How are $L^P$ spaces defined for complex valued functions?

Question

I know that for real valued functions $f\in L^p(\mathbb R^N)$, $p\in [0,\infty)$ if the following integral converges: $$\int_{\mathbb R^N}|f(x)|^p dx$$

How is it for complex valued functions (of real variable)? Do I take modulus or absolute value of real and imaginary part? Does it matter?

Answer 1

For a complex number $z$, the expression $|z|$ means the complex modulus of $z$. So for a complex-valued (measurable) function $f:{\bf R}^n\to{\bf C}$, it makes sense to ask if $f$ is such that $$ \int_{{\bf R}^n}|f(x)|^p\ dx<\infty\tag{1} $$ which gives the definition of $L^p$ space of complex-valued functions. Note that $|f(x)|$ here means the complex modulus of the complex number $f(x)$.

See also the definition of $L^p$ spaces in https://www.encyclopediaofmath.org/index.php/Lp_spaces

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Note that for any real numbers $a$ and $b$, $$ \max\{|a|,|b|\} \leq|a+bi| \leq |a|+|b| \leq 2\cdot\max\{|a|,|b|\} $$ So an equivalent condition to (1) could be $$ \int |\text{Re}\,f(x)|^p\ dx<\infty\quad\text{and } \int |\text{Im}\,f(x)|^p\ dx<\infty\tag{2} $$

Answer 2

For $f:{\bf{R}}^{N}\rightarrow{\bf{C}}$, we let $f=u+iv$ for real functions $u,v$, if $|f|_{m}=\sqrt{u^{2}+v^{2}}$ is such that \begin{align*} \int_{{\bf{R}}^{N}}|f(x)|_{m}^{p}dx<\infty, \end{align*} then both \begin{align*} \int_{{\bf{R}}^{N}}|u(x)|^{p}dx,~\int_{{\bf{R}}^{N}}|v(x)|^{p}dx\leq\int_{{\bf{R}}^{p}}|f(x)|_{m}^{p}dx<\infty. \end{align*} On the other hand, if \begin{align*} \int_{{\bf{R}}^{N}}|u(x)|^{p}dx,~\int_{{\bf{R}}^{N}}|v(x)|^{p}dx<\infty, \end{align*} then \begin{align*} \int_{{\bf{R}}^{n}}|f(x)|_{m}^{p}&\leq\int_{{\bf{R}}^{n}}(|u(x)|+|v(x)|)^{p}dx\\ &\leq 2^{p-1}\left(\int_{{\bf{R}}^{N}}|u(x)|^{p}dx+\int_{{\bf{R}}^{N}}|v(x)|^{p}dx\right)\\ &<\infty. \end{align*}

Answer 3

Do I take modulus or absolute value of real and imaginary part?

As it's already said in the other answers and comments: the usual definition takes modulus but a similar definition with absolute value of real and imaginary parts makes sense as well.

Let $L^p_1(\mathbb R^N)$ be the space obtained when you take modulus. The norm in this space is given by $$\|f\|_1=\left(\int \Big[Re(f)^2+Im(f)^2\Big]^{p/2}\;dx\right)^{1/p}$$

Let $L^p_2(\mathbb R^N)$ be the space obtained when you take absolute value of real and imaginary parts. The norm in this space is given by $$\|f\|_2=\left(\int_{\mathbb R^n}\Big[|Re(f)|+|Im(f)|\Big]^{p}\;dx\right)^{1/p}$$

Does it matter?

The answer depends on the point of view.

We could say "no" because in both case we obtain exactly the same set, i.e., $f\in L_1^p(\mathbb R^N)$ if and only if $f\in L_2^p(\mathbb R^N)$.

But we also could say "yes" because, in general, $\|f\|_1\neq \|f\|_2$.

For many purposes, probably we can say "it doesn't matter at all" because the map $L_1^p(\mathbb R^N)\ni f\mapsto f\in L_2^p(\mathbb R^N)$ is a homeomorphism and thus, from the topological point of view, in both cases we obtain the same space.

Remark. If we define $L^2$ with any other "non standard" norm on $\mathbb C^2$, all the above comments remain valid (the only change is the expression of $\|f\|_2$).

Remark 2. In general, we can define $L^p(\mathbb R^N)$ not only for real-valued and complex-valued functions but also for Banach-valued functions. In other words: we can work with functions $f:\mathbb R^N\to X$, where $X$ stands for a given Banach space. The case $X=\mathbb R$ is the case which you know. Your question refers to the case $X=\mathbb C$ (equipped with different norms).