Strong fractional integration fails at p equal to one

The density and its norm. By [F3] the ball $B(0,ϵ)$ is measurable with $0<\lambda (B(0,ϵ))<\mathrm{\infty }$, so $f_{ϵ}=\lambda (B(0,ϵ){)}^{-1}{1}_{B(0,ϵ)}$ is a well-defined nonnegative measurable function with $\mathrm{\mid }{f}_{ϵ}{\mathrm{\mid }}^1$ integrable and ${\int }_{{\mathbb{R}}^n}{f}_{ϵ}d\lambda =\frac{\lambda (B(0,ϵ))}{\lambda (B(0,ϵ))}=1;$ in particular $\mathrm{\parallel }{f}_{ϵ}{\mathrm{\parallel }}_1=1$ and $f_{ϵ}\in L^1({\mathbb{R}}^n;\mathbb{C})$.

The pointwise lower bound. Fix $x$ with $\mathrm{\mid }x\mathrm{\mid }>2ϵ$. For every $y\in B(0,ϵ)$ the triangle inequality for the Euclidean norm gives $\mathrm{\mid }x-y\mathrm{\mid }\le \mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }<\mathrm{\mid }x\mathrm{\mid }+ϵ<3\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$; since $\alpha -n<0$, raising the positive numbers to the negative power reverses the inequality and $K_{\alpha }(x-y)=\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha -n}\ge {\left(\frac{3\mathrm{\mid }x\mathrm{\mid }}{2}\right)}^{\alpha -n}={\left(\frac{3}{2}\right)}^{\alpha -n}\mathrm{\mid }x{\mathrm{\mid }}^{\alpha -n}\mathrm{.}$ As $f_{ϵ}\ge 0$ and $\int f_{ϵ}=1$, monotonicity and the scalar rule of [F6] applied to the definition [F1] give $I_{\alpha }{f}_{ϵ}(x)=\int K_{\alpha }(x-y)f_{ϵ}(y)dy\ge {\left(\frac{3}{2}\right)}^{\alpha -n}\mathrm{\mid }x{\mathrm{\mid }}^{\alpha -n}\int f_{ϵ}={\left(\frac{3}{2}\right)}^{\alpha -n}\mathrm{\mid }x{\mathrm{\mid }}^{\alpha -n},$ the pointwise absolute convergence being a consequence of the same finite upper bound since $K_{\alpha }(x-y)\le (\mathrm{\mid }x\mathrm{\mid }-ϵ{)}^{\alpha -n}$ on the support for the upper estimate.

Computation of the tail. The function $x\mapsto \mathrm{\mid }x{\mathrm{\mid }}^{-n}{\mathbf{1}}_{\{\mathrm{\mid }x\mathrm{\mid }>2ϵ\}}$ is nonnegative and Borel, so polar coordinates [F4] give ${\int }_E\mathrm{\mid }x{\mathrm{\mid }}^{-n}d{\lambda }_n(x)=\sigma (S^{n-1}){\int }_{2ϵ}^{\mathrm{\infty }}{r}^{-1}d{\lambda }_1(r)\mathrm{.}$ To see that the radial Lebesgue integral is infinite, set $a=2ϵ$ and $J_j=[2^{j}a,2^{j+1}a)$ for $j\ge 0$. These disjoint intervals partition $[a,\mathrm{\infty })$; on $J_j$, $r^{-1}\ge (2^{j+1}a{)}^{-1}$ and ${\lambda }_1(J_j)=2^{j}a$ by [F5]. Thus each ${\int }_{J_j}{r}^{-1}d{\lambda }_1\ge 1\mathrm{/}2$, using [F5] and [F6]. Finite additivity and monotonicity imply ${\int }_a^{\mathrm{\infty }}{r}^{-1}d{\lambda }_1\ge N\mathrm{/}2$ for every positive integer $N$, so it is infinite. Finally $0<\sigma (S^{n-1})<\mathrm{\infty }$: applying [F4] to ${\mathbf{1}}_{B(0,1)}$ gives $\lambda (B(0,1))=\sigma (S^{n-1})\mathrm{/}n$, and [F3] makes the ball measure positive and finite. Hence ${\int }_E\mathrm{\mid }x{\mathrm{\mid }}^{-n}d{\lambda }_n(x)=+\mathrm{\infty }$.

The far tail diverges. Since $q_0=n\mathrm{/}(n-\alpha )$ we have $(\alpha -n)q_0=-n$, so on the measurable set $E:=\{\mathrm{\mid }x\mathrm{\mid }>2ϵ\}$ the lower bound of step 1.2 gives $\mathrm{\mid }{I}_{\alpha }{f}_{ϵ}(x){\mathrm{\mid }}^{q_0}\ge {\left(\frac{3}{2}\right)}^{(\alpha -n)q_0}\mathrm{\mid }x{\mathrm{\mid }}^{(\alpha -n)q_0}={\left(\frac{3}{2}\right)}^{-n}\mathrm{\mid }x{\mathrm{\mid }}^{-n}$. If $I_{\alpha }{f}_{ϵ}$ belonged to $L^{q_0}$, then applicability of [F6] to the nonnegative functions $\mathrm{\mid }{I}_{\alpha }{f}_{ϵ}{\mathrm{\mid }}^{q_0}{1}_E$ and $\mathrm{\mid }x{\mathrm{\mid }}^{-n}{1}_E$ would give ${\int }_E\mathrm{\mid }x{\mathrm{\mid }}^{-n}dx\le {\left(\frac{3}{2}\right)}^n{\int }_E\mathrm{\mid }{I}_{\alpha }{f}_{ϵ}{\mathrm{\mid }}^{q_0}\le {\left(\frac{3}{2}\right)}^n\mathrm{\parallel }{I}_{\alpha }{f}_{ϵ}{\mathrm{\parallel }}_{q_0}^{q_0}<\mathrm{\infty }\mathrm{.}$

Approximate point mass. Let $\phi \in C_c({\mathbb{R}}^n;\mathbb{C})$ be continuous and compactly supported, and fix $\eta >0$. Continuity of $\phi$ at the origin gives $\delta >0$ with $\mathrm{\mid }\phi (y)-\phi (0)\mathrm{\mid }<\eta$ whenever $\mathrm{\mid }y\mathrm{\mid }<\delta$. For every $0<ϵ<\delta$ linearity of the integral [F7] together with the normalization $\int f_{ϵ}=1$ gives $\int \phi f_{ϵ}d\lambda -\phi (0)={\int }_{B(0,ϵ)}(\phi (y)-\phi (0))f_{ϵ}(y)dy,$ and the triangle inequality [F7] and monotonicity of the nonnegative integral [F6] bound its modulus by $\eta \int f_{ϵ}=\eta$. Hence $\int \phi f_{ϵ}\to \phi (0)$ as $ϵ\to 0^{+}$: the normalized balls converge to the point mass at the origin against continuous compactly supported tests.

No strong endpoint estimate. Steps 2.1 and 1.3 are contradictory: if $I_{\alpha }{f}_{ϵ}\in L^{q_0}$ then ${\int }_E\mathrm{\mid }x{\mathrm{\mid }}^{-n}dx<\mathrm{\infty }$, but that integral equals $+\mathrm{\infty }$. Hence $I_{\alpha }{f}_{ϵ}\mathrm{\notin }{L}^{q_0}({\mathbb{R}}^n)$ for every $ϵ>0$. Since $\mathrm{\parallel }{f}_{ϵ}{\mathrm{\parallel }}_1=1$ by step 1.1, no constant $C$ satisfies $\mathrm{\parallel }{I}_{\alpha }f{\mathrm{\parallel }}_{q_0}\le C\mathrm{\parallel }f{\mathrm{\parallel }}_1$ for all $f$ in $L^1({\mathbb{R}}^n;\mathbb{C})$: the family $\{f_{ϵ}{\}}_{ϵ>0}$ alone refutes the estimate.

Conclusion. The normalized ball density $f_{ϵ}$ has unit $L^1$ norm, is an approximate point mass, and its potential has the radial lower bound ${\left(\frac{3}{2}\right)}^{\alpha -n}\mathrm{\mid }x{\mathrm{\mid }}^{\alpha -n}$ outside $B(0,2ϵ)$, whose $q_0$-th power is a nonzero multiple of the divergent tail $\mathrm{\mid }x{\mathrm{\mid }}^{-n}$; therefore the strong $L^1\to L^{q_0}$ endpoint fails. The argument exhibits the failure at fixed $ϵ$ without any limit or Fatou step, and no endpoint case is silently substituted into the strict-range theorem. Countable Choice is used only through the polar and measure interfaces [F3]-[F5] and [F8].