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.


*

*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 $x.$

*Then I would like to prove that $P_x(x) = g(x)$ defines a $\alpha$ - Hölder function.

*Thus we would get that $g \in \mathcal{D}'.$

*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?
 A: Your steps are the right ones, in this answer I'll fill in the details.
Firstly, let's check that $g$ defines an $\alpha$-Holder function. Let $\rho$ be a smooth probability density with support in $B(0,1)$.
As you observed, we can assume that $P_x$ is constant for each $x$ (and I write $P_x$ now to denote that constant so that $g(x) = P_x$). This means that since $\int \rho(x) dx = 1$ we have that $P_x = \langle P_x, \rho_y^\lambda \rangle$ for any $x,y$.
I then write $|g(x) - g(y)| = \langle P_x - P_y, \rho_x^\lambda \rangle$ for $\lambda = |x-y|$. Notice that this means that there exists $\psi$ such that $\rho_x^\lambda = c \psi_y^{2\lambda}$ where $c$ is bounded independently of $x,y$ and $\lambda$.
Therefore
\begin{align*}
|g(x) - g(y)| \le |\langle f - P_x, \rho_x^\lambda \rangle| + c|\langle f - P_y, \psi_y^\lambda \rangle| \lesssim |x-y|^\alpha.
\end{align*}
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.
