I am reading a paper that makes reference to the following fact:
Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is Lipschitz continuous of some positive order $\alpha$. Let $H(x,y)$ be the extension of $f$ to a bounded function continuous function in $\mathbb{R}^{2}$, harmonic in the upper and lower half-planes; $H(x,y)$ is given by the Poisson integral in the upper and lower-half planes for $y>0$ and $y<0$, respectively. That is, for $y>0$,
\begin{align*}H(x,y)=\frac{1}{\pi}\int_{\mathbb{R}}f(x-t)\frac{y}{t^{2}+y^{2}}dt,\end{align*}and similarly for $y<0$.
Then the Lipschitz continuity of $f$ of order $\alpha$ implies the existence of a positive sequence $\{C_{n}\}_{n=1}^{\infty}$ such that:
\begin{align*}|\partial^{\beta}H(x,y)|\le C_{n}|y|^{\alpha-n},\end{align*}for all $x\in\mathbb{R}$, all $y\neq 0$, and every partial derivative $\partial^{\beta}$ of order $n$.
No reference is given for this fact; can anyone guide me to one, or produce a proof sketch? Thanks.
It must be said that $\alpha\in (0,1)$; the statement is false for $\alpha=1$.
Also, it suffices to consider the upper halfplane only; the lower half is no different.
Fix a point $(x_0,y_0)$ in the upper halfplane. Actually, we can assume $x_0=0$ by translating the picture. Then $|f(x)-f(x_0)|\le C|x|^\alpha$ for some $C$, by the continuity assumption.
Using polar coordinates $r,\theta$, define $$U(r,\theta) = \frac{1}{\cos (\alpha \pi/2)} r^\alpha \cos (\alpha (\theta-\pi/2))$$ This is a harmonic function on the upper halfplane, with boundary values $|x|^\alpha$.
By the maximum principle (more precisely, Phragmén–Lindelöf) we have $H-f(x_0)\le CU$ in the upper halfplane. Hence, $H(0,y_0)-f(x_0)\le C' y_0^\alpha$ with $C'=C /\cos (\alpha \pi/2)$. The same applies to $-(H(0,y_0)-f(x_0))$, by reversing the sign of $f$.
Let's summarize the above: $$|H(x_0,y_0)-f(x_0)| \le C' y_0^\alpha \tag1$$ where $C'$ depends only on $\alpha$ and the Lipschitz (Hölder) constant of $f$.
Next step is to amplify (1) to: $$|H(x,y)-f(x_0)|\le C'' y_0^\alpha,\qquad \text{when } \ |(x,y)-(x_0,y_0)|\le y_0/2 \tag{2}$$ This is done by observing that $f(x)-f(x_0)$ is bounded by a multiple of $y_0^\alpha$; thus, applying the estimate (1) in the form $|H(x ,y )-f(x )|\le C'y^\alpha\le C' 2^{\alpha} y_0^\alpha$ we arrive at (2). Note that $C''$ depends only on $\alpha$ and the Lipschitz (Hölder) constant of $f$.
The final step is the interior gradient estimate for harmonic function: if a harmonic function is bounded by $M$ (from both sides) in a disk of radius $R$, then its derivatives of order $n$ at the center of the disk are bounded by $MR^{-n}$. Applying this on the disk of radius $y_0/2$ centered at $(x_0,y_0)$, you'll get the desired bound of $y_0^{\alpha-n}$, times a constant.
Reference (for the more general result, in arbitrary dimensions): Singular integrals and differentiability properties of functions by Stein, page 149 (the proof appears later in section 5.4). I also think any reasonably thorough harmonic analysis book should include this result, since it is a prototype for various function spaces related to the behavior of convolution integrals. I did not check, but Real variable methods in harmonic analysis by Torchinsky should emphasize this approach.