About the polynomial characterization of $C^{1,\alpha}(\bar{\Omega})$ Hölder space in Lipschitz domain

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.

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

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.

• Thank you. If a domain is Lipschitz, or $C^1$ boundary, is there anything to take care of as we approach to the boundary? Commented Mar 11 at 6:03
• 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. Commented Mar 11 at 15:27