Every continuous function is homotopic to a locally Lipschitz one

I would like to know for which category/class/set of metric spaces the following holds: for any two metric spaces $X$, $Y$, for any continuous function $f:X\to Y$ there exists a locally Lipschitz continuous function $g:X\to Y$ which is homotopic to $f$.

EDIT: One could also ask a class of metrizable topological spaces such that each one of them can be given a metric so that the above property holds. Actually, I am more interested in the underlying topological space than in the actual metric space.

In general, the metric spaces I am considering are complete and weakly separable (there exists a sequence $(\phi_h)$ of $1$-Lipschitz functions such that for any two point $x,\ y$ $d(x,y)=\sup_h|\phi_h(x)-\phi_h(y)|$).

I don't know if this is a known fact among experts or not; in that case, I apologize for the standard question and would ask only for a reference.

ADDENDUM: Although I also have an interest for the general question as it is posed above, I could try to highlight some classes of metrizable spaces I have particular interest in knowing if they fulfill the request or not: manifolds, singular spaces (which singularities are allowed), spaces which are manifolds outside a "small" (in some sense) set, compact manifolds of infinite dimension or manifolds modeled on some "nice" linear space (Banach, Hilbert, Fréchet, ...).

Thanks.

• If $Y$ be a convex subset of a topological vector field,then every two maps $f,g$ from $X$ to $Y$ are homotopic because the map $(x,t)$ to $tf(x)+(1-t)g(x)$ is continuous,now a constant map is locally lipschitz continuous. – R Salimi Apr 6 '13 at 20:04
• Samuele: It should work if domain is a metric simplicial complex (or an Alexandiv space) and range is CAT(k) with $k<\infty$. In general, it would be good if you were more specific about the classes of metric spaces you are interested in. – Misha Apr 7 '13 at 4:50
• Well, I assume my metric space to be complete and with a weak property of separability (for any two point $x,\ y$ there exists a sequence of $1$-Lip maps $(\phi_h)$ such that $d(x,y)=\sup_h |\phi_h(x)-\phi_h(y)|$). But obviously I don't expect that every of each spaces has a metric such that my request is satisfied. I think I could ask the following: is it true for manifolds? does it remain true if we allow singularities? which ones? is it true for infinite dimensional manifolds (maybe compact)? for Banachian or Hilbertian compact manifolds, at least? – Samuele Apr 7 '13 at 6:45
• Samuele: Your last condition is satisfied for all metric spaces, since you can use distance function to x as your 1-Lispchitz function. Riemannian manifolds and manifolds with singularities belong to the class from my comment, provided dimension of the domain is finite, dimension of the range could be infinite. – Misha Apr 7 '13 at 13:31
• Samuele: If you have a compact manifold modeled on a Banach space, then the Banach space has to be finite-dimensional. – Misha Apr 7 '13 at 13:48

A modest start.

Consider two finite geometric simplicial complexes with reasonable metrics, e.g. inherited from the ambient Euclidean (or Banach) space (where simplices are affine). Then every continuous function $f$ between them is uniformly approximated by the simplicial maps of iterated baricentric subdivisions of the first complex into the second complex. All these simplicial maps are Lipschitz. When approximation is close enough to $f$ then it is homotopically equivalent to $f$. This gives a positive answer to your question for finite geometric simplicial complexes.

REMARK 0 For the sake of obtaining a Lipschitz map homotopic to a given continuous map one does not need to subdivide the second complex.

On the other hand, it is not difficult to provide two metric functions (distance functions) for the unit circle $S^1$ (thus let's talk about two metric spaces anyway) such that the identity map from one of them to another is not homotopic to any locally Lipschitz function. Indeed, there will not exist any locally Lipschitz function at all (not even at any inverse image of any non-empty open set) from the first space onto the second one (under the fixed but properly selected metric functions; the first one can be the standard metrics).

REMARK 1 Instead of $S^1$ we could consider a space consisting of a convergent sequence and its limit, endowed with two distance functions such that the identity is not Lipschitz (at the limit point). The only map homotopic to the identity is the identity, hence another instance of the negative answer. But $S^1$ is nicer :-)

• First of all, thank you. So, finite (geometric) simmplicial complexes work. Why do you need them to be finite? About the counterexample, that's interesting! But I was kind of expecting something like that: that's why I asked for a class of metric spaces, which come together with their distances. Another way to put the question could be to ask for a class of topological spaces which can be endowed with a distance (inducing their topology, hence metrizable spaces) so that the property holds. Could it be the case that CW-complexes do work? – Samuele Apr 7 '13 at 2:01

Such approximation is possible under some mild assumptions about domain and range.

For the domain you want to have structure of a finite dimensional structure of a metric simplicial complex of locally bounded geometry. For example, a Riemannian manifold or Alexandrov space would do. For the target you should impose some conditions implying local linear contractibility, for instance, a space which is locally CAT(k), where $k<\infty$ would suffice. The proof is based on barycentric maps of smplices, which you can find in the paper of Bruce Kleiner, "The local structure of length spaces of curvature bounded above", Math. Z. 1999.

The construction of Lipschitz approximation is the same as cellular approximation in algebraic topology. First, approximate your map on the set of vertices. Then extend to simplices by induction on skeleta, using barycentric simplices as in Kleiner's paper.

Some of this might even work if domain is infinite dimensional, but you would need to control the Lipschitz constant for the barycentric maps.