# $C^{1,\alpha}$ estimate for Newton potential of $L^\infty$ function

Theorem 13.1.1 in Jost's Partial Differential Equations asserts that if $$f \in L^\infty(\Omega)$$, with $$\Omega$$ a bounded open set in $$\mathbb{R}^2$$, then $$u(x) = \int_\Omega \log |x-y| f(y)\ dy$$ is in $$C^{1,\alpha}(\Omega)$$.

But I think there is an error in the proof. Equation 13.1.7 says that for fixed $$x_1$$ and $$x_2$$ there exist a constant $$c_3$$ and a point $$x_3$$ on the line connecting $$x_1$$ and $$x_2$$ such that

$$\begin{equation*} \left| \frac{x^i_1 - y^i}{|x_1-y|^2} - \frac{x^i_2 - y^i}{|x_2-y|^2} \right| \leq c_3 \frac{|x_1-x_2|}{|x_3-y|^2} \end{equation*}$$

Here, $$x_1^i$$ refers to the $$i^{th}$$ component of $$x_1 \in \mathbb{R}^2$$. However, either the constant or $$x_3$$ must depend on $$y$$. Otherwise, we could send $$y$$ to $$x_1$$ and get the left-hand side of the inequality to blow up, while keeping the right-hand side finite.

But in the next step, Jost splits an integral in $$y$$ over a domain $$D$$ into an integral over $$D \setminus B_\delta(x_3)$$ and $$B_\delta(x_3)$$ and treats $$c_3$$ as a constant independent of $$y$$. I don't see how this works if one of $$c_3$$ or $$x_3$$ depends on $$y$$.

My questions are:

(1) Is this an error in Jost or am I missing something? Perhaps there is an obvious way to fix $$x_3$$ that I don't see.

(2) If it is an error in Jost, is the result still true? I don't see it in, for instance, Gilbarg and Trudinger. They have a similar estimate but require Holder continuity for $$f$$.

• I don't have Jost's book to check so I cannot answer (1) but for (2) the result you want is indeed true. It's sort of in GT but hard to find. See Problem 4.8(a) and also the discussion surrounding equation (4.44). Note that GT use the notation $\mathbf{f}$ when the mean to study an equation $\Delta u = \textrm{div} \mathbf{f}$ and then it is true that $\mathbf{f} \in C^\alpha$ implies $u \in C^{1,\alpha}$. However, note that $\mathbf{f} \in C^\alpha$ is like "$\textrm{div} \mathbf{f} \in C^{-1,\alpha}$," so this is again saying that inverting the Laplacian allows for two derivatives Commented Jan 6, 2022 at 21:01
• Thank you! That's extremely helpful. Commented Jan 7, 2022 at 2:09

The point is that this inequality is only used away from $$x_1$$ and $$x_2$$.
Set $$\delta=\sqrt7 \left|x_2-x_1\right|$$ -- $$\sqrt7$$ being an aribtrary constant bigger than $$1$$. Set $$x_3 = \frac12 x_1 +\frac12 x_2$$, extend the integral to $$B_R(x_3)$$ for $$R=(\sqrt7 + 1)$$diam$$(\Omega)$$, so that the original domain is contained within that large ball, as well as the ball $$B_\delta(x_3)$$. We have
\begin{align*} \int_{B_R(x_3)} \left| \frac{x^i_1 - y^i}{|x_1-y|^{d}} - \frac{x^i_2 - y^i}{|x_2-y|^{d}} \right| dy =& \int_{B_\delta(x_3)} \left|\frac{x^i_1 - y^i}{|x_1-y|^{d}} - \frac{x^i_2 - y^i}{|x_2-y|^{d}}\right| dy\\ &+\int_{B_R\setminus{B_\delta(x_3)}} \left|\frac{x^i_1 - y^i}{|x_1-y|^{d}} - \frac{x^i_2 - y^i}{|x_2-y|^{d}}\right| dy\\ \leq &\int_{B_\delta(x_3)} \frac{1}{|x_1-y|^{d-1}} dy +\int_{B_\delta(x_3)} \frac{1}{|x_2-y|^{d-1}} dy \\ &+\int_{B_R\setminus{B_\delta(x_3)}} \left|\frac{x^i_1 - y^i}{|x_1-y|^{d}} - \frac{x^i_2 - y^i}{|x_2-y|^{d}}\right| dy \end{align*}
For the first integral, since $$B_\delta(x_3) \subset B_{2\delta} (x_1)$$, there holds $$\int_{B_\delta(x_3)} \frac{1}{|x_1-y|^{d-1}} dy \leq \int_{B_{2\delta}(x_1)} \frac{1}{|x_1-y|^{d-1}} dy = 2\omega_d\delta.$$ Similarly, $$\int_{B_\delta(x_3)} \frac{1}{|x_2-y|^{d-1}} dy \leq \int_{B_{2\delta}(x_1)} \frac{1}{|x_1-y|^{d-1}} dy = 2\omega_d\delta.$$ Finally, for the third integral, which is away from $$x_1$$ and $$x_2$$, you can use the intermediate value theorem or algebraic manipulations to obtain that for $$y\in B_R(x_3)\setminus B_\delta(x_3)$$ there holds $$\begin{equation*} \left| \frac{x^i_1 - y^i}{|x_1-y|^d} - \frac{x^i_2 - y^i}{|x_2-y|^d} \right| \leq c_d \frac{|x_1-x_2|}{|x_3-y|^d} \end{equation*}$$ therefore $$\int_{B_R(x_3)\setminus{B_\delta(x_3)}} \left|\frac{x^i_1 - y^i}{|x_1-y|^{d}} - \frac{x^i_2 - y^i}{|x_2-y|^{d}}\right| dy \leq c \delta \int_{B_R(x_3)\setminus{B_\delta(x_3)}} \frac{1}{|x_3-y|^{d}} dy =c^\prime \delta\log \frac R\delta,$$ which gives the desired estimate, as it shows (as written in the book) that $$|D_i u(x_1) -D_i u(x_2)| \leq C |x_1-x_2| \log \frac1{|x_1-x_2|}$$ so $$C^{1,\alpha}$$ for all $$\alpha<1$$.