Complex Lp classes and Euclidean test-function conventions

Let $(X,\mathcal{A},\mu )$ be a measure space. A finite-valued function $f=u+iv:X\to \mathbb{C}$ is measurable when $u,v$ are real measurable functions. Write $N_p(f)={\left({\int }_X\mathrm{\mid }f{\mathrm{\mid }}^{p}d\mu \right)}^{1\mathrm{/}p}(1\le p<\mathrm{\infty }),N_{\mathrm{\infty }}(f)=\inf \{M\in [0,\mathrm{\infty }]:\mathrm{\mid }f\mathrm{\mid }\le M\text{ a.e.}\}\mathrm{.}$ The infimum of the empty set of finite bounds is $\mathrm{\infty }$; equivalently allow $M=\mathrm{\infty }$. Define ${\mathcal{L}}^p(\mu ;\mathbb{C})=\{f:f\text{ measurable},N_p(f)<\mathrm{\infty }\}$ and its set quotient $L^p(\mu ;\mathbb{C})={\mathcal{L}}^p\mathrm{/}\text{\hspace{0.17em}⁣}\sim$, where $f\sim g$ means $f=g$ a.e. This extends The space $L^p(\mu )$ as the quotient by null functions and The essential supremum of a measurable function with respect to a measure. Norm and vector-space assertions are established separately.

Here $\mathrm{\mid }u+iv\mathrm{\mid }=\sqrt{u^2+v^2}$ and $\overline{u+iv}=u-iv$ as in Real and imaginary parts, complex conjugation, and modulus. These operations preserve measurability: $u^2+v^2$ is measurable by Arithmetic and lattice operations preserve measurability whenever they are defined, and $\{\mathrm{\mid }f\mathrm{\mid }>a\}=\{\begin{array}{ll}\{u^2+v^2>a^2\},& a\ge 0,\\ X,& a<0.\end{array}$ Thus Threshold characterisations of real-valued and extended-real-valued measurability applies. For $g=a+ib$, the formulas $fg=(ua-vb)+i(ub+va)$ and $\bar{f}=u-iv$ prove the remaining claims by real arithmetic closure. Nonnegative powers are measurable because $\{r^t>b\}=\{r>b^{1\mathrm{/}t}\}$ for $r\ge 0,t>0,b\ge 0$, with the negative thresholds automatic.

A finite simple complex function has finite range and measurable fibers. Its finite-measure support condition is $\mu (\{s\mathrm{\ne }0\})<\mathrm{\infty }$. This concerns the nonzero set, not compactness of its closure. The zero function is an admissible finite simple function, including when $X=\mathrm{\varnothing }$ or $\mu (X)=0$.

Integration is componentwise: for integrable $f=u+iv$, $\int fd\mu =\int ud\mu +i\int vd\mu \mathrm{.}$ This is the earlier convention of Integrable real and complex functions, and their integrals and The class $L^1(\mu )$ of integrable functions. The inequalities $\mathrm{\mid }u\mathrm{\mid },\mathrm{\mid }v\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }+\mathrm{\mid }v\mathrm{\mid }$ show that integrability of the modulus and integrability of both components are equivalent.

For $n\ge 1$, define complex $C_c({\mathbb{R}}^n)$, $C_0({\mathbb{R}}^n)$ and $C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ by requiring both components to lie in the corresponding real spaces (The spaces $C_c({\mathbb{R}}^n)$ and $C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$, The space $C_0({\mathbb{R}}^n)$ of continuous functions vanishing at infinity). Derivatives are componentwise, with the multi-index and all-ordered-partials conventions of $C^k$ maps and multi-index derivative notation in Euclidean space. The union of the two compact component supports is compact. A measurable complex $f$ is locally integrable for Lebesgue measure if ${\int }_K\mathrm{\mid }f\mathrm{\mid }<\mathrm{\infty }$ on every compact $K$. The component inequalities give the equivalent componentwise condition. By Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, closed bounded balls are compact and each compact set is bounded. Hence compact integrability implies integrability on every bounded open ball by restriction from its closure; conversely each compact set is contained in a bounded open ball. These prove both directions of the ball formulation.

For a topological $X$ with Borel sets contained in $\mathcal{A}$, define ${\mathrm{ess\hspace{0.17em}supp}}_{\mu }f=X\setminus \bigcup \{U:U\text{ open and }f=0\text{ a.e. on }U\}\mathrm{.}$ Changing $f$ on one measurable null set preserves the zero-a.e. property on every $U$, since the union of that set and the old exceptional set is null. Thus essential support depends only on the a.e. class. Ordinary support means $\overline{\{f\mathrm{\ne }0\}}$ and can change with the representative.

For the following illustration assume The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$). For $f={\mathbf{1}}_{\mathbb{Q}}$ on $\mathbb{R}$, Every at most countable subset of ${\mathbb{R}}^n$ is Lebesgue null; in particular ${\lambda }_1(\mathbb{Q})=0$ gives $f=0$ Lebesgue-a.e.; therefore $N_{\mathrm{\infty }}(f)=0$ and $\mathrm{ess\hspace{0.17em}supp}f=\mathrm{\varnothing }$, since $U=\mathbb{R}$ occurs in the union. But Both $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable gives $\overline{\{f\mathrm{\ne }0\}}=\mathbb{R}$. For the Dirac probability measure at zero (The Dirac set function at a point, A Dirac set function is a probability measure), the bound $1$ holds everywhere and every $0\le M<1$ fails on $\{0\}$, of measure one. Thus $N_{\mathrm{\infty },{\delta }_0}(f)=1$.

The raw $L^2$ pairing convention is $⟨f,g⟩=\int f\bar{g}$, linear in the first variable. Its integrability and class invariance are obligations of the later pairing items. Bilinear tests instead use $\int fs$, with no conjugation of $s$.