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I was reading Evans’ PDE book and got stuck on the proof of Theorem 10 (§2.2.1 pg. 31). Namely,

Assume $u$ is harmonic in $U$. Then, $u$ is analytic in $U$.

Now, for any fixed point $x_0,$ we have the goal of showing $u$ can be written as a convergent power series in some neighborhood of $x_0.$ We define $$r \triangleq \frac{1}{4} \mathrm{dist}(x_0,\partial u), \\ M \triangleq \frac{1}{\alpha(n)r^n}\|u\|_{L^1(B(x_0,2r))}<\infty.$$ Since $B(x,r) \subset B(x_0,2r) \subset U$ for every $x\in B(x_0,r),$ Theorem 7 provides the bound $$\tag{1} \|D^{\alpha}u\|_{L^{\infty}(B(x_0,r))}\le M \left(\frac{2^{n+1}n}{r} \right)^{|\alpha|}|\alpha|^{|\alpha|}.$$ Theorem 7 states that for $u$ harmonic in $U,$ $$|D^{\alpha}u(x_0)|\le\frac{C_k}{r^{n+k}}\|u\|_{L^1(B(x_0,r))} \tag{2}.$$ I’m not sure how the above statement follows from theorem 7 because the RHS involves an $L^{\infty}$ norm, not a p-norm. I’m surprised that this isn’t on stack exchange anywhere, honestly. This proof is very difficult to follow, but so is most of this chapter. I also looked for some online math notes which didn’t help since they were basically just notes on Evans’ book. Now, for $k\in\mathbb{N}, \frac{k^k}{k!}\lt e^k.$ Thus, $$|\alpha|^{\alpha}\le e^{|\alpha|}|\alpha|! \tag{3}$$ for all multi-indices $\alpha.$ I’m not sure where this came from either. I wish he would show more steps. Next, we apply the Multinomial theorem to attain $$n^k = (1+\mathrm{...}+1)^k = \sum_{|\alpha|=k} \frac{|\alpha|!}{\alpha!}$$ Thus, we have $$|\alpha|!\le n^{|\alpha|}\alpha! \tag{4}$$ I believe this follows from an algebraic manipulation of the equation preceding it. Combining this with multiple equalities above, we get $$\|D^{\alpha}u\|_{L^{\infty}(B(x_0,r))}\le M\left(\frac{2^{n+1}n^2e}{r}\right)^{|\alpha|}\alpha! $$ With the understanding of the portion of the proof the proof that I covered, I think I should be able to reason the rest. To sum it up,

  • Why does equation (1) follow from (2)? Why does the RHS of one involve an $L^{\infty}$ norm and not a p-norm.
  • Where did (3) come from? Where did the restriction $\frac{k^k}{k!}\lt e^k$ come from?
  • Is (4) an algebraic manipulation of the sum with $k = \alpha$? I’m pretty sure it is, but I just wanted to check.

Thanks in advance.

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First, the fact that is $\dfrac{k^k}{k!}<e^k$ is a simple consequence of the Stirling's formula $k! = \Big(\dfrac{k}{e}\Big)^k\sqrt{2\pi k}e^{\theta_k}$, where $\theta_k\in(0,1)$. Or one can even prove the assertion using just induction and basic calculus.

This, then, directly implies $(3)$ by taking $k = |\alpha|.$ Note that you forgot to write absolute value outside the exponent of $|\alpha|^{\alpha}$.

$(4)$ is almost trivial - since $n^{|\alpha|} = \sum\limits_{\beta:|\beta|=|\alpha|}\dfrac{|\beta|!}{\beta!}\geq \dfrac{|\alpha|!}{\alpha!}$. Simply forget everything but the case where $\beta = \alpha$ and this is equivalent to $(4).$

Finally, going from $(2)$ to $(1)$ is nothing but an application of the mean value theorem for $u$ (since $u$ is harmonic) and the inequalities we have just established. Try writing them down explicitly, if you don't see it immediately.

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  • $\begingroup$ Why is it \le and not \lt in inequality (3). I’ve also seen, in integral expressions, that instead of =, \le is written instead. I believe this is because when integrating over a ball, the radius varies from 0 to the \mathrm{dist}(x_0,\partial B). $\endgroup$ – user516557 Jan 14 '18 at 14:03
  • $\begingroup$ well, in my version of Evans', the inequality is written as $|\alpha|^{|\alpha|}\leq C|\alpha|!e^{|\alpha|}$, where $C$ is a constant less than 1. This is certainly true and stronger with the $\leq$ sign. Your inequality $(3)$ is weaker and you can replace it with the strict $<$ sign. As for the mean-value theorem, what you saw was probably the version of the theorems for subharmonic or superharmonic functions. In that case, the equality sign is replaced by $\leq$ and $\geq$, respectively. $\endgroup$ – dezdichado Jan 14 '18 at 21:51

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