# Sufficient criteria for $X \subset \mathcal{H}$ to be a Lipschitz (or unif. cont.) retract of $\mathcal{H}$

I am interested in sufficient criteria which ensure that a subset $X$ of a Hilbert space $\mathcal{H}$ is a Lipschitz (or at least uniformly continuous) retract of $\mathcal{H}$.

Under which conditions on $X \subset \mathcal{H}$ is there a (nonlinear) Lipschitz (or uniformly continuous) map $\pi : \mathcal{H} \to X$ with $\pi x = x$ for all $x \in X$?

Note: It is a crucial requirement here that the range of $\pi$ be (contained in) $X$! Thus, this cannot be resolved by an application of Kirszbraun's theorem, and really depends on the properties of the set $X$.

My hope is that this is a known fact to people working in the geometry of Hilbert spaces.

In fact, it would be enough for my purposes if for each $x_0 \in X$, there is a small ball $B_r (x_0)$, and a Lipschitz (or uniformly continuous) map $\pi : B_r (x_0) \to X$ with $\pi x = x$ for $x \in X \cap B_r (x_0)$. In this case, it would be very nice if size $r$ of the ball and the modulus of continuity of $\pi$ can be bounded independently of $x_0$, as long as $\| x_0 \| \leq R$.

I know that something like the above condition is true for closed convex sets, but unfortunately, the set $X$ that I am interested in is not convex.

To explain what type sets I am looking at, and what kind of conditions I am interested in, let me introduce a bit of notation: We are given a family $(\psi_x)_{x \in \mathbb{R}^{2d}}$ of functions $\psi_x \in L^2 (\mathbb{R}^d)$, such that $x \mapsto \psi_x$ is continuous, and such that the map $$V : L^2 (\Bbb{R}^d) \to L^2 (\Bbb{R}^{2d}), f \mapsto V f \quad \text{with} \quad V f(x) = \langle f, \psi_x \rangle$$ is a well-defined isometry (but $V$ is not surjective). We then consider the set $$X := \{ \, |V f| \,:\, f \in L^2 (\Bbb{R}^d) \, \},$$ i.e., we take the (pointwise) absolute values of the range of the operator $V$.

From the properties stated above, one can derive the following:

• $X$ is "star-shaped" with respect to the origin, and thus simply connected.
• $X$ is weakly sequentially closed. In fact, $X \cap \overline{B_r} (x)$ is weakly compact for each $r > 0$, $x \in L^2 (\Bbb{R}^{2d})$.
• More precisely, for a weakly convergent sequence $(F_n)_n$ in $X$, say $F_n \rightharpoonup F$, it follows that $F_n \to F$ pointwise, and that $F = |V f|$ for some $f \in L^2$.

The set $X$ probably enjoys quite a lot more nice properties, but since I don't know what type of property would imply the "Lipschitz retract property", it would be helpful to have some kind of list of sufficient conditions, so that we might try to check these :)

• Inspect the proof of the Kirszbraun Theorem and isolate the property of intersectiong balls used in the proof. Then check this property on your particular example. More information on Lipschitz retractions of Hilbert spaces can be found in the book of Benyamini and Lindenstrauss. May 19 '18 at 6:57

• Thank you for your answer. Unfortunately, I don't think this is correct (or I don't see why it would be), since I require the range of $\pi$ to be contained in $X$, please see also the edited version of my question. Dec 19 '17 at 11:36