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Tony Huynh
Tony Huynh
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Why do we need the concept of Fourier mesurabilitymeasurability with grouthgrowth 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$M$ if it can be expressed as $$f(n) = \sum_{m=1}^{M'}c_me(\alpha_mn)$$ for some $M\leq M$$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$N$.

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 mesurablemeasurable functions.

Then, it defines that a function $f:[N]\to \mathbb C$ is Fourier mesurablemeasurable 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 fourierFourier 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 (witchwhich doesn't depend on $N$) if every function has finite Fourier complexity and we don't need to approximate it by this kindthese kinds of functions. What do I not understand here?

Why we need the concept of Fourier mesurability with grouth 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 mesurable functions.

Then, it defines that a function $f:[N]\to \mathbb C$ is Fourier mesurable 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 (witch doesn't depend on $N$) if every function has finite Fourier complexity and we don't need to approximate it by this kind of functions. What do I not understand here?

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?

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El Tudey
El Tudey
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Why we need the concept of Fourier mesurability with grouth 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 mesurable functions.

Then, it defines that a function $f:[N]\to \mathbb C$ is Fourier mesurable 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 (witch doesn't depend on $N$) if every function has finite Fourier complexity and we don't need to approximate it by this kind of functions. What do I not understand here?