$C^{1,\alpha}$ 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 $f$.
 A: I asked Jürgen Jost what he meant, and what I write below follows the logic of his answer.
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 $\alpha<1$.
