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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.

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  • $\begingroup$ Thanks Willie for pointing out the typos in the definition. It has been modified. $\endgroup$ Commented Mar 11 at 6:10

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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:

$\exists M$ such that $\forall x_0\in \Omega$, there exists an affine function $P_{x_0}$ and a neighborhood $N_{x_0}\ni x_0$ such that for every $x\in N_{x_0}$, $$ |u(x) - P_{x_0}(x)| \leq M |x - x_0|^{1+\alpha}$$

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

$\exists M$ such that $\forall x_0\in \mathbb{R}$, there exists an affine function $P_{x_0}$ such that for every $x\in (x_0 - 1, x_0+1)$, $$ |u(x) - P_{x_0}(x)| \leq M |x - x_0|^{1+\alpha}$$

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 $\partial\Omega$.

$\exists M$ such that $\forall x_0\in \mathbb{R}^d$, there exists an affine function $P_{x_0}$ such that for every $x\in B_1(x_0)$, $$ |u(x) - P_{x_0}(x)| \leq M |x - x_0|^{1+\alpha}$$

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.

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  • $\begingroup$ Thank you. If a domain is Lipschitz, or $C^1$ boundary, is there anything to take care of as we approach to the boundary? $\endgroup$ Commented Mar 11 at 6:03
  • $\begingroup$ I'm not sure. The proof given above only works if you can find $x,y,z$ as described with bounds $|x-y| \approx |y-z| \approx |z-x|$. So it would work for interior estimates (at points that are $\delta$ away from the boundary). As written it doesn't work for points near the boundary, but I haven't been able to come up with a counterexample. $\endgroup$ Commented Mar 11 at 15:27

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