evans; estimate on derivatives I have troubles understanding a certain necessity in Evans' proof of a theorem on the estimates for derivatives of harmonic functions. So consider the following, (this is the same theorem as discussed here, however i am asking a different question)


Theorem 7 (Estimates on derivatives).  Assume u is harmonic in U. Then
  \begin{align}
|D^\alpha u(x_0)| \le \frac{C_k}{r^{n+k}} \|u\|_{L^1(B(x_0,r))} \tag{18}
\end{align}
  for each ball $B(x_0,r) \subseteq U$ and each multiindex $\alpha$ of order $|\alpha| = k$.
Here 
  \begin{align}
C_0 = \frac{1}{\alpha(n)}, C_k = \frac{(2^{n+1}nk)^k}{\alpha (n)} \text{ for } k=1,2,\ldots \tag{19}
\end{align}

Proof.


*

*We establish $(\text{18}), (\text{19})$ by induction on $k$, with the case $k=0$ being immediate from the mean value formula $u(x) = \frac{1}{\alpha(n)r^n} \int_{B(x_0,r)} u \, dx = \frac{1}{\alpha(n)r^{n-1}} \int_{\partial B(x,r)} u \, dS$ (which denote average values of $u$ over the ball and sphere, respectively). $$ \, $$ For $k = 1$, we note upon differentiating Laplace's equation that $u_{x_i}$ (for $i=1,...n$) is harmonic. Consequently,
\begin{align}
\left|u_{x_i}(x_0)\right| 
&\le \frac{2n}{r} \|u\|_{L^\infty(\partial B(x_0,\frac{r}{2})}
\end{align}
Now if $x \in \partial B(x_0,\frac{r}{2})$, then $B(x,\frac{r}{2}) \subseteq B(x_0,r) \subseteq U$, and so
\begin{align}
|u(x)| \le \frac{1}{\alpha(n)} \left(\frac{2}{r}\right)^n \|u\|_{L^1(B(x_0,r))}
\end{align}
by (18), (19) for $k=0$. Combining the inequalities above, we get
\begin{align}|D^\alpha u(x_0)| &\le \frac{2^{n+1}n}{\alpha(n)} \frac{1}{r^{n+1}} \|u\|_{L^1(B(x_0,r))} \\
&= \frac{2^{n+1}n}{r^{n+1}} \|u\|_{L^1(B(x_0,r))} \\
\end{align}
if $|\alpha| = 1$. This verifies $(\text{18})$ and $(\text{19})$ for $k = 1$. $$\,$$

*Now in the induction step we proceed analogously: assume $k \geq 2$ and (18), (19) are valid for all balls in $U$ and each multiindex of order less than or equal to $k-1$. Fix $B(x_0,r) \subset U$ and let $\alpha$ be a multiindex with $|\alpha| = k$. Then $D^\alpha u = (D^\beta u)_{x_i}$ for some $i \in \{1, \cdots,n\}$, $|\beta|=k-1$. By calculations similar to those in (20) we establish that $$|D^\alpha u(x)| \leq \frac{n\,k} r \, \|D^\beta u\|_{L^\infty(\partial B(x_0,\frac r k))}.$$ If $x \in \partial B(x_0,\frac r k)$ then $B(x,\frac {k-1} k \, r) \subset B(x_0,r) \subset U$. Thus (18), (19) for $k-1$ imply $$|D^\beta u(x)| \leq \frac{(2^{n+1} \, n \, (k-1) )^{k-1} } {\alpha(n) \, (\frac{k-1} k \, r)^{n+k-1} } \, \|u\|_{L^1(B(x_0,r))}.$$
Now combine the two results to conclude the theorem.

Question


*

*Why don't we just use $\frac r 2$ (as before) instead of $\frac r k$ in the induction step?

*What is the use of $\frac r k$?


I don't see any reason why $\frac r 2$ wouldn't work.
Im very glad for any thoughts on this!
 A: Assuming I didn't miscalculate, both options work.
But before you can conclude that both options work, you will actually have to figure out the estimates they lead to.
Such choices of radii can be crucial in some estimates, but less important in others.
Philosophically, it appears that each of the $k$ derivatives is given a step of length $r/k$.
Therefore the $k-1$ derivatives in $\partial^\beta u$ are given the interval $[r/k,r]$ and the one derivative between $\partial^\beta u$ and $\partial^\alpha u$ is given the interval $[0,r/k]$.
The estimate is proven via studying $\partial^\beta u$ on $\partial B(x_0,r/k)$.
Combining the two estimates you have gives
$$
|D^\alpha u(x)| \leq \frac{n\,k} r \, \frac{(2^{n+1} \, n \, (k-1) )^{k-1} } {\alpha(n) \, (\frac{k-1} k\,\color{red}{r})^{n+k-1} } \, \|u\|_{L^1(B(x_0,r))}.
$$
You were missing an $r$, indicated by red.
There is a factor $r^{-n-k}$ as required and simplifying the constant gives
$$
\frac12\left(\frac{k}{2(k-1)}\right)^n\times
\frac{(2^{n+1}nk)^k}{\alpha (n)}.
$$
This yields the desired estimate.
Now, if instead you study $\partial^\beta u$ on $\partial B(x_0,r/2)$, you get
$$
|D^\alpha u(x)| \leq \frac{2n} r \, \|D^\beta u\|_{L^\infty(\partial B(x_0,\frac r 2))}
$$
and
$$
|D^\beta u(x)| \leq \frac{(2^{n+1} \, n \, (k-1) )^{k-1} } {\alpha(n) \, (\frac12r)^{n+k-1} } \, \|u\|_{L^1(B(x_0,r))}.
$$
Combining these two leads to an otherwise similar estimate, but the factor that comes with $r^{-n-k}$ is
$$
2^{1-2n}(k-1)^{-1}\left(\frac{k-1}{2k}\right)^k
\times
\frac{(2^{n+1}nk)^k}{\alpha (n)}.
$$
This also leads to the desired estimate.
