- #1

Shackleford

- 1,656

- 2

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.

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.

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.

Last edited: