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https://www.encyclopediaofmath.org/index.php/Fourier_transform_of_a_generalized_function

For a periodic generalized function $f$, I see from the link that $$f(x) = \sum_{|k|=0}^\infty c_k(f)e ^{i\langle kw,x\rangle}, ~\left| c_k(f)\right|\le A \frac{1}{(1+k)^m}.$$

However, the definition or description of $m$ and $A$ are not given... Can anyone help me a little bit on this? Explain a little bit to me. Actually, if one can provide a reference, it will be very nice (not necessarily generalized functions, but only functions with one variable).

I am thinking a question. This result could be of some help.

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1 Answer 1

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Actually, here $m$ is a non-negative integer depending on the generalized function, and the correct formula is $$|c_k(f)|\le A(1+|k|)^m,\ \ \ \ (*)$$ where $k=(k_1,...,k_n)$ is a vector of integers, $|k|:=|k_1|+...+|k_n|$ and $A$ is a finite positive constant depending on the generalized function. The result basically states that the Fourier coefficients of a generalized functions can grow at most polynomially in k.

For a lucid discussion of this topic in one variable, see: Alberto Torchinsky, Real-Variable Methods in Harmonic Analysis, chapter 1, paragraph 4, Theorem 4.5.

Edit: in order to keep things simple, assume that the dimension is $1$. A more symmetric characterization than that given by $(*)$ would be: $$\sum_{k\in\mathbb{Z}}(1+|k|)^{-2M}|c_k(f)|^2<+\infty.\ (**)$$ As you can show, $(*)$ holds true for some $A>0$ and $m\in\mathbb{N}$ if and only if $(**)$ holds true for some $M\in\mathbb{N}$. The gain in using $(**)$ instead of $(*)$ is that the least $M\in\mathbb{N}$ such that $(**)$ holds, also gives you the largest Sobolev space $H^M(\mathbb{T})$ where you can extend by continuity your generalized function (intially defined only on $C^\infty(\mathbb{T})$). You can also allows negative integer values for $M$ in $(**)$. In this case, i.e. if $N\ge0$ is an integer and if $(**)$ holds with $M=-N$, it can be proved that your distribution is represented (by integral pairing) by an element of $H^N({\mathbb{T}})$. Finally, $(**)$ holds true for every $M\in\mathbb{Z}$, if and only if your distribution is represented by a $C^\infty(\mathbb{T})$ function. So, basically, the least $M\in\mathbb{Z}$ such that $(**)$ holds true is a measure of the regularity of your generalized function: smaller means more regular (a similar statement also holds true for $m$ in $(*)$, but in a less nitid way).

Edit edit: some examples.

  • The Dirac $\delta :C^\infty(\mathbb{T})\rightarrow\mathbb{C}, \varphi\mapsto \varphi(0)$ is such that $\forall k\in\mathbb{Z}, c_k(\delta)=1$ (here, as in the following examples, the constant actually depends on the normalization constant you have chosen in the definition of Fourier transform, so maybe you have seen that $\forall k\in\mathbb{Z}, c_k(\delta)=\frac{1}{2\pi}$). Then $(**)$ holds true with $M=1$ but fails with $M=0$. Then the largest Sobolev space $H^M(\mathbb{T})$ where you can extend $\delta$ is $H^1(\mathbb{T})$, meaning that you can't extend continuosly $\delta$ to $H^0(\mathbb{T})=L^2(\mathbb{T})$. Notice however that $\delta$ can be extended continuously to the whole $C^0(\mathbb{T})$ and $H^1(\mathbb{T})\subset C^0(\mathbb{T})\subset L^2(\mathbb{T})$. The point is that $C^0(\mathbb{T})$ isn't one of the Sobolev spaces in the scale spaces $H^m(\mathbb{T})$ for any $m\in\mathbb{Z}$. In other words, the characterization of $(**)$ doesn't tell you that $H^M(\mathbb{T})$ is the largest Banach space where you can extend your distribution continuosly, but only that it is the largest Sobolev space of the form $H^M(\mathbb{T})$ for some $M\in\mathbb{Z}$. Also, notice that the least $m\in\mathbb{N}$ such that $(*)$ holds is $m=0$.
  • The derivative of the Dirac, i.e. $\delta':C^\infty(\mathbb{T})\rightarrow\mathbb{C}, \varphi\mapsto- \varphi'(0)$ is such that $\forall k\in\mathbb{Z}, c_k(\delta')=ik$. Then $(**)$ holds true with $M=2$ but fails with $M=1$. So the largest Sobolev $H^M(\mathbb{T})$ where you can extend $\delta'$ is $H^2(\mathbb{T})$, meaning that you can't extend $\delta'$ continuously to the whole $H^1(\mathbb{T})$. Notice that the derivative takes away one "degree" of regularity from your distribution. Here again you can extend the distribution continuously to the whole $C^1(\mathbb{T})$ and $H^2(\mathbb{T})\subset C^1(\mathbb{T})\subset H^1(\mathbb{T})$. Notice that the least $m\in\mathbb{N}$ such that $(*)$ holds is $m=1$... so, by these last two examples, you may wonder that, if $m\in\mathbb{N}$ is the least integer such that $(*)$ holds, then your distribution can be extended continuosly to the whole $C^m(\mathbb{T})$, but:
  • The principal value distribution $f : C^\infty(\mathbb{T})\rightarrow\mathbb{C}, \varphi \mapsto \lim_{\varepsilon\rightarrow0^+}\frac{1}{2\pi}\int_{\mathbb{T}\backslash[-\varepsilon,\varepsilon]}\varphi(t)\cot(\frac{t}{2})\operatorname{d}t$ is such that $\forall k\in\mathbb{Z}, c_k(f)=-i \operatorname{sgn}(k)$. So, again, noticing that $(**)$ holds true with $M=1$ but not with $M=0$, you can extend $f$ continuosly to the Sobolev $H^1(\mathbb{T})$ but not to $L^2(\mathbb{T})$. However, notice that the least $m\in\mathbb{N}$ such that $(*)$ holds is $m=0$, but this time doesn't exists a continuous extension of $f$ to the whole $C^0(\mathbb{T})$. Then, the conjecture done in the previous example simply doesn't hold true. This is an istance of what I meant when I said that $(**)$ is "more symmetric" then $(*)$.
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