Ngô Quốc Anh

February 25, 2015

Continuous functions on subsets can be extended to the whole space: The Kirzbraun-Pucci theorem

Filed under: Uncategorized — Ngô Quốc Anh @ 1:22

Let f be a continuous function defined on a set E \subset \mathbb R^N with values in \mathbb R and with modulus of continuity

\displaystyle \omega_f (s) := \sup_{|x-y|\leqslant s,x,y\in E} |f(x) - f(y)| \quad s>0.

Obviously, the function s \mapsto \omega_f(s) is nonnegative and nondecreasing in [0,+\infty).

Our first assumption is that \omega_f is bounded from above in [0, \infty) by some increasing, affine function; that is to say there exists some a,b \in \mathbb R^+ such that

\displaystyle \omega_f (s) \leqslant a s +b \quad \forall s \geqslant 0.

Associated with \omega_f having the above first assumption is the concave modulus of continuity of f, i.e. some smallest concave function c_f lies above \omega_f. Such the function c_f can be easily constructed using the following

\displaystyle c_f (s) = \inf_\ell \{\ell(s) : \ell \text{ is affine and } \ell \geqslant \omega_f \text{ in } [0,+\infty)\}.

As can be easily seen, once \omega_f can be bounded from above by some affine function, the concave modulus of continuity of f exists and is well-defined.

By definition and the monotonicity of \omega_f, we obtain

\displaystyle |f(x)-f(y)| \leqslant \omega_f (|x-y|) \leqslant c_f (|x-y|).

In this note, we prove the following extension theorem.

Theorem (Kirzbraun-Pucci). Let f be a real-valued, uniformly continuous function on a set E \subset \mathbb R^N with modulus of continuity \omega_f satisfying the first assumption. There exists a continuous function \widetilde f defined on \mathbb R^N that coincides with f on E. Moreover, f and \widetilde f have the same concave modulus of continuity c_f and

\displaystyle \sup_{\mathbb R^N} \widetilde f = \sup_E f, \quad \inf_{\mathbb R^N} \widetilde f = \inf_E f.

For the sake of simplicity, we only sketch a proof for \sup_{\mathbb R^N} \widetilde f = \sup_E f and \inf_{\mathbb R^N} \widetilde f = \inf_E f and omit a proof for the fact that f and \widetilde f have the same concave modulus of continuity c_f.

Proof. For each x \in \mathbb R^N, we set

\displaystyle g(x)=\inf_{y \in E} \{f(y) + c_f( |x-y| )\}.

If we only need an extension for f, the function g constructed above is the desired function since, as we shall see later g \equiv f on E. However, if we further want to pursue the two identities \sup_{\mathbb R^N} \widetilde f = \sup_E f and \inf_{\mathbb R^N} \widetilde f = \inf_E f, we need another step.

We shall prove that our (really) extension \widetilde f is

\displaystyle\widetilde f(x)=\min\{g(x), \sup_E f\}.

Why \sup_{\mathbb R^N} \widetilde f = \sup_E f? To see this, first, if x\in E, then we obtain

\displaystyle f(y)+c_f (|x-y|) \geqslant f(x) +c_f (|x-y|)-|f(x)-f(y)| \geqslant f(x)

for all y\in E; hence g \geqslant f in E. In particular, g=f in E since we can select y=x as thanks to c_f(0)=0. From this, we know that \widetilde f(x) = \sup_E f for any x \in E. Also by definition, \widetilde f(x) \leqslant \sup_E f for all x \in \mathbb R^N; hence \sup_{\mathbb R^N} \widetilde f = \sup_E f as claimed.

Next, for all x\in \mathbb R^N and for all y \in E, we clearly have

\displaystyle \inf_E f+\inf_{y \in E}c_f(|x-y|) \leqslant g(x) \leqslant f(y)+c_f(|x-y|).

Thus, we first have \inf_E f \leqslant \widetilde f and \inf_E f \leqslant \inf_{\mathbb R^N} g; hence \inf_E f \leqslant \inf_{\mathbb R^N} \widetilde f. To see the equality, we only need to show that \inf_E f = \inf_{\mathbb R^N} g. Once we have this, from the estimate \inf_E f \leqslant \widetilde f \leqslant g and by taking the infimum all over \mathbb R^N, we would have \inf_E f = \inf_{\mathbb R^N} \widetilde f. However, this is trivial since g \equiv f in E making \inf_E g = \inf_E f.

As indicated in Real Analysis by Emmanuele DiBenedetto, Kirzbraun first proved the theorem when \omega_f is of linear. The general modulus of continuity was proved by Pucci.

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