Estimates on Derivatives

Question

I have trouble in filling in the details of the proof on Estimates on derivates, from page 29 of PDE Evans, 2nd edition. Namely, I am lost at some steps. The book gives:

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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. 1. 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| &= \left|\frac{1}{\alpha(n) (\frac{r}{2})^n} \int_{B(x_0,\frac{r}{2})} u_{x_i} dx\right| \tag{20} \\ &= \left|\frac{2^n}{\alpha(n) r^n} \int_{B(x_0,\frac{r}{2})} u_{x_i} dx\right| \\ &= \left|\frac{2^n}{\alpha(n) r^n} \int_{\partial B(x_0,\frac{r}{2})} u \nu_i dS\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 $k = 1$.

(There is a second part of this proof for $k \ge 2$, but I think I can understand that on my own, once I fully understand $k=1$.)

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I had major trouble understanding all the steps that involve inequalities (with the $\le$), though I completely understand the equality steps (=). May I seek help in understanding how these inequalities are obtained and how they are true? (I think Evans likes to skip a lot of steps...)

• $\displaystyle \left|\frac{2^n}{\alpha(n) r^n} \int_{\partial B(x_0,\frac{r}{2})} u \nu_i dS \right| \le \frac{2n}{r} \|u\|_{L^\infty(\partial B(x_0,\frac{r}{2})}$
• $\displaystyle |u(x)| \le \frac{1}{\alpha(n)} \left(\frac{2}{r}\right)^n \|u\|_{L^1(B(x_0,r))}$
• $\displaystyle |D^\alpha u(x_0)| \le \frac{2^{n+1}n}{\alpha(n)} \frac{1}{r^{n+1}} \|u\|_{L^1(B(x_0,r))}$

I would love to show my work attempt in filling in the details for these relations, but they are a mess and probably won't be helpful here. But hopefully what's given above will suffice. (As far as only equalities are concerned, I tried to add a little more detail to the steps in the above proof than what is originally presented by the textbook.)

Answer 1

1. Inequality. All we need to observe is that $|\nu_i|\leq 1$ and $|\partial B_{(x_0,r/2)}|=\alpha(n)n(r/2)^{n-1}$

2. Inequality is the maximum principle for $B(x_0,r/2)$ followed by $\|u\|_{L^1(B(x_0,r/2))}\leq \|u\|_{L^1(B(x_0,r))}$.

3. Inequality comes from application of Gauss theorem to the LHS of 1st inequality

Answer 2

For the second inequality first use the mean value theorem for harmonic functions. Then some integral properties and the last inequality is that you are integrating non negative functions over a bigger set.

\begin{equation} |u(x)| = \left| \rlap{-}\!\!\int_{B\left(x,\frac{r}{2}\right)}{u(y)dy}\right| = \left| \frac{2^n}{\alpha(n) r^n}\int_{B\left(x,\frac{r}{2}\right)}{u(y)dy}\right| \leq \\ \frac{2^n}{\alpha(n) r^n}\int_{B\left(x,\frac{r}{2}\right)}{\vert u(y) \vert dy} \leq \frac{2^n}{\alpha(n) r^n}\int_{B\left(x_{0},r\right)}{\vert u(y) \vert dy} \end{equation}

Now as

\begin{equation} \int_{B\left(x_{0},r\right)}{\vert u(y) \vert dy} = \Vert u \Vert _{L^1 B \left(x_0,r\right) } \end{equation}

we have

\begin{equation} |u(x)| \leq \frac{2^n}{\alpha(n) r^n} \Vert u \Vert _{L^1 B \left(x_0,r\right) } \end{equation}

For the third inequality the second one tells us that

\begin{equation} \Vert u \Vert _{L^\infty \partial B \left(x_0,r/2 \right)} \leq \frac{2^n}{\alpha(n) r^n} \Vert u \Vert _{L^1 B \left(x_0,r\right) } \end{equation}

so by the first inequality combined with the second and third

\begin{equation} \vert D^\alpha u(x_0) \vert \leq \frac{2n}{r} \Vert u \Vert _{L^\infty \partial B(x_0, r/2)} \leq \frac{2n}{r} \frac{2^n}{\alpha(n) r^n} \Vert u \Vert _{L^1 B \left(x_0,r\right) } = \frac{2^{n+1}n}{\alpha(n) r^{n+1}} \Vert u \Vert _{L^1 B \left(x_0,r\right) } \end{equation}