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I'm having trouble in using a rigorous proof that starts from defining the Fourier transform on Schwartz spaces to arrive at the end defining the inverison formula in $L^1$ and in $L^2$. So I'm okay we can define the Fourier transform in $S(\mathbb{R})$ and obtain its inverse formula by lot of calculus and definitions (gaussian transform, kernels, multiplication formula through Fubini's theorem etc..) But how can I do a step forward and pass to $L^1$ and then $L^2$ ? I've a lot of references that start from considering a function directly in $L^1$ but I want to reach this point from a function in a Schwartz space. I know the arguments would be to consider the fact $C^{\infty}_0 \subset S(\mathbb{R}) \subset L^{1}$ where $C_0^{\infty}$ is the set of smooth functions with compact support but I don't understand how to use this fact honestly. Any reference or helps would be really appreciated !

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Let us define the Fourier Transform of a function $f\in\mathcal{S}(\mathbb R)$ by $$\hat{f}(x)=c\int_\mathbb Rf(y)e^{-iyx}\;dy,$$ where $c=\frac{1}{\sqrt{2\pi}}$. You are okay with the following result.

(1) Inversion formula in $\mathcal{S}(\mathbb R)$. If $f\in\mathcal{S}(\mathbb R)$, then $$f(x)=c\int_\mathbb R\hat{f}(y)e^{iyx}\;dy,\quad\forall\ x\in\mathbb {R}.$$

You want to pass to $L^1$. In other words, you want to prove the following result.

(2) Inversion formula in $L^1$. If $f\in L^1$ and $\hat{f}\in L^1$, then $$f(x)=c\int_\mathbb R\hat{f}(y)e^{iyx}\;dy,\quad \text{a.e.}\ x\in\mathbb {R}.$$

You know that you should use the inclusions $C^{\infty}_0 \subset S(\mathbb{R}) \subset L^{1}$ but don't know how. Here is some details, from Hörmmander's book, p. 164 (which proves the result for functions in $\mathcal{S}'(\mathbb R)$, the space of tempered distributions).

Proof of (2): From (1), given $\varphi\in\mathcal S(\mathbb R)$, $$\varphi(-x)=c\int_\mathbb R\hat{\varphi}(y)e^{iy(-x)}\;dy=\hat{\hat{\varphi}}(x),\quad\forall\ x\in\mathbb {R}.$$ So, $$\int_\mathbb R \hat{\hat{f}}(x)\varphi(x)\;dx\overset{(*)}{=}\int_\mathbb R f(x)\hat{\hat{\varphi}}(x)\;dx=\int_\mathbb R f(x)\varphi(-x)\;dx=\int_\mathbb R f(-x)\varphi(x)\;dx,\quad\forall\ \varphi\in C_0^\infty$$ and thus $\hat{\hat{f}}(x)=f(-x)$ a.e. $x\in\mathbb R$ (by the du Bois Reymond Lemma), which implies $$f(x)=\hat{\hat{f}}(-x)=c\int_\mathbb R \hat{f}(y)e^{-iy(-x)}\;dy=c\int_\mathbb R \hat{f}(y)e^{iyx}\;dy,\quad\text{a.e. } x\in \mathbb R.$$

Note that the equality $(*)$ is valid because $f,\hat{f},\varphi,\hat{\varphi}\in L^1$. $\square$

Remark 1: The inversion formulas (1) and (2) allows us to define the Inverse Fourier Transform in $\mathcal{S}(\mathbb R)$ and $L^1$. In $L^2$, the Inverse Fourier Transform (as well as the the Fourier Transform) is defined by extension. So, it is not clear for me what you call "inversion formula in $L^2$".

Remark 2: In $\mathbb R^n$, the same argument works.

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  • $\begingroup$ And you are using that if everything converges then $\int_\mathbb{R} \hat{g}(x) h(x)dx=\int_\mathbb{R} \int_\mathbb{R}e^{-i x y}g(y)dy h(x)dx=\int_\mathbb{R} g(y)(\int_\mathbb{R}e^{-i x y} h(x)dx)dy = \int_\mathbb{R} g(y) \hat{h}(y)dy$ $\endgroup$ – reuns Jan 3 '17 at 1:36
  • $\begingroup$ For $L^2$, we need to say the Fourier transform is unitary $L^2 \to L^2$ and $L^1 \cap L^2$ is dense in $L^2$ $\endgroup$ – reuns Jan 3 '17 at 1:39
  • $\begingroup$ @user1952009 Yes, I'm using it in the proof of $(*)$. And I did not add details about how the Fourier Transform is defined in $L^2$ because it seems for me that it is not what the OP is asking for. $\endgroup$ – Pedro Jan 3 '17 at 1:51
  • $\begingroup$ @user1952009 the (*) equality is justified by the first comment you added applied twice ? For $L^2$ yes it need extensions and it's not my purpose here. $\endgroup$ – Xaler Jan 5 '17 at 15:50
  • $\begingroup$ @user1952009 Does can I be more formal and introduce the definition of Fourier transform in $L^1$ through extension from $S$ given the fact it's dense in $L^1$ ? $\endgroup$ – Xaler Jan 5 '17 at 15:54

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