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.