Here is a proof of the analyticity of harmonic functions. This is a bit hard to do from scratch, so I will assume various facts and use notation that is standard (at least in Evans). We denote the volume of the unit ball in \(\mathbb{R}^n\) as \(\alpha(n)\). The surface measure of the unit sphere in \(\mathbb{R}^n\) is then \(n\alpha(n)\). We will also denote multi-indices using \(\alpha\).
We begin with the assumption that \(u\) is harmonic in the open subset \(U\subseteq\mathbb{R}^n\). We will employ the following bound on the derivatives of \(u\). Let \(\alpha\) be a multi-index of order \(|\alpha|=k\), and let \(B_r(x_0)\subseteq U\). Then, \[|D^{\alpha}u(x_0)|\leq\frac{C_k}{r^{n+k}}||u||_{L^1(B_r(x_0))},\] where the constants \(C_k\) are given by \(C_0=\frac{1}{\alpha(n)}\) and \(C_k=\frac{(2^{n+1}nk)^k}{\alpha(n)}\) for \(k>0\). This estimate follows from strong induction on \(k\) and the mean-value property of harmonic functions. Fix \(x_0\in U\). If we set \(r=\frac{1}{4}\inf_{x\in\partial U}{|x-x_0|}\) (which is positive since \(U\) is open) and \(M=\frac{||u||_{L^1(B_{2r}(x_0))}}{\alpha(n)r^n}<\infty\), and apply the above estimate on the derivative, we obtain for any \(x\in B_r(x_0)\) \[|D^{\alpha}u(x)|\leq\frac{(2^{n+1}n|\alpha|)^{|\alpha|}}{r^{n+|\alpha|}\alpha(n)}||u||_{L^1(B_r(x))}=\left[\frac{||u||_{L^1(B_r(x))}}{\alpha(n)r^n}\right]\left(\frac{2^{n+1}n|\alpha|}{r^n}\right)^{|\alpha|}.\] By the triangle inequality, \(B_r(x)\subseteq B_{2r}(x_0)\), so we can modify the bracketed factor on the right hand side of the above inequality to \(M\). This gives us \[|D^{\alpha}u(x)|\leq M\left(\frac{2^{n+1}n}{r}\right)^{|\alpha|}|\alpha|^{|\alpha|}.\] Now, for every \(k\in\mathbb{N}\), notice that \(\frac{k^k}{k!}\) is merely one term in the Taylor expansion of \(e^k\), so \(k^k<e^k\). It follows that \(|\alpha|^{|\alpha|}<e^{|\alpha|}\). Moreover, the multinomial theorem states that \((x_1+\dots+x_n)^k=\sum_{|\alpha|=k}{\binom{|\alpha|}{\alpha}x^{\alpha}}\), where \(x^{\alpha}=\prod_{j=1}^{n}{x_j^{\alpha_j}}\) and \(\binom{|\alpha|}{\alpha}=\frac{|\alpha|!}{\alpha!}=\frac{|\alpha|!}{\prod_{j=1}^{n}{\alpha_j!}}\). It follows that \(n^k=(1+\dots+1)^k=\sum_{|\alpha|=k}{\frac{|\alpha|!}{\alpha!}}\). So for any fixed \(\alpha\), \(\frac{|\alpha|!}{\alpha!}\) is a single term in the expansion of \(n^{|\alpha|}\), and thus \(\frac{|\alpha|!}{\alpha!}\leq n^{|\alpha|}\) or \(|\alpha!|\leq n^{|\alpha|}\alpha!\). Combining all of these estimates, we have \[||D^{\alpha}u||_{L^{\infty}(B_r(x_0))}\leq\sup_{x\in B_r(x_0)}{|D^{\alpha}u(x)|}\leq M\left(\frac{2^{n+1}n}{r}\right)^{|\alpha|}|\alpha|^{|\alpha|}<M\left(\frac{2^{n+1}ne}{r}\right)^{|\alpha|}.\] Since \(1\leq|\alpha|!\leq n^{|\alpha|}\alpha!\), so \(\left(\frac{2^{n+1}ne}{r}\right)^{|\alpha|}\leq\left(\frac{2^{n+1}n^2e}{r}\right)^{|\alpha|}\alpha!\) and we obtain the estimate \[||D^{\alpha}u||_{L^{\infty}(B_r(x_0))}<M\left(\frac{2^{n+1}n^2e}{r}\right)^{|\alpha|}\alpha!\] The Taylor expansion of \(u\) about \(x_0\) is given by \[\sum_{\alpha}{\frac{D^{\alpha}u(x_0)}{\alpha!}(x-x_0)^{\alpha}}.\] We wish to show that this power series converges in some neighborhood. To do this, we study the remainder term, \[R_N(x)=u(x)-\sum_{k=0}^{N-1}{\sum_{|\alpha|=k}{\frac{D^{\alpha}u(x_0)}{\alpha!}(x-x_0)^{\alpha}}}.\] Consider the function \(g\colon\mathbb{R}\to\mathbb{R}\) given by \(g(t)=u(x_0+t(x-x_0))\). By applying the mean-value remainder form of Taylor's theorem to the Taylor expansion of \(g\) about \(0\) evaluated at \(t=1\), we obtain \[R_N(x)=g(1)-\sum_{k=0}^{N-1}{\sum_{|\alpha|=k}{\frac{D^{\alpha}u(x_0)}{\alpha!}(x-x_0)^{\alpha}}}=\sum_{|\alpha|=N}{\frac{D^{\alpha}u(x_0+\lambda(x-x_0))}{\alpha!}(x-x_0)^{\alpha}}\] for some \(\lambda\in[0,1]\). Now, suppose that we are considering \(x\) such that \(|x-x_0|<\frac{r}{2^{n+2}n^3e}\). In particular, we will have \(x\in B_r(x_0)\), so by the convexity of the ball, \(x_0+\lambda(x-x_0)\in B_r(x_0)\) and we will have access to our estimate of the \(L^{\infty}\) norm of \(D^{\alpha}u\). Combining all of the bounds, we obtain \[|R_n(x)|\leq M\sum_{|\alpha|=N}{\left(\frac{2^{n+1}n^2e}{r}\right)^N\left(\frac{r}{2^{n+2}n^3e}\right)^N}=M\sum_{|\alpha|=N}{\frac{1}{(2n)^N}}.\] The relevant multi-indices with order \(N\) that we are summing over are \(N\)-tuples of integers from the set \(\{1,2,\dots,n\}\), because it is these integers that correspond to the components that can exist in \(\mathbb{R}^n\). Hence, there are \(n^N\) multi-indices we are summing over, so \[M\sum_{|\alpha|=N}{\frac{1}{(2n)^N}}=\frac{Mn^N}{(2n)^N}=\frac{M}{2^N}.\] Thus, the remainder vanishes as we take \(N\to\infty\) and the Taylor series converges in the neighborhood we have defined.
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