The fundamental Hessian is not absolutely locally integrable

Put $r=\mathrm{\mid }x\mathrm{\mid }$ and $\omega =x\mathrm{/}r$ for $x\mathrm{\ne }0$. On $(0,\mathrm{\infty })$ the power and logarithm profiles in [F1] are $C^2$ by [F13]–[F15]; $r=({\sum }_k{x}_k^2{)}^{1\mathrm{/}2}$ is $C^2$ on the punctured space by [F8]–[F10]. The chain and product rules [F11]–[F15] give ${\mathrm{\partial }}_{i}q(r)=q^{\mathrm{\prime }}(r)x_i\mathrm{/}r$ and ${\mathrm{\partial }}_j{\mathrm{\partial }}_{i}q(r)=q^{\mathrm{\prime }\mathrm{\prime }}(r)x_i{x}_j\mathrm{/}{r}^2+q^{\mathrm{\prime }}(r)({\delta }_{ij}\mathrm{/}r-x_i{x}_j\mathrm{/}{r}^3)$. For $n\ge 3$ set $c_n=1\mathrm{/}{\omega }_{n-1}$; for $n=2$ set $c_2=1\mathrm{/}(2\pi )$. In both cases $q^{\mathrm{\prime }}=-c_n{r}^{1-n}$ and $q^{\mathrm{\prime }\mathrm{\prime }}=(n-1)c_n{r}^{-n}$, hence ${\mathrm{\partial }}_{ij}\mathrm{\Phi }(x)=c_n\mathrm{\mid }x{\mathrm{\mid }}^{-n}{A}_{ij}(\omega )$ where $A_{ij}(\omega )=n{\omega }_i{\omega }_j-{\delta }_{ij}$. In particular $c_n>0$ and this derivative is even under $x\mapsto -x$.

Fix $\epsilon >0$ and let $h(r)=1\mathrm{/}r$ on $(0,\epsilon )$. For $k\ge 1$, put $I_k=(\epsilon 2^{-k},\epsilon 2^{1-k})$. Its length is $\epsilon 2^{-k}$ by [F21], and $h(r)\ge 2^{k-1}\mathrm{/}\epsilon$ on $I_k$, so [F18] gives ${\int }_{I_k}h(r)dr\ge 1\mathrm{/}2$. The measure $E\mapsto {\int }_{E}h(r)dr$ is countably additive by [F19]–[F20]; the intervals are disjoint, so the integral over $(0,\epsilon )$ is at least $m\mathrm{/}2$ for every finite union of the first $m$ intervals. Letting $m$ increase proves ${\int }_0^{\epsilon }dr\mathrm{/}r=+\mathrm{\infty }$. The interval-measure supplier uses the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ assumption [A1].

On $S^{n-1}$, $A_{ij}$ is bounded and continuous. If $i\mathrm{\ne }j$, reflection of coordinate $i$ preserves $\sigma$ and sends $A_{ij}$ to $-A_{ij}$, so [F23] gives $\int A_{ij}d\sigma =0$. If $i=j$, coordinate permutations make all $\int {\omega }_i^{2}d\sigma$ equal; since ${\sum }_i{\omega }_i^2=1$, linearity [F22] gives each value $\sigma (S^{n-1})\mathrm{/}n$ and again $\int A_{ii}d\sigma =0$. For $i=j$, $A_{ii}(e_i)=n-1>0$ and $A_{ii}(e_k)=-1<0$ for $k\mathrm{\ne }i$; for $i\mathrm{\ne }j$, $A_{ij}((e_i\pm e_j)\mathrm{/}\sqrt{2})=\pm n\mathrm{/}2$. Thus each sign occurs on a nonempty relative-open cap. Choose such a cap $E=\{\omega \in S^{n-1}:\mathrm{\mid }\omega -{\omega }_0\mathrm{\mid }<\eta \}$ small enough that one sign of $A_{ij}$ has magnitude greater than some $c>0$ throughout $E$. For $0<\rho <\min (1\mathrm{/}8,\eta \mathrm{/}8)$ and $z\in B({\omega }_0\mathrm{/}2,\rho )$, one has $0<\mathrm{\mid }z\mathrm{\mid }<1$ and $\mathrm{\mid }z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }-{\omega }_0\mathrm{\mid }\le \mathrm{\mid }1-2\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }+2\mathrm{\mid }z-{\omega }_0\mathrm{/}2\mathrm{\mid }\le 4\rho <\eta$. Hence this Euclidean ball is contained in the cone $\{r\omega :\omega \in E,0<r\le 1\}$. The positive ball measure [F5] and monotonicity applied to indicators [F18] give $\sigma (E)>0$ by [F2]. Since $\sigma$ is finite by [F25], $a_{ij}:={\int }_{S^{n-1}}\mathrm{\mid }{A}_{ij}\mathrm{\mid }d\sigma$ is finite and strictly positive, and the positive and negative angular parts each have positive integral.

Extend the formula in step 1.1 by setting its value at $0$ to $0$; [F24] makes this choice immaterial to its integral. The punctured space is open because it is the union of the open balls $B(x,\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2)$ over $x\mathrm{\ne }0$ [F6], so $\{0\}$ is closed. The extension is continuous on that open set. For every open $V\subseteq \mathbb{R}$, its preimage is the open preimage under the restriction to ${\mathbb{R}}^n\setminus \{0\}$, together with $\{0\}$ exactly when $0\in V$; thus it is Borel. Since open sets generate the Borel sigma-algebra [F7] and inverse images preserve sigma-algebra operations, the extension is Borel measurable. For $B_{\epsilon }(0)$, [F4] and step 2.1 give ${\int }_{B_{\epsilon }(0)}\mathrm{\mid }{\mathrm{\partial }}_{ij}\mathrm{\Phi }(x)\mathrm{\mid }dx=c_n{a}_{ij}{\int }_0^{\epsilon }\frac{dr}{r}=+\mathrm{\infty }$ by step 1.2. Since $c_n>0$ and $a_{ij}>0$, [F16] proves ${\mathrm{\partial }}_{ij}\mathrm{\Phi }\mathrm{\notin }{L}^1(B_{\epsilon }(0))$ for every $\epsilon >0$, and [F17] says it is not locally integrable at the pole. This uses ${\mathrm{A}\mathrm{C}}_{\omega }$ [A1] through the polar formula [F4].

For $0<a<b<\mathrm{\infty }$, the derivative in step 1.1 is integrable on the annulus $a<\mathrm{\mid }x\mathrm{\mid }<b$, since it is bounded there and the annulus has finite measure. Applying [F4] separately to its positive and negative parts and using [F19] and [F22], the signed integral equals $c_n{\int }_a^b{r}^{-1}dr{\int }_{S^{n-1}}{A}_{ij}(\omega )d\sigma (\omega )=0$ by step 2.1. The radial integral is finite because $1\mathrm{/}r$ is bounded on $[a,b]$ and that interval has finite measure [F21]. Thus every concentric annular truncation cancels, although absolute integrability fails at the removed pole.

Let $f(y)={\mathbf{1}}_{B_1(0)}(y)$, a Borel bounded function with $f(0)=1$ by [F6]–[F7]. At $x=0$, evenness from step 1.1 and step 3.1 give $\int \mathrm{\mid }{\mathrm{\partial }}_{ij}\mathrm{\Phi }(-y)f(y)\mathrm{\mid }dy=+\mathrm{\infty }$. Write $A_{ij}^{+}=\max (A_{ij},0)$ and $A_{ij}^{-}=\max (-A_{ij},0)$. More specifically, [F4] on the positive and negative parts gives respectively $c_n\left({\int }_{S^{n-1}}{A}_{ij}^{+}d\sigma \right){\int }_0^1\frac{dr}{r}=+\mathrm{\infty },c_n\left({\int }_{S^{n-1}}{A}_{ij}^{-}d\sigma \right){\int }_0^1\frac{dr}{r}=+\mathrm{\infty },$ since both angular factors are positive by the sign caps in step 2.1 and the radial factor diverges by step 1.2. Each concentric annular truncation inside $B_1$ has signed integral zero by step 3.2. Thus the ordinary Lebesgue integral is undefined there by [F26], and the formal absolute-convergence claim fails at a point where $f$ is nonzero. Countable Choice [A1] is used only through the named measure suppliers [F2]–[F5], [F21], and [F24]; no full AC is invoked.