Hölder continuity definition through distributions.

I am trying to prove that for a given Hölder parameter $\alpha \in (0, 1)$ and a distribution $f \in \mathcal{D}'(\mathbb{R}^d)$ the following are equivalent:

• $f \in C^{\alpha}$
• For any $x$ there exists a polynomial $P_x$ such that $| \langle f - P_x, \phi_x^{\lambda} \rangle | \le C \lambda^{\alpha.}$

Where the latter estimate holds uniformly over all$x$ and $\phi \in \mathcal{D}$ with compact support in the unit ball, and: $$\phi_x^{\lambda}(\cdot) = \lambda^{-d} \phi\left( \frac{\cdot \ - \ x}{\lambda} \right)$$

Proving that the first implies the second is fairly easy. The other way around gives me some problems.

1. The first step I took is to realize that we care only about the order zero term of $P_x,$ since all other terms vanish at a order higher than $\alpha.$ Here we assume that the polynomial is centered in $x.$
2. Then I would like to prove that $P_x(x) = g(x)$ defines a $\alpha$ - Hölder function.
3. Thus we would get that $g \in \mathcal{D}'.$
4. Eventually I would like to prove that $g = f$ in $\mathcal{D}'.$

My problem is that I can't prove any implications between 2,3,4 nor any of 2,3,4 starting from 1. Any hints, help, suggestions?

Solution 1:

Your steps are the right ones, in this answer I'll fill in the details. Throughout this answer, I will call $\phi_x^\lambda$ a test function of scale $\lambda$ at $x$ (where $\phi \in \mathcal{D}$ has support in $B(0,1)$).

Firstly, let's check that $g$ defines an $\alpha$-Holder function. Let $\rho$ be a smooth probability density with support in $B(0,1)$. Notice here that $\rho_x^\lambda$ is essentially just a recentered mollifier. This means that we have $$|g(x) - g(y)| = \lim_{\lambda \to 0} |\langle P_x, \rho_x^\lambda \rangle - \langle P_y, \rho_y^\lambda \rangle|$$ Now we seek to bound the right hand side. \begin{align*} |\langle P_x, \rho_x^\lambda \rangle - \langle P_y, \rho_y^\lambda \rangle| \leq& |\langle P_x - f, \rho_x^\lambda \rangle| + |\langle f, \rho_x^\lambda - \rho_y^\lambda \rangle | + | \langle f - P_y, \rho_y^\lambda \rangle | \\ \leq & C_1 \lambda^\alpha + |\langle f, \rho_x^\lambda - \rho_y^\lambda \rangle | \end{align*}

We are left to control $|\langle f, \rho_x^\lambda - \rho_y^\lambda \rangle |$. To do this, we notice that $\Phi_{x,y}^\lambda = \rho_x^\lambda - \rho_y^\lambda$ is a test function at scale $\frac{|x-y|}{2} + \lambda$ centered at $\frac{x+y}{2}$ such that $\langle C, \Phi_{x,y}^\lambda \rangle = 0$ for any constant $C$. This means that \begin{align*} |\langle f, \Phi_{x,y}^\lambda \rangle| =& |\langle f - P_{\frac{x+y}{2}}\big(\frac{x+y}{2}\big), \Phi_{x,y}^\lambda \rangle| \\ \leq & |\langle f - P_{\frac{x+y}{2}}, \Phi_{x,y}^\lambda \rangle| + |\langle P_{\frac{x+y}{2}} - P_{\frac{x+y}{2}}\big(\frac{x+y}{2}\big), \Phi_{x,y}^\lambda \rangle| \\ \leq& C_2 (|x-y| + \lambda)^\alpha \end{align*} where we used $\alpha$-Holder continuity of $P_{\frac{x+y}{2}}$ to bound the second term in the second to last line. In total, $$|g(x) - g(y)| = \lim_{\lambda \to 0} |\langle P_x, \rho_x^\lambda \rangle - \langle P_y, \rho_y^\lambda \rangle| \leq \lim_{\lambda \to 0} C_1 \lambda^\alpha + C_2(|x-y| + \lambda)^\alpha = C_2 |x-y|^\alpha$$ so $g$ is $\alpha$-Holder. This already implies that $g$ defines a regular distribution.

Finally, we must show that $f = g$ in $\mathcal{D}'$. This will follow from the fact that any distribution that vanishes at order $\alpha > 0$ must be the zero distribution.

Lemma: Suppose $F \in \mathcal{D}'(\mathbb{R}^d)$ and for every compact set $k \subseteq \mathbb{R}^d$, $$|\langle F, \phi_x^\lambda \rangle| \leq C \lambda^\alpha$$ for $\alpha>0$ where $C$ is uniform over $x \in k$, $\gamma \in [0,1)$ and $\phi \in \mathcal{D}(\mathbb{R}^d)$ with $\operatorname{supp} \phi \subseteq B(0,1)$. Then $F = 0$.

Proof of Lemma: Fix a smooth probability density $\rho$ with compact support in $B(0,1)$. Then for any test function $\varphi \in \mathcal{D}'$, by standard facts about mollification, $$\langle \varphi, \rho_x^\lambda \rangle = \int_{\mathbb{R}^d} \varphi(y) \rho_x^\lambda(y) dy \to \varphi(x)$$ as $\lambda \to 0$ and in fact, the convergence takes place in $\mathcal{D}$ where $\Phi^\lambda(x) := \langle \varphi, \rho_x^\lambda \rangle$ is considered as a function of $x$. Hence \begin{align*} |\langle F, \varphi \rangle| =& \lim_{\lambda \to 0} |\langle F, \Phi^\lambda \rangle| \\ =& \lim_{\lambda \to 0} \bigg |\int_{\mathbb{R}^d} \langle F, \rho_x^\lambda \rangle \varphi(y) dy \bigg|\\ \leq& \lim_{\lambda \to 0} C\lambda^\alpha \int_{\mathbb{R}^d} |\varphi(y)| dy = 0 \end{align*}

In our particular case, we have \begin{align*} |\langle f-g, \phi_x^\lambda \rangle \leq& |\langle f - P_x, \phi_x^\lambda \rangle| + |\langle P_x - g(x), \phi_x^\lambda \rangle| + |\langle g - g(x), \phi_x^\lambda \rangle | \\ \leq& C_3 \lambda^\alpha \end{align*} by our assumption on $f$ and $\alpha$-Holder continuity of $g$ and $P_x$ (since $P_x - g(x) = P_x-P_x(x)$). Hence, by the lemma, $f=g$ as desired.