Complex Lp Spaces and Test-Function Conventions

By F1 the real functions $\mathrm{\mid }f\mathrm{\mid },\mathrm{\mid }g\mathrm{\mid }$ are measurable with finite respective norms. Applying F3 to these functions and using $\mathrm{\mid }fg\mathrm{\mid }=\mathrm{\mid }f\mathrm{\mid }\mathrm{\mid }g\mathrm{\mid }$ gives $\int \mathrm{\mid }fg\mathrm{\mid }\le N_p(f)N_{p^{\mathrm{\prime }}}(g)<\mathrm{\infty }$, for $1<p<\mathrm{\infty }$ and also $(p,p^{\mathrm{\prime }})=(1,\mathrm{\infty }),(\mathrm{\infty },1)$. Thus $fg$ is complex integrable and F5 gives the asserted integral bound.

For finite $p$, F2 and F8 give $N_p(f+g)\le N_p(\mathrm{\mid }f\mathrm{\mid }+\mathrm{\mid }g\mathrm{\mid })$. Real Minkowski on the two real nonnegative functions gives $N_p(f+g)\le N_p(f)+N_p(g)$, so addition preserves the finite-functional class.

For $p=\mathrm{\infty }$ put $a=N_{\mathrm{\infty }}(f)$ and $b=N_{\mathrm{\infty }}(g)$. Every $a+\eta$ and $b+\eta$ with $\eta >0$ is an essential bound: an essential bound smaller than it exists by the infimum property. Off the union of two null sets, $\mathrm{\mid }f+g\mathrm{\mid }\le a+b+2\eta$. Taking infima and then $\eta \downarrow 0$ gives $N_{\mathrm{\infty }}(f+g)\le a+b$.

If $f=f_1$ a.e. and $g=g_1$ a.e., then $\alpha f+\beta g=\alpha f_1+\beta g_1$ off the union of their measurable disagreement sets. For finite $p$, $\mathrm{\mid }f{\mathrm{\mid }}^p=\mathrm{\mid }{f}_1{\mathrm{\mid }}^p$ a.e., so F7 makes their integrals equal. At infinity the sets of essential bounds agree. Thus both operations and $N_p$ descend. Pointwise complex vector identities descend as well; scalar closure and $N_p(cf)=\mathrm{\mid }c\mathrm{\mid }{N}_p(f)$ follow from F2 and F8 for finite $p$, and scaling essential bounds for infinity. For $c=0$ this equality is immediate without dividing by $c$.

For finite $p$, $N_p(f)=0$ iff $\mathrm{\mid }f{\mathrm{\mid }}^p=0$ a.e. by F6, iff $f=0$ a.e. by F2. For infinity, if $N_{\mathrm{\infty }}(f)=0$, each measurable set $E_m=\{\mathrm{\mid }f\mathrm{\mid }>1\mathrm{/}m\}$ is null by F9. Since $\{f\mathrm{\ne }0\}={\bigcup }_{m\ge 1}{E}_m$, F10 gives $f=0$ a.e. The converse follows since zero is then an essential bound. This proves positive definiteness, including zero measure spaces.

Finally $\mathrm{\mid }\bar{f}\mathrm{\mid }=\mathrm{\mid }f\mathrm{\mid }$, and pointwise $\mathrm{\mid }\mathrm{Re}f\mathrm{\mid },\mathrm{\mid }\mathrm{Im}f\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }\le \mathrm{\mid }\mathrm{Re}f\mathrm{\mid }+\mathrm{\mid }\mathrm{Im}f\mathrm{\mid }$. Monotonicity of finite integrals or of essential bounds gives the two lower component bounds; the triangle inequality applied to $f=\mathrm{Re}f+i\mathrm{Im}f$ gives the upper bound. Together with homogeneity, definiteness and the descended operations, this proves all norm assertions.

For $u_n=\mathrm{Re}[f_n]$ and $v_n=\mathrm{Im}[f_n]$, F1 gives $\mathrm{\parallel }{u}_n-u_m{\mathrm{\parallel }}_p,\mathrm{\parallel }{v}_n-v_m{\mathrm{\parallel }}_p\le \mathrm{\parallel }[f_n]-[f_m]{\mathrm{\parallel }}_p$. Both real sequences are therefore Cauchy. F2 supplies real classes $u,v$ with $u_n\to u$, $v_n\to v$.

For a sequence already converging to $[f]$, its real components converge to $\mathrm{Re}[f]$ by F1. Apply F3 to obtain indices $n_k$ and real representatives $a_k\to U$ off a measurable null set. Its imaginary components still converge in norm; apply F3 to that subsequence to obtain further indices $k_j$ and imaginary representatives $b_j\to V$ off a second measurable null set. Then $a_{k_j}+i{b}_j$ represents $[f_{n_{k_j}}]$ and converges to $U+iV$ outside the union of those two null sets, which is null by F5.

Choose measurable representatives $U,V$ of these two classes and put $f=U+iV$. F1 shows $f\in {\mathcal{L}}^p(\mu ;\mathbb{C})$ and $\mathrm{\parallel }[f_n]-[f]{\mathrm{\parallel }}_p\le \mathrm{\parallel }{u}_n-u{\mathrm{\parallel }}_p+\mathrm{\parallel }{v}_n-v{\mathrm{\parallel }}_p\to 0$. Thus every Cauchy sequence converges, at infinity as well as at finite $p$.

The simultaneous representative selections used by the real results are permitted by F4; selection of the two limit representatives requires only two choices. All functions can be assigned zero on the measurable exceptional sets: a function pieced from a measurable function on a measurable set and zero on its complement is measurable. If predetermined measurable representatives are desired, their disagreement sets with the selected representatives are themselves measurable and null; F5 applied to their countable union preserves the a.e. convergence. Hence the assertions hold on incomplete measures without prescribing arbitrary, possibly nonmeasurable, values on null sets.

By F1, $u,v\in L^p(\mu ;\mathbb{R})$. F2 supplies real simple $a,b$ with $\mathrm{\parallel }u-a{\mathrm{\parallel }}_p<\eta \mathrm{/}2$ and $\mathrm{\parallel }v-b{\mathrm{\parallel }}_p<\eta \mathrm{/}2$, each with finite-measure nonzero set. The finite intersections of their fibers form a finite measurable partition on which $s=a+ib$ is constant, and $\{s\mathrm{\ne }0\}\subseteq \{a\mathrm{\ne }0\}\cup \{b\mathrm{\ne }0\}$ has finite measure. F1 gives $\mathrm{\parallel }f-s{\mathrm{\parallel }}_p<\eta$. This uses only two approximation choices for the specified tolerance, not a simultaneous choice function.

In Euclidean Lebesgue space, under the countable-choice hypothesis F5, apply F3 to $u,v$ with errors $\eta \mathrm{/}2$ to obtain $a,b\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n;\mathbb{R})$. Their sum $a+ib$ is smooth componentwise and is supported in the union of the two compact supports, hence is complex $C_c^{\mathrm{\infty }}$. F1 again bounds its error by $\eta$. The inclusion $C_c^{\mathrm{\infty }}\subseteq C_c$ proves continuous compact-support density as well.

If a complex L-infinity class $f$ is in the closure of complex $C_c$, approximating $f$ to any positive tolerance and projecting its approximants gives, by F1, real $C_c$ approximations to both component classes. F4 therefore supplies $U,V\in C_0({\mathbb{R}}^n;\mathbb{R})$ representing them. The complex function $U+iV$ represents $f$ and vanishes at infinity: outside the union of two compact sets where the separate component errors are below $\eta \mathrm{/}2$, its modulus is below $\eta$.

Conversely, if $f$ has a $C_0$ representative $U+iV$, F4 supplies real $C_c$ approximants to $U,V$ with essential-norm errors below $\eta \mathrm{/}2$. Their complex sum lies in $C_c$ and has error below $\eta$ by F1. Thus precisely the stated classes form the closure. The constant-one class is excluded: any continuous representative equal to one a.e. must equal one everywhere, since a nonzero continuous discrepancy persists on an open ball of positive Lebesgue measure; that constant does not vanish at infinity. The finite-p assertions therefore have no such infinity extension.

F2 makes $f\bar{g}$ integrable. Replacing $f,g$ by a.e.-equal representatives changes their product only on the union of the two measurable null disagreement sets. F4 therefore leaves the integral in F1 unchanged. This proves representative independence.

For an integrable $h=u+iv$, F7 gives $\int \bar{h}=\int u-i\int v=\overline{\int h}$. Hence F6 implies $⟨g,f⟩=\overline{⟨f,g⟩}$. F3 applied to $(af+bk)\bar{g}$ gives $⟨af+bk,g⟩=a⟨f,g⟩+b⟨k,g⟩$, and applied to $f\overline{ag+bk}=\bar{a}f\bar{g}+\bar{b}f\bar{k}$ gives conjugate-linearity in the second variable. All products are integrable by F2.

F6 gives $⟨f,f⟩=\int \mathrm{\mid }f{\mathrm{\mid }}^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2\ge 0$. By F5 this number is zero iff $\mathrm{\mid }f{\mathrm{\mid }}^2=0$ a.e., which is equivalent to $f=0$ as a class. Thus the form is positive definite and its norm is exactly the modulus $L^2$ norm.

For $g\mathrm{\ne }0$ put $a=⟨f,g⟩$, $b=\mathrm{\parallel }g{\mathrm{\parallel }}_2^2>0$ and $c=a\mathrm{/}b$. Sesquilinearity yields $\mathrm{\parallel }f-cg{\mathrm{\parallel }}_2^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2-\bar{c}a-c\bar{a}+\mathrm{\mid }c{\mathrm{\mid }}^{2}b=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2-\mathrm{\mid }a{\mathrm{\mid }}^2\mathrm{/}b$. Nonnegativity proves $\mathrm{\mid }a{\mathrm{\mid }}^2\le \mathrm{\parallel }f{\mathrm{\parallel }}_2^{2}b$, hence Cauchy–Schwarz. Equality implies $\mathrm{\parallel }f-cg{\mathrm{\parallel }}_2=0$, so $f=cg$ a.e. Conversely, if $f=dg$ a.e., then $⟨f,g⟩=db$ and $\mathrm{\parallel }f{\mathrm{\parallel }}_2=\mathrm{\mid }d\mathrm{\mid }\mathrm{\parallel }g{\mathrm{\parallel }}_2$, giving equality. For $g=0$, both sides of the inequality are zero for every $f$.

Finite summation preserves the linearity and symmetry identities. Also $B(F,F)={\sum }_{j<m}\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_2^2\ge 0$, and a finite sum of nonnegative reals is zero iff every summand is zero; step 1.3 then gives definiteness. For $G\mathrm{\ne }0$, set $c=B(F,G)\mathrm{/}B(G,G)$. Expanding the finite sum using step 1.2 gives $\mathrm{\parallel }F-cG{\mathrm{\parallel }}^2=\mathrm{\parallel }F{\mathrm{\parallel }}^2-\mathrm{\mid }B(F,G){\mathrm{\mid }}^2\mathrm{/}\mathrm{\parallel }G{\mathrm{\parallel }}^2$. Nonnegativity gives $\mathrm{\mid }B(F,G)\mathrm{\mid }\le \mathrm{\parallel }F\mathrm{\parallel }\mathrm{\parallel }G\mathrm{\parallel }$. The expansion $\mathrm{\parallel }F+G{\mathrm{\parallel }}^2=\mathrm{\parallel }F{\mathrm{\parallel }}^2+2\mathrm{Re}B(F,G)+\mathrm{\parallel }G{\mathrm{\parallel }}^2\le (\mathrm{\parallel }F\mathrm{\parallel }+\mathrm{\parallel }G\mathrm{\parallel }{)}^2$ gives the triangle inequality; scalar homogeneity follows by scaling each squared component norm. Thus this square root is indeed a norm. As above, equality in Cauchy–Schwarz is equivalent to each $f_j-c{g}_j$ having norm zero, with this same $c$ for all $j$; conversely a common scalar multiple gives equality by homogeneity. For $G=0$ both sides are zero. If $m=0$, the tuple space has just its zero element and all sums are zero, so the same axioms and zero case apply.

The measure space and exponent satisfy F1, and the assumed countable choice is exactly its additional hypothesis. It therefore supplies completeness for every $1\le p\le \mathrm{\infty }$ and a.e.-convergent subsequences with the specified limit class.

For $p<\mathrm{\infty }$, F2 applies to the same measure space and gives finite-simple approximants with finite-measure nonzero sets. In the Euclidean clause its Lebesgue and countable-choice hypotheses also hold, so it supplies smooth compactly supported approximants.

For $p=2$, F3 proves that the representative formula descends to an inner product and that its squared norm equals $\int \mathrm{\mid }f{\mathrm{\mid }}^2$. Thus its induced norm is the same $L^2$ norm used in step 1.1. F3 also supplies precisely the nonzero-second-argument scalar-multiple criterion and the zero-second-argument exception, and its finite-sum proof supplies the common-scalar tuple version. This collects the asserted interfaces without any additional analytic hypothesis.

Put $a=\mathrm{\mid }g\mathrm{\mid }$ and $\theta =\bar{g}\mathrm{/}a$ on $\{a>0\}$, zero elsewhere. The function $b=1\mathrm{/}a$ on $\{a>0\}$, zero elsewhere, is measurable: for $t\ge 0$, its strict upper level set is $\{0<a<1\mathrm{/}t\}$ when $t>0$ and $\{a>0\}$ when $t=0$; negative upper thresholds give $X$. Real positive powers of $a$ have upper sets $\{a>t^{1\mathrm{/}\gamma }\}$ for $\gamma >0,t\ge 0$. Thus F9–F10 and F1 show the phase, the powers and all ensuing products are measurable. F8 gives $\mathrm{\mid }\theta \mathrm{\mid }\le 1$ and $g\theta =a$.

Let $E$ be measurable of finite measure and let $h$ be a bounded measurable function vanishing outside $E$, with $\mathrm{\parallel }h{\mathrm{\parallel }}_r\le 1$ and ${\int }_E\mathrm{\mid }g\mathrm{\mid }<\mathrm{\infty }$. Round each coordinate of $h$ down to an integer multiple of $1\mathrm{/}m$ on $E$ and set the result $t_m$ to zero off $E$. This has finite range, measurable fibers, and $\mathrm{\mid }{t}_m-h\mathrm{\mid }\le \sqrt{2}\mathrm{/}m$. When $r=\mathrm{\infty }$ and $\mathrm{\mid }h\mathrm{\mid }\le 1$, replace each of its finitely many values $z$ by $z\mathrm{/}\max (1,\mathrm{\mid }z\mathrm{\mid })$; these values lie in the closed unit disk and the error is at most $2\sqrt{2}\mathrm{/}m$, because the displacement of $z$ is at most $\mathrm{\mid }z-h\mathrm{\mid }$. For finite $r$, the uniform error $e_m$ gives $\mathrm{\parallel }{t}_m-h{\mathrm{\parallel }}_r\le e_m\mu (E{)}^{1\mathrm{/}r}\to 0$; for infinity it gives $\mathrm{\parallel }{t}_m-h{\mathrm{\parallel }}_{\mathrm{\infty }}\le e_m\to 0$. Put $d_m=\max (1,\mathrm{\parallel }{t}_m{\mathrm{\parallel }}_r)$ and $s_m=t_m\mathrm{/}{d}_m$. F2 gives $1\le d_m\le 1+\mathrm{\parallel }{t}_m-h{\mathrm{\parallel }}_r$, hence $d_m\to 1$ and $s_m\to h$ uniformly, since $h$ is bounded. Each $s_m$ is an admissible finite simple test, and F7 gives $\mathrm{\mid }\int g(s_m-h)\mathrm{\mid }\le {\sup }_E\mathrm{\mid }{s}_m-h\mathrm{\mid }{\int }_E\mathrm{\mid }g\mathrm{\mid }\to 0$. Therefore $S(g)\ge \mathrm{\mid }\int gh\mathrm{\mid }$. If $\mu (E)=0$, the integral is zero and the zero test already suffices.

Let $E$ have finite measure with $a\le B<\mathrm{\infty }$ on $E$, and put $A=\mathrm{\parallel }g{\mathbf{1}}_E{\mathrm{\parallel }}_q$ for finite $q$. If $A=0$, $S(g)\ge A$ from the zero test. If $A>0$ and $1<q<\mathrm{\infty }$, set $h={\mathbf{1}}_E\theta a^{q-1}\mathrm{/}{A}^{q-1}$. This is bounded, and $(q-1)r=q$ gives $\int \mathrm{\mid }h{\mathrm{\mid }}^r=A^{-q}{\int }_E{a}^q=1$. Moreover $\int gh=A^{-(q-1)}{\int }_E{a}^q=A$. For $q=1$ set $h={\mathbf{1}}_E\theta$ instead; it is bounded by one and $\int gh={\int }_{E}a=A$. In both cases ${\int }_E\mathrm{\mid }g\mathrm{\mid }\le B\mu (E)<\mathrm{\infty }$, so step 1.2 applies and gives $S(g)\ge \mathrm{\parallel }g{\mathbf{1}}_E{\mathrm{\parallel }}_q$.

Under sigma-finiteness take a covering $H_j$ of finite measure and put $E_n=({\bigcup }_{j\le n}{H}_j)\cap \{1\mathrm{/}n\le a\le n\}$ for $n\ge 1$. These finite-measure sets increase and cover $\{a>0\}$. For finite $q$, $a^q{\mathbf{1}}_{E_n}\uparrow a^q$ everywhere, so F6 yields $\mathrm{\parallel }g{\mathbf{1}}_{E_n}{\mathrm{\parallel }}_q\uparrow \mathrm{\parallel }g{\mathrm{\parallel }}_q$, possibly infinitely. Step 2.1 proves $S(g)\ge \mathrm{\parallel }g{\mathrm{\parallel }}_q$. If that norm is finite, F2 bounds every admissible test by $\mathrm{\parallel }g{\mathrm{\parallel }}_q$; if it is infinite the lower bound already gives equality at infinity. The assumed test integrability ensures every integral in the defining supremum is meaningful.

If instead $g\in L^q$ with $q<\mathrm{\infty }$ on an arbitrary measure space, use $E_n=\{1\mathrm{/}n\le a\le n\}$. F11 gives $n^{-q}\mu (E_n)\le \int a^q<\mathrm{\infty }$, hence $\mu (E_n)<\mathrm{\infty }$. The same increasing limit and step 2.1 give $S(g)\ge \mathrm{\parallel }g{\mathrm{\parallel }}_q$. Hölder gives the reverse bound and integrability of every $gs$: a finite simple function of finite-measure support belongs to every finite-exponent Lr, and is bounded when $r=\mathrm{\infty }$. No sigma-finiteness of $X$ is needed.

Now let $q=\mathrm{\infty }$ and $M=\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}>0$, possibly infinite under the sigma-finite hypothesis. For any $0<t<M$, the measurable set $D=\{a>t\}$ has positive measure, since otherwise $t$ would be an essential bound. In the sigma-finite case, $D={\bigcup }_n(D\cap E_n)$ with the sets from step 3.1. F12 implies some $B=D\cap E_n$ has positive measure, and $0<\mu (B)<\mathrm{\infty }$ with $a\le n$ there. In the semifinite already-L-infinity case, $M<\mathrm{\infty }$; the set $\{a>M+1\}$ is null by F5. Semifiniteness applied to $D\cap \{a\le M+1\}$ gives a measurable $B$ of finite positive measure. In either case $h=\theta {\mathbf{1}}_B\mathrm{/}\mu (B)$ is bounded, $\mathrm{\parallel }h{\mathrm{\parallel }}_1=1$, and $\int gh=\mu (B{)}^{-1}{\int }_{B}a\ge t$. Step 1.2 now yields $S(g)\ge t$. Letting $t\uparrow M$, or taking arbitrarily large $t$ if $M=\mathrm{\infty }$, proves $S(g)\ge M$. For finite $M$, F2 gives $S(g)\le M$. If $M=0$, F2 and the zero test give $S(g)=0$.

The measure on $X=\{x\}$ with $\mu (\mathrm{\varnothing })=0$, $\mu (X)=\mathrm{\infty }$ is countably additive: a disjoint family contains at most one nonempty member. It is not semifinite. For $g(x)=1$, its essential norm is one, yet the only finite-measure-supported simple function is zero, so $S(g)=0$. This verifies the necessity of a measure hypothesis at the infinity endpoint. Finally the zero test makes every stated supremum nonempty; on zero measure spaces it and every other integral have value zero. The previous steps prove all the claimed identities and therefore the finite-bound membership conclusion.

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 $\mathrm{\parallel }{\tau }_{h}f-{\tau }_{h_0}f{\mathrm{\parallel }}_p=\mathrm{\parallel }{\tau }_{h-h_0}f-f{\mathrm{\parallel }}_p$.

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 $(x,y)\mapsto K(y)f(x-y)$ Borel. Apply F4 to the nonnegative real inputs $\mathrm{\mid }K\mathrm{\mid }\in L^1$ and $\mathrm{\mid }f\mathrm{\mid }\in L^p$: the envelope $A(x)=\int \mathrm{\mid }K(y)\mathrm{\mid }\mathrm{\mid }f(x-y)\mathrm{\mid }dy$ is finite a.e. and $\mathrm{\parallel }A{\mathrm{\parallel }}_p\le \mathrm{\parallel }K{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_p$. On this set the complex integral exists absolutely and its modulus is at most $A$ by F16. For measurability, expand $K=a+ib$, $f=u+iv$ and write $K\ast f=(a\ast u-b\ast v)+i(a\ast v+b\ast u)$ wherever all integrals converge. Each real convolution is an Lp measurable function by F4 since $a,b\in L^1$ and $u,v\in L^p$ by F2. Define the output to be zero on the measurable exceptional set. F2 and monotonicity now give the claimed sharp norm bound, including $p=\mathrm{\infty }$.

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.

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 1.2 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 $\mathrm{\parallel }{K}_{\epsilon }\ast f-f{\mathrm{\parallel }}_p\le \int \mathrm{\mid }{K}_{\epsilon }(y)\mathrm{\mid }\mathrm{\parallel }{\tau }_{y}f-f{\mathrm{\parallel }}_{p}dy$.