Boundary-scale derivative blowup despite bounded ball data

Work under [F8], fix $n\ge 3$ and $k\ge 2$, and define $u_k:{\mathbb{R}}^n\to \mathbb{R}$ by $u_k(x):=\mathrm{Re}((x_1+i{x}_2{)}^k)$, so that $u_k(x)=p_k(x_1,x_2)$ for the polynomial $p_k$ of [F2]; in particular $u_k$ is a polynomial, hence of class $C^{\mathrm{\infty }}$ on ${\mathbb{R}}^n$. It is harmonic on ${\mathbb{R}}^n$: it does not depend on $x_3,\dots ,x_n$, so those second partial derivatives vanish by [F3], while ${\mathrm{\partial }}_1^2{u}_k$ and ${\mathrm{\partial }}_2^2{u}_k$ are the corresponding partial derivatives of $p_k$ evaluated at $(x_1,x_2)$, so $\mathrm{\Delta }{u}_k={\mathrm{\partial }}_1^2{u}_k+{\mathrm{\partial }}_2^2{u}_k+{\sum }_{i\ge 3}{\mathrm{\partial }}_i^2{u}_k={\mathrm{\partial }}_1^2{p}_k+{\mathrm{\partial }}_2^2{p}_k=0$ by [F2] and [F3].

Bounded continuous trace. The restriction $g_k=u_k{\mathrm{\mid }}_{\mathrm{\partial }{B}_1(0)}$ of the polynomial $u_k$ is continuous on $\mathrm{\partial }{B}_1(0)$. For $y\in \mathrm{\partial }{B}_1(0)$ put $w:=y_1+i{y}_2$, so that $g_k(y)=\mathrm{Re}(w^k)$; by [F4] $(\mathrm{Re}{w}^k{)}^2\le \mathrm{\mid }{w}^k{\mathrm{\mid }}^2=\mathrm{\mid }w{\mathrm{\mid }}^{2k}=(y_1^2+y_2^2{)}^k$ and $y_1^2+y_2^2\le \parallel y{\parallel }_2^2=1$, so $(y_1^2+y_2^2{)}^k\le 1$ by the last two clauses of [F4]; taking nonnegative square roots with [F4] gives $\mathrm{\mid }{g}_k(y)\mathrm{\mid }\le 1$. Hence $\parallel g_k{\parallel }_{\mathrm{\infty }}\le 1$. Moreover the same computation with $\parallel x{\parallel }_2\le 1$ in place of $\parallel y{\parallel }_2=1$ gives ${\sup }_{B_1(0)}\mathrm{\mid }{u}_k\mathrm{\mid }\le 1$.

The partial derivative at $x_k$. Write $t_0:=1-1\mathrm{/}k>0$, so $x_k=t_0{e}_1$. By [F3] the partial derivative ${\mathrm{\partial }}_1{u}_k(x_k)$ is the derivative at $t=0$ of the one-variable map $t\mapsto u_k(x_k+t{e}_1)=\mathrm{Re}(((t_0+t)+i\cdot 0{)}^k)=(t_0+t{)}^k$, for $t$ near $0$ (where $t_0+t>0$); by [F5] this derivative equals $k(t_0+t{)}^{k-1}$ at $t=0$, so ${\mathrm{\partial }}_1{u}_k(x_k)=k(1-1\mathrm{/}k{)}^{k-1}>0$.

Poisson representation. By [F1] with $R=1$, $a=0$ and datum $g_k\in C(\mathrm{\partial }{B}_1(0))$, the Poisson integral $U_{g_k}$ is smooth and harmonic on $B_1(0)$, continuous on $\overline{B_1(0)}$ with trace $g_k$, and is the unique such function. Steps 1.1 and 2.1 show that $u_k\in C^2(B_1(0))\cap C(\overline{B_1(0)})$ is harmonic on $B_1(0)$ with trace $g_k$; hence $u_k=U_{g_k}$: each $u_k$ is exactly the Poisson extension of its boundary datum.

Divergence at boundary scale. By [F7] applied to $x=-1$, the sequence $a_k:=(1-1\mathrm{/}k{)}^k=(1+(-1)\mathrm{/}k{)}^k$ converges to $\exp (-1)>0$; choose $K$ with $a_k\ge \exp (-1)\mathrm{/}2$ for all $k\ge K$. Since $0<1-1\mathrm{/}k\le 1$ for $k\ge 2$, one has $(1-1\mathrm{/}k{)}^{k-1}=a_k\mathrm{/}(1-1\mathrm{/}k)\ge a_k\ge \exp (-1)\mathrm{/}2$ for all $k\ge K$, and therefore step 2.2 gives $\mid {\mathrm{\partial }}_1{u}_k(x_k)\mid \ge k\mathrm{/}(2e)$ for $k\ge K$; given any real $M$, every $k\ge \max \{K,2eM\}$ satisfies $k\mathrm{/}(2e)\ge M$, so $\mid {\mathrm{\partial }}_1{u}_k(x_k)\mid \to +\mathrm{\infty }$.

The correct estimate carries the inverse distance, and no distance-free bound can hold. First, $x_k$ lies in $B_1(0)$ with $\parallel x_k{\parallel }_2=1-1\mathrm{/}k$, and for $y\in \mathrm{\partial }{B}_1(0)$ the triangle inequality of [F4] gives $\parallel x_k-y{\parallel }_2\ge \parallel y{\parallel }_2-\parallel x_k{\parallel }_2=1\mathrm{/}k$, with equality for $y=e_1$; so the distance from $x_k$ to the boundary is exactly $1\mathrm{/}k$, and the ball $B_{1\mathrm{/}k}(x_k)$ is contained in $B_1(0)$ because $\parallel z{\parallel }_2\le \parallel z-x_k{\parallel }_2+\parallel x_k{\parallel }_2<1\mathrm{/}k+1-1\mathrm{/}k=1$ for $z\in B_{1\mathrm{/}k}(x_k)$. Second, $u_k$ and $f:=0$ satisfy the hypotheses of [F6] with centre $x_k$ and radius $r=1\mathrm{/}k$, so $\mid {\mathrm{\partial }}_1{u}_k(x_k)\mid \le \parallel D{u}_k{\parallel }_{\mathrm{\infty };B_{1\mathrm{/}(2k)}(x_k)}\le C_{n}k\parallel u_k{\parallel }_{\mathrm{\infty };B_{1\mathrm{/}k}(x_k)}\le C_{n}k$, using $\parallel u_k{\parallel }_{\mathrm{\infty };B_1(0)}\le 1$ from step 2.1: the scale-aware bound grows like $1\mathrm{/}r=k$, exactly as the family $u_k$ does, so [F6] is not contradicted. Third, a bound with a constant depending only on $n$ and on the boundary supremum norm would give $\mid {\mathrm{\partial }}_1{u}_k(x_k)\mid \le C(n)$ for every $k\ge 2$, since $\parallel g_k{\parallel }_{\mathrm{\infty }}\le 1$ by step 2.1 and $u_k$ is the harmonic extension of $g_k$ by step 3.1; that is impossible because step 3.2 makes the left-hand side tend to $+\mathrm{\infty }$. Hence any interior gradient estimate for harmonic functions must degenerate as the distance to the boundary tends to zero.

Summary. The harmonic polynomials $u_k(x)=\mathrm{Re}((x_1+i{x}_2{)}^k)$ on the unit ball have Poisson boundary data $g_k$ with $\parallel g_k{\parallel }_{\mathrm{\infty }}\le 1$ and satisfy $\mid {\mathrm{\partial }}_1{u}_k(x_k)\mid =k(1-1\mathrm{/}k{)}^{k-1}\to +\mathrm{\infty }$ at points $x_k$ of distance $1\mathrm{/}k$ from the boundary, so the interior gradient bound cannot be extended to points whose distance to the boundary tends to zero with a constant depending only on the fixed ball and the boundary supremum norm.