I have trouble proving the following statement regarding a characterization of $C^{1,\alpha}$:
Let $\Omega$ be a Lipschitz domain. $u$ is pointwise $C^{1,\alpha}$ at all points with the same constant $M$: That is, $\forall x_0 \in \Omega$, $\exists P_{x_0}(x)$ of degree at most 1 such that $\lvert u(x) – P_{x_0}(x) \rvert\leq M\lvert x-x_0 \rvert^{1+\alpha}$ around some neighborhood $B_{\delta}(x_0)$, where $\delta$ is a universal constant works for all $x_0 \in \Omega$. Show that under this condition, $f$ is a $C^{1,\alpha}$ function.
My thought: With the definition, it's not hard to show partial derivative exists (Showing continuity is by the similar idea): Take $\delta$ small, then $$ \frac{u(x_0+\delta e_1) – u(x_0)}{\delta} = \frac{u(x_0+\delta e_1) – P_{x_0}(x_0+\delta e_1)}{\delta}+ \frac{P_{x_0}(x_0+\delta e_1) – P_{x_0}(x_0)}{\delta}+ \frac{P_{x_0}(x_0) – u(x_0)}{\delta}$$ where the first term is controlled by $M \lvert \delta \rvert^{\alpha}$ and third terms is 0 by definition. Hence $\frac{\partial u}{\partial x_1}(x_0) = \frac{\partial P_{x_0}}{\partial{x_1}}(x_0)$ which shows the existence and it applies to all other $x_i$'s. What's more, by above analysis, the polynomial $P_{x_0}(x) = u(x_0)+\nabla u^{T}(x_0)(x-x_0)$ near $x_0$.
For Hölder continuity of the derivatives, I tried to first prove the Hölder continuity within a tiny neighborhood of each $x_0$ (once we have this, in general we may use finite balls to cover the curve joining $x_0$, $y_0$ and sum up). The case for $\mathbb{R}$ is well-solved by the answer here: Characterization of $C^{k,\alpha}$ (functions with Hölder continuous derivatives) through Taylor estimates.
I tried to copy the method and apply it to $\mathbb{R}^n$, $n\geq2$: Fix $x_0 \in \Omega$, WLOG $h$ small and $d$ is a unit vector of any direction. By polynomial criterion, $$\lvert u(x_0+hd)-u(x_0)-\nabla u^{T}(x_0)hd \rvert \leq M \lvert h \rvert^{1+\alpha}.$$
Replacing by $-hd$:
$$\lvert u(x_0-hd)-u(x_0)+\nabla u^{T}(x_0)hd \rvert \leq M \lvert h \rvert^{1+\alpha}.$$
Replacing by $x_0-hd$ in the first equation:$$\lvert u(x_0)-u(x_0-hd)-\nabla u^{T}(x_0-hd)hd \rvert \leq M \lvert h \rvert^{1+\alpha}.$$
Summing up the second and the third equation, apply triangle inequality and divide by $h$ on both sides:
$$\lvert (\nabla u^{T}(x_0)-\nabla u^{T}(x_0-hd))\cdot d \rvert \leq 2M \lvert h \rvert^{\alpha}.$$
In $\mathbb{R}^1$, the vector $d$ doesn't matter. However in $\mathbb{R}^n$, $n\geq 2$ we cannot directly get the desired form. I'm stuck here since the difference of the gradient also depends on $d$ and it's hard to get rid of it. Any idea or help is appreciated.
Best Answer
Firstly, your definition of pointwise $C^{1,\alpha}$ is strictly speaking "incorrect" in that it doesn't give the desired conclusion. My interpretation of what you wrote is, with all the quantifiers included:
But note that this statement is satisfied for any $x_0$ at which the second derivative is well-defined, since $N_{x_0}$ is allowed to depend on $x_0$. In particular, even in one dimension the given statement doesn't imply a function is $C^{1,\alpha}$.
Example: (This is for $\mathbb{R}\to\mathbb{R}$ functions.) Let $f(x) = x^2 \sin(x^{-100})$, extended to $f(0) = 0$, satisfies the given statement; when $x_0 \neq 0$ the function is $C^{\infty}$ and Taylor's remainder theorem tells you that the statement holds. At $x_0 = 0$ one can simply take $P_0 \equiv 0$. But $f'(x)$ is unbounded in a neighborhood of $0$.
The key to Hairer's claim in the linked question is that some degree of uniformity is required. His statement is in fact
In the proof you linked to, this uniformity is required to ensure that one can simultaneously estimate $$ f(x) - f(y) \approx f'(y)\cdot (x-y) $$ and $$ f(y) - f(x) \approx f'(x) \cdot (y-x) $$
In view of that, let's consider the "corrected" claim. Note that here I set $\Omega = \mathbb{R}^d$; for domains more care will be needed for points near $\partial\Omega$.
I claim that this implies $f$ is in $C^{1,\alpha}$.
For convenience we will write $A \approx_r B$ for $|A-B| \leq M r^{1+\alpha}$, where $M$ is some universal constant that may change from line to line.
Let $x,y,z\in \Omega$, with pairwise distance $< r$. Then our hypothesis implies
$$ f(x) - f(y) \approx_{r} Df(y) \cdot (x-y) $$
and
$$ f(y) - f(x) \approx_r Df(x) \cdot (y-x) $$
so
$$ Df(x) \cdot (x-y) \approx_r Df(y) \cdot (y-x) $$
which provides that $|(Df(x) - Df(y)) \cdot (y-x) | \lesssim |y-x|^{1+\alpha}$ as you observed.
To handle directions not parallel to $y-x$, consider now
$$ f(z) - f(x) \approx_r Df(x) \cdot (z-x) , \qquad f(z) - f(y) \approx_r Df(y) \cdot (z-y) $$
and hence, subtracting the two expressions we find
$$ f(y) - f(x) \approx_r Df(x) \cdot (z-x) - Df(y) \cdot (z-y) $$
as we already know that the left hand side is approximately $Df(x) \cdot (y-x)$ we find
$$ 0 \approx_r Df(x) \cdot (z-y) - Df(y) \cdot (z-y) \tag{*}$$
Now given $x, y$, given any direction orthogonal to $x-y$, we can take $z$ to be such that $|x-y| = |z-y|$ and $z-y$ is in the specified direction. Then equation (*) implies
$$ |(Df(x) - Df(y))\cdot (z-y) | \lesssim |x-y|^{1 + \alpha}$$
This concludes the argument.