Complex translation, convolution, approximate identities, and mollification

Write $f=u+iv$. Translation invariance F8 gives $\int \mathrm{\mid }f(x-h){\mathrm{\mid }}^{p}dx=\int \mathrm{\mid }f(x){\mathrm{\mid }}^{p}dx$ for finite $p$: first substitute translated level sets for a nonnegative simple function, then take the supremum defining its nonnegative integral. For infinity the sets $\{\mathrm{\mid }f(\cdot -h)\mathrm{\mid }>a\}$ have the same measure as $\{\mathrm{\mid }f\mathrm{\mid }>a\}$, so the essential bounds coincide. Null disagreement sets also translate to null sets, so translations act on classes. For finite $p$, F3 and F2 give $\mathrm{\parallel }{\tau }_{h}f-f{\mathrm{\parallel }}_p\le \mathrm{\parallel }{\tau }_{h}u-u{\mathrm{\parallel }}_p+\mathrm{\parallel }{\tau }_{h}v-v{\mathrm{\parallel }}_p\to 0$. The isometry gives continuity at every $h_0$ from $\mathrm{\parallel }{\tau }_{h}f-{\tau }_{h_0}f{\mathrm{\parallel }}_p=\mathrm{\parallel }{\tau }_{h-h_0}f-f{\mathrm{\parallel }}_p$.

A complex $f\in C_0$ is bounded: its components are bounded on a sufficiently large closed ball by F17 and F19 and are small outside by definition. It is uniformly continuous: for a given $\eta >0$, take $R$ with $\mathrm{\mid }f(x)\mathrm{\mid }<\eta \mathrm{/}2$ for $\mathrm{\mid }x\mathrm{\mid }>R$ and use F18 on the closed ball of radius $R+2$ to choose $0<\delta <1$ making differences below $\eta$ there. If $\mathrm{\mid }x-z\mathrm{\mid }<\delta$, either both points are in that ball or both have radius greater than $R$, in which case their values differ by less than $\eta$. This proves uniform continuity. Hence $K\ast f$ exists everywhere and $\mathrm{\mid }(K\ast f)(x+z)-(K\ast f)(x)\mathrm{\mid }\le \mathrm{\parallel }K{\mathrm{\parallel }}_1{\sup }_w\mathrm{\mid }f(w+z)-f(w)\mathrm{\mid }\to 0$. As $\mathrm{\mid }x\mathrm{\mid }\to \mathrm{\infty }$, $f(x-y)\to 0$ for fixed $y$, dominated by the integrable function $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }K(y)\mathrm{\mid }$; F11 gives $(K\ast f)(x)\to 0$. Thus $K\ast f\in C_0$.

For a fixed $K\in L^1$ of integral one, substitution $y=x\mathrm{/}\epsilon$ using F9 gives $\int K_{\epsilon }=\int K=1$, $\mathrm{\parallel }{K}_{\epsilon }{\mathrm{\parallel }}_1=\mathrm{\parallel }K{\mathrm{\parallel }}_1$, and ${\int }_{\mathrm{\mid }x\mathrm{\mid }\ge \delta }\mathrm{\mid }{K}_{\epsilon }(x)\mathrm{\mid }dx={\int }_{\mathrm{\mid }y\mathrm{\mid }\ge \delta \mathrm{/}\epsilon }\mathrm{\mid }K(y)\mathrm{\mid }dy\to 0$ by F11, dominated by $\mathrm{\mid }K\mathrm{\mid }$. Thus scaled integrable mass-one kernels satisfy all three conditions, whether or not they are nonnegative.

For locally integrable $f=u+iv$, F1 gives locally integrable real components. Apply F12 to each with the real kernel ${\rho }_{\epsilon }$. By F15 their recombination gives all ordered partial derivatives, equal to the integrals with the corresponding kernel derivatives. To check their continuity explicitly, fix $x_0$ and a closed unit ball of x-values about it. The y-supports of ${\mathrm{\partial }}^{\alpha }{\rho }_{\epsilon }(x-y)$ for those x-values lie in a fixed closed bounded ball $Q$, compact by F17. The derivative of the kernel is globally bounded, say by $C_{\alpha }$, by F19 on a ball containing its compact support. Thus the integrands are dominated by $C_{\alpha }\mathrm{\mid }f\mathrm{\mid }{\mathbf{1}}_Q$, integrable by local integrability. As $x\to x_0$ their pointwise limits are the integrands at $x_0$, so F11 gives continuity of every derivative integral. The change of variables between the two convolution orders follows from F8–F9. Hence the convolution is smooth with the stated formula.

Under the given countable-choice hypothesis F14, F6 applied to each component supplies Borel representatives of $K,f$; any infinite component values in the base representative occur on a Borel null set and can be set to zero there. F7 makes $(x,y)\mapsto K(y)f(x-y)$ jointly Borel. Both Euclidean factors are sigma-finite, covered by the finite-measure boxes of F20. For $p<\mathrm{\infty }$, apply F5 to $F(x,y)=K(y)f(x-y)$: step 1.1 gives $\mathrm{\parallel }F(\cdot ,y){\mathrm{\parallel }}_p=\mathrm{\mid }K(y)\mathrm{\mid }\mathrm{\parallel }f{\mathrm{\parallel }}_p$, so the envelope $A(x)=\int \mathrm{\mid }F(x,y)\mathrm{\mid }dy$ belongs to $L^p$ and $\mathrm{\parallel }A{\mathrm{\parallel }}_p\le \mathrm{\parallel }K{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_p$. For $p=\mathrm{\infty }$, modify the Borel representative of $f$ on the Borel null set where $\mathrm{\mid }f\mathrm{\mid }>\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, setting it to zero. Then $A(x)\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }K{\mathrm{\parallel }}_1$ for every $x$. By F21, $A$ and the positive and negative parts of the real and imaginary integrands have measurable section integrals. Where $A(x)<\mathrm{\infty }$, their finite differences define a measurable complex integral with modulus at most $A(x)$ by F16; set it to zero on the measurable exceptional set. These bounds prove absolute a.e. existence, measurability and the sharp $L^p$ norm estimate independently of real Young.

For $f\in C_0$, put ${\omega }_f(\delta )={\sup }_{\mathrm{\mid }y\mathrm{\mid }<\delta ,x}\mathrm{\mid }f(x-y)-f(x)\mathrm{\mid }$. Step 1.2 gives ${\omega }_f(\delta )\to 0$. Taking pointwise absolute values in the normalized error integral and then the supremum gives $\mathrm{\parallel }{K}_{\epsilon }\ast f-f{\mathrm{\parallel }}_{\mathrm{s}\mathrm{u}\mathrm{p}}\le M{\omega }_f(\delta )+2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{s}\mathrm{u}\mathrm{p}}{\int }_{\mathrm{\mid }y\mathrm{\mid }\ge \delta }\mathrm{\mid }{K}_{\epsilon }(y)\mathrm{\mid }dy$. First send $\epsilon \downarrow 0$, then $\delta \downarrow 0$. This proves the uniform assertion by a direct supremum estimate.

For any other measurable representatives let $N_K,N_f$ be their measurable null disagreement sets. For each fixed $x$, the integrands coincide outside $N_K\cup (x-N_f)$, a measurable null set by F8–F9. The same is true of their absolute values, so absolute integrability holds for either pair exactly when it holds for the other. At those points F10 gives equal integrals. Thus the measurable class in step 2.1 is representative-independent throughout $L^1$ times Lp, without restricting both inputs to $L^1$.

For finite $p$, normalization and F15 give $K_{\epsilon }\ast f-f=\int K_{\epsilon }(y)(f(\cdot -y)-f(\cdot ))dy$ a.e. Its absolute value is bounded by the envelope with integrand $\mathrm{\mid }{K}_{\epsilon }(y)\mathrm{\mid }\mathrm{\mid }f(x-y)-f(x)\mathrm{\mid }$ by F16. This integrand is measurable by F7 applied to Borel representatives and ordinary products. Its section norm is $\mathrm{\mid }{K}_{\epsilon }(y)\mathrm{\mid }\mathrm{\parallel }{\tau }_{y}f-f{\mathrm{\parallel }}_p$, measurable by step 1.1 and bounded by $2\mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{\mid }{K}_{\epsilon }(y)\mathrm{\mid }$, which is integrable. Euclidean Lebesgue measure is sigma-finite, since $[-j,j{]}^n$ for positive integers $j$ cover it and have finite measure $(2j{)}^n$ by F20. Therefore F5 applies and gives $\mathrm{\parallel }{K}_{\epsilon }\ast f-f{\mathrm{\parallel }}_p\le \int \mathrm{\mid }{K}_{\epsilon }(y)\mathrm{\mid }\mathrm{\parallel }{\tau }_{y}f-f{\mathrm{\parallel }}_{p}dy$.

For any $\eta >0$, step 1.1 gives $\delta >0$ with $\mathrm{\parallel }{\tau }_{y}f-f{\mathrm{\parallel }}_p<\eta$ whenever $\mathrm{\mid }y\mathrm{\mid }<\delta$. Splitting the last integral yields $\mathrm{\parallel }{K}_{\epsilon }\ast f-f{\mathrm{\parallel }}_p\le M\eta +2\mathrm{\parallel }f{\mathrm{\parallel }}_p{\int }_{\mathrm{\mid }y\mathrm{\mid }\ge \delta }\mathrm{\mid }{K}_{\epsilon }(y)\mathrm{\mid }dy$. The tail tends to zero, so the limit superior is at most $M\eta$. Let $\eta \downarrow 0$ to obtain convergence. No separate normalization of the real and imaginary kernel parts has been used.

If the locally integrable input has compact support $S$, then $\int \mathrm{\mid }f\mathrm{\mid }={\int }_S\mathrm{\mid }f\mathrm{\mid }<\mathrm{\infty }$. A Borel representative can be made zero outside the closed set $S$ while preserving its class by step 3.1. F13 gives output support inside $\overline{S+\mathrm{supp}{\rho }_{\epsilon }}$. Both input supports are bounded, so this closed sum closure is bounded and hence compact by F17. The smooth output has closed support inside it, therefore compact support. Zero input or zero convolution yields the empty support; neither requires a nonempty support choice. All estimates above apply at p=1; the only general infinity assertion is the isometry and Young bound, with uniform approximation restricted to C0.