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The Fourier transform $\hat u$ is defined on the Schwartz space $\mathscr S(\mathbb R^n)$ by $ \hat u(\xi)=\int e^{-2iπ x\cdot \xi} u(x) dx. $ It is an isomorphism of $\mathscr S(\mathbb R^n)$ and the inversion formula is $ u(x)=\int e^{2iπ x\cdot \xi} \hat u(\xi) d\xi. $ The Fourier transformation can be extended to the tempered distributions $\mathscr S'(\mathbb R^n)$ with the formula $$ \langle \hat T,\phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)} = \langle T, \hat \phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}. $$ Below we note $\mathcal F$ the Fourier transformation on $\mathscr S'(\mathbb R^n)$. We find easily that $\mathcal F^4=Id$, so that if $T\in\mathscr S'(\mathbb R^n)$ is such that $\mathcal F T=\lambda T$, then $\lambda$ is a fourth root of unity.

Question: Are all the tempered distributions $T$ such that $\mathcal FT=T$ known? Two examples are very classical: first the Gaussians $e^{-π\vert x\vert^2}, e^{-π\langle Ax,x\rangle } $ where $A$ is a positive definite matrix with determinant 1, second the case $$ T_0=\sum_{k\in \mathbb Z^n}\delta_k, $$ where the equality $\mathcal FT_0=T_0$ is the Poisson summation formula.

I think that, thanks to your answers and a reference, I found the answer: using What are fixed points of the Fourier Transform which deals with the $L^2$ case, one may guess that the fixed points of $\mathcal F$ in $\mathscr S'(\mathbb R^n)$ are $$ \Bigl\{S+\mathcal F S+\mathcal F^2 S+\mathcal F^3 S \Bigr\}_{S\in \mathscr S'(\mathbb R^n)}=(Id+\mathcal F+\mathcal F^2+\mathcal F^3)(\mathscr S'(\mathbb R^n)). $$ Since $\mathcal F^4=Id$ on $\mathscr S'(\mathbb R^n)$, the above distributions are indeed fixed points and if $\mathcal F T=T$, then $$ T=\frac14\bigl(T+\mathcal F T+\mathcal F^2 T+\mathcal F^3 T\bigr), $$ conluding the proof.

The Fourier transform $\hat u$ is defined on the Schwartz space $\mathscr S(\mathbb R^n)$ by $ \hat u(\xi)=\int e^{-2iπ x\cdot \xi} u(x) dx. $ It is an isomorphism of $\mathscr S(\mathbb R^n)$ and the inversion formula is $ u(x)=\int e^{2iπ x\cdot \xi} \hat u(\xi) d\xi. $ The Fourier transformation can be extended to the tempered distributions $\mathscr S'(\mathbb R^n)$ with the formula $$ \langle \hat T,\phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)} = \langle T, \hat \phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}. $$ Below we note $\mathcal F$ the Fourier transformation on $\mathscr S'(\mathbb R^n)$. We find easily that $\mathcal F^4=Id$, so that if $T\in\mathscr S'(\mathbb R^n)$ is such that $\mathcal F T=\lambda T$, then $\lambda$ is a fourth root of unity.

Question: Are all the tempered distributions $T$ such that $\mathcal FT=T$ known? Two examples are very classical: first the Gaussians $e^{-π\vert x\vert^2}, e^{-π\langle Ax,x\rangle } $ where $A$ is a positive definite matrix with determinant 1, second the case $$ T_0=\sum_{k\in \mathbb Z^n}\delta_k, $$ where the equality $\mathcal FT_0=T_0$ is the Poisson summation formula.

I think that, thanks to your answers and a reference, I found the answer: using What are fixed points of the Fourier Transform which deals with the $L^2$ case, one may guess that the fixed points of $\mathcal F$ in $\mathscr S'(\mathbb R^n)$ are $$ \Bigl\{S+\mathcal F S+\mathcal F^2 S+\mathcal F^3 S \Bigr\}_{S\in \mathscr S'(\mathbb R^n)}=(Id+\mathcal F+\mathcal F^2+\mathcal F^3)(\mathscr S'(\mathbb R^n)). $$ Since $\mathcal F^4=Id$ on $\mathscr S'(\mathbb R^n)$, the above distributions are indeed fixed points and if $\mathcal F T=T$, then $$ T=\frac14\bigl(T+\mathcal F T+\mathcal F^2 T+\mathcal F^3 T\bigr), $$ conluding the proof.

The Fourier transform $\hat u$ is defined on the Schwartz space $\mathscr S(\mathbb R^n)$ by $ \hat u(\xi)=\int e^{-2iπ x\cdot \xi} u(x) dx. $ It is an isomorphism of $\mathscr S(\mathbb R^n)$ and the inversion formula is $ u(x)=\int e^{2iπ x\cdot \xi} \hat u(\xi) d\xi. $ The Fourier transformation can be extended to the tempered distributions $\mathscr S'(\mathbb R^n)$ with the formula $$ \langle \hat T,\phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)} = \langle T, \hat \phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}. $$ Below we note $\mathcal F$ the Fourier transformation on $\mathscr S'(\mathbb R^n)$. We find easily that $\mathcal F^4=Id$, so that if $T\in\mathscr S'(\mathbb R^n)$ is such that $\mathcal F T=\lambda T$, then $\lambda$ is a fourth root of unity.

Question: Are all the tempered distributions $T$ such that $\mathcal FT=T$ known? Two examples are very classical: first the Gaussians $e^{-π\vert x\vert^2}, e^{-π\langle Ax,x\rangle } $ where $A$ is a positive definite matrix with determinant 1, second the case $$ T_0=\sum_{k\in \mathbb Z^n}\delta_k, $$ where the equality $\mathcal FT_0=T_0$ is the Poisson summation formula.

I think that, thanks to your answers and a reference, I found the answer: using What are fixed points of the Fourier Transform which deals with the $L^2$ case, one may guess that the fixed points of $\mathcal F$ in $\mathscr S'(\mathbb R^n)$ are $$ \Bigl\{S+\mathcal F S+\mathcal F^2 S+\mathcal F^3 S \Bigr\}_{S\in \mathscr S'(\mathbb R^n)}=(Id+\mathcal F+\mathcal F^2+\mathcal F^3)(\mathscr S'(\mathbb R^n)). $$ Since $\mathcal F^4=Id$ on $\mathscr S'(\mathbb R^n)$, the above distributions are indeed fixed points and if $\mathcal F T=T$, then $$ T=\frac14\bigl(T+\mathcal F T+\mathcal F^2 T+\mathcal F^3 T\bigr), $$ conluding the proof.

The Fourier transform $\hat u$ is defined on the Schwartz space $\mathscr S(\mathbb R^n)$ by $ \hat u(\xi)=\int e^{-2iπ x\cdot \xi} u(x) dx. $ It is an isomorphism of $\mathscr S(\mathbb R^n)$ and the inversion formula is $ u(x)=\int e^{2iπ x\cdot \xi} \hat u(\xi) d\xi. $ The Fourier transformation can be extended to the tempered distributions $\mathscr S'(\mathbb R^n)$ with the formula $$ \langle \hat T,\phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)} = \langle T, \hat \phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}. $$ Below we note $\mathcal F$ the Fourier transformation on $\mathscr S'(\mathbb R^n)$. We find easily that $\mathcal F^4=Id$, so that if $T\in\mathscr S'(\mathbb R^n)$ is such that $\mathcal F T=\lambda T$, then $\lambda$ is a fourth root of unity.

Question: Are all the tempered distributions $T$ such that $\mathcal FT=T$ known? Two examples are very classical: first the Gaussians $e^{-π\vert x\vert^2}, e^{-π\langle Ax,x\rangle } $ where $A$ is a positive definite matrix with determinant 1, second the case $$ T_0=\sum_{k\in \mathbb Z^n}\delta_k, $$ where the equality $\mathcal FT_0=T_0$ is the Poisson summation formula.

I think that, thanks to your answers and a reference, I found the answer: using What are fixed points of the Fourier Transform which deals with the $L^2$ case, one may guess that the fixed points of $\mathcal F$ in $\mathscr S'(\mathbb R^n)$ are $$ \Bigl\{S+\mathcal F S+\mathcal F^2 S+\mathcal F^3 S \Bigr\}_{S\in \mathscr S'(\mathbb R^n)}=(Id+\mathcal F+\mathcal F^2+\mathcal F^3)(\mathscr S'(\mathbb R^n)). $$ Since $\mathcal F^4=Id$ on $\mathscr S'(\mathbb R^n)$, the above distributions are indeed fixed points and if $\mathcal F T=T$, then $$ T=\frac14\bigl(T+\mathcal F T+\mathcal F^2 T+\mathcal F^3 T\bigr), $$ conluding the proof.

The Fourier transform $\hat u$ is defined on the Schwartz space $\mathscr S(\mathbb R^n)$ by $ \hat u(\xi)=\int e^{-2iπ x\cdot \xi} u(x) dx. $ It is an isomorphism of $\mathscr S(\mathbb R^n)$ and the inversion formula is $ u(x)=\int e^{2iπ x\cdot \xi} \hat u(\xi) d\xi. $ The Fourier transformation can be extended to the tempered distributions $\mathscr S'(\mathbb R^n)$ with the formula $$ \langle \hat T,\phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)} = \langle T, \hat \phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}. $$ Below we note $\mathcal F$ the Fourier transformation on $\mathscr S'(\mathbb R^n)$. We find easily that $\mathcal F^4=Id$, so that if $T\in\mathscr S'(\mathbb R^n)$ is such that $\mathcal F T=\lambda T$, then $\lambda$ is a fourth root of unity.

Question: Are all the tempered distributions $T$ such that $\mathcal FT=T$ known? Two examples are very classical: first the Gaussians $e^{-π\vert x\vert^2}, e^{-π\langle Ax,x\rangle } $ where $A$ is a positive definite matrix with determinant 1, second the case $$ T=\sum_{k\in \mathbb Z^n}\delta_k, $$ where the equality $\mathcal FT=T$ is the Poisson summation formula. A discussion of the $L^2$ solutions is given in What are fixed points of the Fourier Transform, but I am interested in all distributions solutions.

The Fourier transform $\hat u$ is defined on the Schwartz space $\mathscr S(\mathbb R^n)$ by $ \hat u(\xi)=\int e^{-2iπ x\cdot \xi} u(x) dx. $ It is an isomorphism of $\mathscr S(\mathbb R^n)$ and the inversion formula is $ u(x)=\int e^{2iπ x\cdot \xi} \hat u(\xi) d\xi. $ The Fourier transformation can be extended to the tempered distributions $\mathscr S'(\mathbb R^n)$ with the formula $$ \langle \hat T,\phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)} = \langle T, \hat \phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}. $$ Below we note $\mathcal F$ the Fourier transformation on $\mathscr S'(\mathbb R^n)$. We find easily that $\mathcal F^4=Id$, so that if $T\in\mathscr S'(\mathbb R^n)$ is such that $\mathcal F T=\lambda T$, then $\lambda$ is a fourth root of unity.

Question: Are all the tempered distributions $T$ such that $\mathcal FT=T$ known? Two examples are very classical: first the Gaussians $e^{-π\vert x\vert^2}, e^{-π\langle Ax,x\rangle } $ where $A$ is a positive definite matrix with determinant 1, second the case $$ T_0=\sum_{k\in \mathbb Z^n}\delta_k, $$ where the equality $\mathcal FT_0=T_0$ is the Poisson summation formula.

I think that, thanks to your answers and a reference, I found the answer: using What are fixed points of the Fourier Transform which deals with the $L^2$ case, one may guess that the fixed points of $\mathcal F$ in $\mathscr S'(\mathbb R^n)$ are $$ \Bigl\{S+\mathcal F S+\mathcal F^2 S+\mathcal F^3 S \Bigr\}_{S\in \mathscr S'(\mathbb R^n)}=(Id+\mathcal F+\mathcal F^2+\mathcal F^3)(\mathscr S'(\mathbb R^n)). $$ Since $\mathcal F^4=Id$ on $\mathscr S'(\mathbb R^n)$, the above distributions are indeed fixed points and if $\mathcal F T=T$, then $$ T=\frac14\bigl(T+\mathcal F T+\mathcal F^2 T+\mathcal F^3 T\bigr), $$ conluding the proof.