \ . $$

You can check easily that $\Phi$ and $\Psi$ are inverses one each other. Moreover, if $F$ is continuous, so is $\Phi (F)$. If $X$ is a locally compact Hausdorff space, the same is true for $\Psi$: $\gamma$ continuous implies $\Psi (\gamma )$ continuous. (See Munkres' "Topology", theorem 46.11.)

So, when $X$ is a locally compact Hausdorff space, you have bijections, inverses one each other:

$$ \Phi : Y^{X \times Z} \longrightarrow (Y^X)^Z \qquad \text{and} \qquad \Psi: (Y^X)^Z \longrightarrow Y^{X \times Z} $$

For instance, take $Z = I$, the unit interval. This bijection tells us that, when $X$ is a locally compact Hausdorff space, paths $\gamma : I \longrightarrow Y^X$ and homotopies $F : X \times I \longrightarrow Y$ are the same.

3. An easy topological exercise: any map $f: X \longrightarrow Y$ is homotopic to a constant map and all the constant maps are homotopic, either if

  1. $X$ is contractile and $Y$ is path-connected, or
  2. $Y$ is contractile.

So, coming back to your situation, we could have said that ${\cal C}^0 (\mathbb{R}, \mathbb{R})$, with the compact-open topology, is always path-connected because $X = \mathbb{R}$ is locally compact, Hausdorff and contractile, and $Y= \mathbb{R}$ is path-connected (or because $X=\mathbb{R}$ is locally compact and Hausdorff and $Y= \mathbb{R}$ is contractile).

(Notice that Jason has used the second possibility: he has constructed a homotopy from the constant function $0$ to $f$. We could use the first one: $F(x,t) = f(xt)$ is a homotopy from the constant function $f(0)$ to $f$.)

Of course, Jason's solution is more straightforward and doesn't need all this elementary machinery of point-set topology. But now we can say that also other function spaces, with the compact-open topology, are path-connected: every time you take as $X$ a locally compact Hausdorff space and any of the two previous possibilities, ${\cal C}^0(X,Y)$ is path-connected. For instance,

$$ {\cal C}^0(\mathbb{R}, Y) ,\ {\cal C}^0([a,b],Y) , \ {\cal C}^0(\mathbb{R}^n,Y) ,\ {\cal C}^0(D^2,Y) \dots $$

for any path-connected space $Y$, are path-connected. Also

$$ {\cal C}^0(X,\mathbb{R}) , \ {\cal C}^0(X, [a,b]) , \ {\cal C}^0(X,\mathbb{R}^n) ,\ {\cal C}^0(X,D^2) \dots $$

for any locally compact Hausdoff space $X$, are path-connected.

share|improve this answer

Any real vector space $V$ with a "reasonable" topology is path connected. Given a vector $v_1\in V$ move along the line spanned by $v_1$ to $0$ and from there move to any other vector $v_2$ moving along le line spanned by it.

Reasonable means that for any $w\in V$ every open $U\ni w$ must contain an open segment around $v$ of the line spanned by $w$, which is certainly true for the case under consideration (topology under the sup norm).

I guess that $L^2({\Bbb R})$ is not reasonable, so maybe my terminology could be improved :-)

*Edit : * Of course $L^2({\Bbb R})$ IS reasonable! $$ \left(\int_{\Bbb R}(f-tf)^2dx\right)^{1/2}=|1-t|\left(\int_{\Bbb R}f^2dx\right)^{1/2} $$ gets as close as needed to 0 as $t\to1$.

share|improve this answer
    
Just realized that my "reasonability" condition is ambiguous at $0$. Then one also needs that every open $U\ni0$ contains an open segment around 0 of every line through 0. This is still true for the sup norm, but maybe "reasonable" spaces are not so rwasonable......... –  Andrea Mori Aug 22 '10 at 23:33
3  
Any "reasonable" condition which is weaker than "the multiplication by scalars is continuous" is quite unreasonable, that the latter is quite enough to do what you want :) –  Mariano Suárez-Alvarez Aug 23 '10 at 1:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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