How Does the Derivation of Harmonic Function Derivatives Work in Evans' PDE?

[Shackleford]

This is from Evans PDE page 29. Assume u is harmonic.

18.
$$ |D^\alpha u(x_0)| \leq \frac{C_k}{r^{n+k}} \|u\|_{L^1(B(x_0,r))} $$

19.
$$ C_0 = \frac{1}{\alpha(n)}, \qquad C_k = \frac{(2^{n+1}nk)^k}{\alpha(n)}, \qquad (k=1,...).
$$

20. \fint is the average integral.
$$
\begin{gather*}
\begin{split}
|u_{x_i}(x_0)| & = \left| \fint_{B(x_0,r/2)} u_{x_i} \; dx\right| \\
& = \left| \frac{2^n}{\alpha(n)r^n}\int_{B(x_0,\frac{r}{2})} u \nu_i \; dS \right| \quad \text{(by Gauss-Green)}\\
& \leq \frac{2n}{r} \|u\|_{L^\infty(\partial B(x_0,\frac{r}{2}))}.
\end{split}
\end{gather*}
$$

$$
\text{If } x \in \partial B(x_0,r/2)), \text{then } B(x,r/2) \subset B(x_0,r) \subset U, \text{ and so}
$$

21.
$$
|u(x)| \leq \frac{1}{\alpha(n)} \left(\frac{2}{r}\right)^n \|u\|_{L^1(B(x_0,r))} \\
$$
by (18), (19) for k = 0. Combining the inequalities above, we deduce

$$ |D^\alpha u(x_0)| \leq \frac{2^{n+1}n}{\alpha(n)} \frac{1}{r^n} \|u\|_{L^1(B(x_0,r))}$$

if |alpha| = 1. Verifies for k = 1.

Questions:

Am I following the derivation correctly?

For k = 0:

$$
\begin{align*}
|D^0 u(x_0)| = &|u(x_0)| \leq \frac{C_0}{r^{n+0}} \|u\|_{L^1(B(x_0,r))} = \frac{1}{\alpha(n)r^{n}} \|u\|_{L^1(B(x_0,r))} \leq \frac{1}{\alpha(n)r^{n}} \cdot 2^n\|u\|_{L^1(B(x_0,r))} \\
& = \frac{1}{\alpha(n)} \cdot \left( \frac{2}{r} \right)^n \|u\|_{L^1(B(x_0,r))}.
\end{align*}
$$

Because the ball is contained in the other, we can state
$$ u(x) \leq \frac{1}{\alpha(n)} \cdot \left( \frac{2}{r} \right)^n \|u\|_{L^1(B(x_0,r))}.$$

I'm not sure how he arrived at the deduction. I thought maybe he's thinking of (20) as this $$|u_{x_i}(x_0)| = |Du(x_0)| $$

and then comparing it to what I'm calling (21) to arrive at the formula when |alpha|=1.

 

[jvicens]

Yes, you are following the derivation correctly. For k = 0, the equation becomes:

$$
|D^0 u(x_0)| = |u(x_0)| \leq \frac{C_0}{r^{n+0}} \|u\|_{L^1(B(x_0,r))} = \frac{1}{\alpha(n)r^{n}} \|u\|_{L^1(B(x_0,r))} \leq \frac{1}{\alpha(n)} \cdot \left( \frac{2}{r} \right)^n \|u\|_{L^1(B(x_0,r))}.
$$

This is because when k = 0, the term $C_k$ becomes $C_0 = \frac{1}{\alpha(n)}$.

The deduction is arrived at by comparing the inequality in (20) to the inequality in (21). In (20), we have the average integral of $u_{x_i}$ over the ball $B(x_0,r/2)$, while in (21) we have the average integral of $u$ over the same ball. By comparing these two inequalities, we can see that they are essentially the same, except for the factor of $2^n$ in front of the integral in (20). This is because when taking the average integral of $u_{x_i}$, we also need to take into account the factor of $2^n$ from the Jacobian of the transformation from $x_i$ to $x$.

Therefore, by comparing these two inequalities, we can deduce that:

$$
|D^\alpha u(x_0)| \leq \frac{2^{n+1}n}{\alpha(n)} \frac{1}{r^n} \|u\|_{L^1(B(x_0,r))}
$$

if $|\alpha| = 1$.