Skip to main content
2 of 2
Fixed typos.
Tony Huynh
  • 32.1k
  • 11
  • 112
  • 187

Why do we need the concept of Fourier measurability with growth function $\mathcal F$?

I'm studying the book Higher Order Fourier Analysis by Terence Tao (https://terrytao.files.wordpress.com/2011/03/higher-book.pdf). There, it defines that a function $f:[N]\to\mathbb{C}$ has Fourier complexity at most $M$ if it can be expressed as $$f(n) = \sum_{m=1}^{M'}c_me(\alpha_mn)$$ for some $M'\leq M$ and $c_1,\ldots,c_{M'}\in\mathbb C$ of magnitude at most 1.

It notes that from the Fourier inversion formula, every function has finite Fourier complexity, but it may grow with $N$.

But then, it says that, in order to work with indicator functions we need to take the $L^1$ closure and work with the Fourier measurable functions.

Then, it defines that a function $f:[N]\to \mathbb C$ is Fourier measurable with growth function $\mathcal F:\mathbb R^+ \to \mathbb R^+$ if, for every $K>1$ one can find a function $f_K:[N]\to\mathbb C$ of Fourier complexity at most $\mathcal F(K)$ such that $E_{n\in[N]}|f(n)-f_K(n)|\leq 1/K$.

I don't understand why we need this concept (which doesn't depend on $N$) if every function has finite Fourier complexity and we don't need to approximate it by these kinds of functions. What do I not understand here?