**Idea of the proof.** Fix $x_0 \in U$ and choose $r > 0$ so that $B(x_0,2r) \subset U$. By the interior derivative estimate from [Theorem 37](/theorems/7), derivatives of order $k$ at any point of $B(x_0,r)$ are controlled by $\|u\|_{L^1(B(x_0,2r))}$ multiplied by a factor $(c/r)^{k}$ with $c$ depending only on $n$. Using Stirling’s formula and the multinomial theorem this yields, for every multiindex $\alpha$, \begin{align*} \|D^\alpha u\|_{L^\infty(B(x_0,r))} \le C\,M \left(\frac{2^{\,n+1} n\,e}{r}\right)^{|\alpha|}\alpha!, \end{align*} with $M$ independent of $\alpha$. This bound implies absolute convergence of the Taylor series for $|x-x_0| < r/(2^{\,n+1} n e)$, and the remainder tends to $0$ by a one–variable Taylor argument along the segment from $x_0$ to $x$.

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**Step 1 (fix the scale and define $M$).** Choose \begin{align*} r:= \tfrac{1}{4}\,\mathrm{dist}(x_0,\partial U) \quad \Rightarrow \quad B(x_0,2r) \subset U. \end{align*} Set \begin{align*} M := \frac{\Gamma\!\big(1 + \tfrac{n}{2}\big)}{\pi^{n/2}} \,\frac{1}{r^{n}} \, \|u\|_{L^1(B(x_0,2r))}, \end{align*} which is finite.

**Step 2 (uniform derivative bounds on $B(x_0,r)$).** Fix a multiindex $\alpha$ with $|\alpha|=k \ge 1$. For any $x \in B(x_0,r)$ we have $B(x,r) \subset B(x_0,2r) \subset U$, hence [Theorem 37](/theorems/37) (with first–order step iterated $k$ times) yields \begin{align*} |D^\alpha u(x)| &\le \frac{(2^{\,n+1} n)^k}{r^{k}} \cdot \frac{\Gamma\!\big(1 + \tfrac{n}{2}\big)}{\pi^{n/2}} \,\frac{1}{r^{n}} \,\|u\|_{L^1(B(x_0,2r))} \\ &= M \left( \frac{2^{\,n+1} n}{r} \right)^{k} \, k^{\,k} \cdot \frac{1}{k^{\,k}}. \tag{A} \end{align*} To pass from $k$–wise iteration to a factorial–type bound we use two combinatorial/analytic facts. First, Stirling’s formula implies \begin{align*} k^{\,k} \le C_{\mathrm{St}}\, e^{\,k} \, k! \quad \text{for all } k \in \mathbb{N}. \tag{B} \end{align*} Second, the multinomial identity \begin{align*} n^k = (1+\cdots+1)^k = \sum_{|\alpha|=k} \frac{k!}{\alpha!} \end{align*} yields, for each $\alpha$ with $|\alpha|=k$, \begin{align*} \frac{k!}{\alpha!} \le n^k \quad \Longrightarrow \quad k! \le n^{k}\, \alpha!. \tag{C} \end{align*} Combining (A), (B), (C) and absorbing harmless constants into a single $C \ge 1$ (independent of $\alpha$) gives the uniform bound \begin{align*} \|D^\alpha u\|_{L^\infty(B(x_0,r))} \le C\, M \left( \frac{2^{\,n+1} n\, e}{r} \right)^{|\alpha|} \alpha!. \tag{22} \end{align*}

**Step 3 (convergence of the Taylor series at $x_0$).** Let \begin{align*} \rho := \frac{r}{2^{\,n+1} n\, e}. \end{align*} For $x$ with $|x-x_0| < \rho$ the series \begin{align*} \sum_{\alpha \in \mathbb{N}_0^n} \frac{D^\alpha u(x_0)}{\alpha!}\,(x-x_0)^\alpha \end{align*} is absolutely convergent by (22): \begin{align*} \sum_{\alpha} \left| \frac{D^\alpha u(x_0)}{\alpha!} (x-x_0)^\alpha \right| &\le \sum_{\alpha} C\,M \left( \frac{2^{\,n+1} n\, e}{r} \right)^{|\alpha|} \alpha! \cdot \frac{|x-x_0|^{|\alpha|}}{\alpha!} \\ &= C\,M \sum_{k=0}^\infty \left( \frac{|x-x_0|}{\rho} \right)^{k} \sum_{|\alpha|=k} 1 \le C'\,M \sum_{k=0}^\infty \left( \frac{|x-x_0|}{\rho} \right)^{k}, \tag{since $\#\{\alpha:|\alpha|=k\}\le C'(n)\,(k+1)^{n-1}$} \end{align*} and the last geometric series converges because $|x-x_0|/\rho < 1$.

**Step 4 (identify $u$ with its Taylor series via a remainder estimate).** Fix $x \in B(x_0,\rho)$. Consider the one–variable map $g(t):=u(x_0+t(x-x_0))$ on $[0,1]$. Taylor’s theorem with integral (or Cauchy) remainder gives \begin{align*} u(x) &= \sum_{|\alpha| < N} \frac{D^\alpha u(x_0)}{\alpha!} (x-x_0)^\alpha + R_N(x), \\ R_N(x) &= \sum_{|\alpha|=N} \frac{D^\alpha u(x_0+t(x)(x-x_0))}{\alpha!}\,(x-x_0)^\alpha \quad \text{for some } t(x) \in (0,1). \tag{Cauchy form along the segment} \end{align*} Because $|x-x_0| < \rho \le r$, the point $x_0+t(x)(x-x_0)$ lies in $B(x_0,r)$, hence (22) gives \begin{align*} |R_N(x)| &\le \sum_{|\alpha|=N} C\,M \left( \frac{2^{\,n+1} n\, e}{r} \right)^{N} \alpha! \cdot \frac{|x-x_0|^{N}}{\alpha!} = C\,M \left( \frac{|x-x_0|}{\rho} \right)^{N} \sum_{|\alpha|=N} 1 \\ &\le C''\,M \left( \frac{|x-x_0|}{\rho} \right)^{N} \to 0 \quad \text{as } N \to \infty, \end{align*} because $|x-x_0|/\rho < 1$. Therefore the Taylor series converges to $u(x)$ on $B(x_0,\rho)$.

Since $x_0 \in U$ was arbitrary, $u$ is analytic in $U$.