[proofplan] We fix $x_0 \in U$, choose a radius $r$ so that $B(x_0, 2r) \subset U$, and use the interior derivative estimate from [Theorem 37](/theorems/37) to bound $|D^\alpha u(x)|$ for $x \in B(x_0, r)$ and any multi-index $\alpha$. Combining Stirling's formula ($k^k \le C\, e^k\, k!$) with the multinomial identity ($k! \le n^k\, \alpha!$) converts the $k$-th order bound into factorial-type Cauchy estimates $|D^\alpha u(x)| \le C\, M\, (A/r)^{|\alpha|}\, \alpha!$. These estimates imply absolute convergence of the Taylor series on a ball of radius $\rho = r/(2^{n+1} n^2 e)$, and a remainder estimate via Taylor's theorem with Cauchy remainder shows the series converges to $u$. [/proofplan] [step:Fix the scale and define the controlling constant $M$] Fix $x_0 \in U$ and choose \begin{align*} r := \tfrac{1}{4}\, \operatorname{dist}(x_0, \partial U), \end{align*} so that $B(x_0, 2r) \subset U$. Define \begin{align*} M := \frac{1}{\omega_n\, r^n}\, \|u\|_{L^1(B(x_0, 2r))}, \end{align*} where $\omega_n = \pi^{n/2}/\Gamma(1 + n/2)$ is the volume of the unit ball. Since $u$ is [continuous](/page/Continuity) on the compact set $\overline{B}(x_0, 2r) \subset U$, the integral is finite, so $M < \infty$. [/step] [step:Derive uniform derivative bounds on $B(x_0, r)$ using the interior estimates] Fix a multi-index $\alpha$ with $|\alpha| = k \ge 1$. For any $x \in B(x_0, r)$, the triangle inequality gives $B(x, r) \subset B(x_0, 2r) \subset U$. Applying [Theorem 37](/theorems/37) (interior [derivative](/page/Derivative) estimates) on $B(x, r)$: \begin{align*} |D^\alpha u(x)| \le \frac{C_k}{r^{n+k}}\, \|u\|_{L^1(B(x, r))} \le \frac{C_k}{r^{n+k}}\, \|u\|_{L^1(B(x_0, 2r))}, \end{align*} where the second inequality uses $B(x, r) \subset B(x_0, 2r)$ to enlarge the domain. The recursive constant from [Theorem 37](/theorems/37) satisfies $C_k \le C_0 \cdot (2^{n+1} n)^k \cdot k^k$. Indeed, the recursion $C_k = nk(k/(k-1))^{n+k-1} C_{k-1}$ and the bound $(k/(k-1))^{n+k-1} \le e^{(n+k-1)/(k-1)} \le e^{n+1}$ (since $k \ge 1$ gives $(n+k-1)/(k-1) \le n+1$) yield $C_k \le 2^{n+1} n k \cdot k^{k-1}/(k-1)^{k-1} \cdot C_{k-1}$; telescoping the product and using $\prod_{j=1}^{k} j \cdot j^{j-1}/(j-1)^{j-1} \le (2^{n+1} n)^k k^k$ gives the stated bound. Therefore \begin{align*} |D^\alpha u(x)| \le M \left(\frac{2^{n+1} n}{r}\right)^k k^k. \end{align*} [guided] The interior derivative estimate from [Theorem 37](/theorems/37) controls $|D^\alpha u(x)|$ in terms of $\|u\|_{L^1}$ over a ball centred at $x$. We need this estimate to be uniform over $x \in B(x_0, r)$, which is why we chose $r = \frac{1}{4} \operatorname{dist}(x_0, \partial U)$: for any $x \in B(x_0, r)$, the ball $B(x, r)$ lies inside $B(x_0, 2r) \subset U$, so the estimate applies on $B(x, r)$, and we can enlarge the domain of the $L^1$ norm from $B(x, r)$ to $B(x_0, 2r)$. Fix $\alpha$ with $|\alpha| = k \ge 1$. For $x \in B(x_0, r)$, [Theorem 37](/theorems/37) on $B(x, r)$ gives \begin{align*} |D^\alpha u(x)| \le \frac{C_k}{r^{n+k}}\, \|u\|_{L^1(B(x, r))} \le \frac{C_k}{r^{n+k}}\, \|u\|_{L^1(B(x_0, 2r))} = C_k \cdot \omega_n \cdot M \cdot r^{-k}. \end{align*} The constant $C_k$ from [Theorem 37](/theorems/37) is defined recursively by $C_k = nk(k/(k-1))^{n+k-1} C_{k-1}$. To convert this into a form suitable for Taylor series estimates, we need an upper bound in terms of $k^k$ or $k!$. The factor $(k/(k-1))^{n+k-1}$ is bounded by $e^{(n+k-1)/(k-1)}$ (using $(1 + 1/(k-1))^{k-1} \le e$), so inductively $C_k \le C_0 (2^{n+1} n)^k k^k$. This gives \begin{align*} |D^\alpha u(x)| \le M \left(\frac{2^{n+1} n}{r}\right)^k k^k. \end{align*} The factor $k^k$ is not yet in factorial form -- the next step converts it. [/guided] [/step] [step:Convert the $k^k$ bound to a factorial bound using Stirling and the multinomial identity] We apply two combinatorial estimates to convert the bound into one involving $\alpha!$. **Stirling bound.** Stirling's formula implies that for all $k \ge 1$, \begin{align*} k^k \le C_{\mathrm{St}}\, e^k\, k!, \end{align*} where $C_{\mathrm{St}} \ge 1$ is an absolute constant. **Multinomial identity.** The multinomial theorem gives $n^k = \sum_{|\alpha| = k} k!/\alpha!$, so for each $\alpha$ with $|\alpha| = k$: \begin{align*} \frac{k!}{\alpha!} \le n^k \quad \implies \quad k! \le n^k\, \alpha!. \end{align*} Combining these with the derivative bound from the previous step and absorbing all constants independent of $\alpha$ into a single $C \ge 1$: \begin{align*} \|D^\alpha u\|_{L^\infty(B(x_0, r))} \le C\, M \left(\frac{2^{n+1} n^2\, e}{r}\right)^{|\alpha|} \alpha!. \end{align*} [guided] The bound $|D^\alpha u(x)| \le M (2^{n+1} n / r)^k k^k$ involves $k^k$, which grows faster than $k!$. For the [Taylor series](/page/Power%20Series) to converge, we need the coefficients $|D^\alpha u(x_0)|/\alpha!$ to be bounded by a geometric sequence. Two estimates bridge this gap. **Stirling's formula** gives $k! \sim \sqrt{2\pi k}\, (k/e)^k$, which implies $k^k \le C_{\mathrm{St}}\, e^k\, k!$ for a universal constant $C_{\mathrm{St}}$. This absorbs the $k^k$ into $e^k \cdot k!$. **The multinomial identity** relates $k!$ to $\alpha!$: from $(1 + \cdots + 1)^k = n^k = \sum_{|\alpha|=k} k!/\alpha!$, each summand satisfies $k!/\alpha! \le n^k$, i.e., $k! \le n^k \alpha!$. This replaces $k!$ by $n^k \alpha!$, which is what the Taylor series denominator contains. Chaining the estimates: $k^k \le C_{\mathrm{St}}\, e^k\, k! \le C_{\mathrm{St}}\, e^k\, n^k\, \alpha! = C_{\mathrm{St}}\, (ne)^k\, \alpha!$. Substituting into the derivative bound: \begin{align*} |D^\alpha u(x)| &\le M \left(\frac{2^{n+1} n}{r}\right)^k \cdot C_{\mathrm{St}}\, (ne)^k\, \alpha! = C_{\mathrm{St}}\, M \left(\frac{2^{n+1} n^2 e}{r}\right)^k \alpha!. \end{align*} Defining $A := 2^{n+1} n^2 e$ and absorbing $C_{\mathrm{St}}$ into a constant $C \ge 1$ depending only on $n$, the final bound is \begin{align*} \|D^\alpha u\|_{L^\infty(B(x_0, r))} \le C\, M \left(\frac{2^{n+1} n^2\, e}{r}\right)^{|\alpha|} \alpha!, \end{align*} with $C \ge 1$ depending only on $n$. [/guided] [/step] [step:Prove absolute convergence of the Taylor series on $B(x_0, \rho)$] Define \begin{align*} \rho := \frac{r}{2^{n+1} n^2\, e}. \end{align*} For $x \in B(x_0, \rho)$, the [Taylor series](/page/Power%20Series) satisfies \begin{align*} \sum_{\alpha \in \mathbb{N}_0^n} \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 \#\{\alpha : |\alpha| = k\}. \end{align*} The number of multi-indices of order $k$ in $n$ variables satisfies $\#\{\alpha \in \mathbb{N}_0^n : |\alpha| = k\} = \binom{n + k - 1}{k} \le C'(n)\, (k+1)^{n-1}$. Since the polynomial $(k+1)^{n-1}$ does not affect convergence of a geometric series with ratio $|x - x_0|/\rho < 1$, the series converges absolutely. [/step] [step:Identify $u$ with its Taylor series via a remainder estimate] Fix $x \in B(x_0, \rho)$ and consider the one-variable function \begin{align*} g: [0,1] &\to \mathbb{R} \\ t &\mapsto u(x_0 + t(x - x_0)). \end{align*} Since $|x - x_0| < \rho \le r$ and $t \in [0,1]$, the point $x_0 + t(x - x_0)$ lies in $B(x_0, r) \subset U$, so $g$ is well-defined and smooth. The chain rule gives $g^{(N)}(t) = \sum_{|\alpha| = N} \frac{N!}{\alpha!}\, D^\alpha u(x_0 + t(x - x_0))\, (x - x_0)^\alpha$, since $(x - x_0)$ is a fixed direction and $u$ is smooth. Taylor's theorem applied to $g$ with the Cauchy form of the remainder gives \begin{align*} u(x) = \sum_{|\alpha| < N} \frac{D^\alpha u(x_0)}{\alpha!}\, (x - x_0)^\alpha + R_N(x), \end{align*} where, for some $t_N \in (0,1)$, \begin{align*} |R_N(x)| \le \sum_{|\alpha| = N} \frac{|D^\alpha u(x_0 + t_N(x - x_0))|}{\alpha!}\, |x - x_0|^N. \end{align*} The point $x_0 + t_N(x - x_0) \in B(x_0, r)$, so the derivative bound from the earlier step applies: \begin{align*} |R_N(x)| &\le \sum_{|\alpha| = N} C\, M \left(\frac{2^{n+1} n^2\, e}{r}\right)^N \alpha! \cdot \frac{|x - x_0|^N}{\alpha!} = C\, M \left(\frac{|x - x_0|}{\rho}\right)^N \cdot \#\{\alpha : |\alpha| = N\}. \end{align*} Since $|x - x_0|/\rho < 1$ and $\#\{\alpha : |\alpha| = N\}$ grows polynomially in $N$, we have $R_N(x) \to 0$ as $N \to \infty$. Therefore \begin{align*} u(x) = \sum_{\alpha \in \mathbb{N}_0^n} \frac{D^\alpha u(x_0)}{\alpha!}\, (x - x_0)^\alpha \quad \text{for all } x \in B(x_0, \rho), \end{align*} with absolute convergence. Since $x_0 \in U$ was arbitrary, $u$ is real-analytic in $U$. [/step]