If $1<p<2$ then $\mathscr{F}: L^p \to L^{p'}$ is not surjective. I had this as a homework problem a week back.

The reason is the bounded inverse theorem: $\mathscr{F}: L^p \to L^{p'}$ is injective, (by fourier inversion on the dense subspace of schwarz functions). If the map were surjective then there would be an inverse that would be continuous, since $\mathscr{F}$ is an open map under this assumption.

Thus we just need to prove that there is no bounded inverse: for $f \in \mathscr{S}$, there is no $c$ such that $||\hat f||_{p} \leq c ||f||_{p'}$ with the constant only depending on $p$. Equivalently we can show that there is no constant $c$ such that $||\hat f||_{p'} \geq c ||f||_p$ for $f \in \mathscr{S}$.

This is easy: The function $f_\lambda=e^{-\pi i \lambda x^2-\pi x^2}$ satisfies $||f_\lambda||_p=c$ independent of $\lambda$, whereas $||\hat f_\lambda||_{p'} \leq c \lambda^{1/p'-1/2}$. But there is no constant such that $\lambda^{1/p'-1/2} \geq c$ for all $\lambda >0$. Therefore the fourier transform is not surjective from $L^p \to L^{p'}$ for $1<p<2$

If $1\leq p<2$ then $\mathscr{F}: L^p \to L^{p'}$ is not surjective. I had this as a homework problem a week back.

The reason is the bounded inverse theorem: $\mathscr{F}: L^p \to L^{p'}$ is injective, (by fourier inversion on the dense subspace of schwarz functions). If the map were surjective then there would be an inverse that would be continuous, since $\mathscr{F}$ is an open map under this assumption.

Thus we just need to prove that there is no bounded inverse: For $f \in \mathscr{S}$, there is no constant $c$ such that $ ||f||_{p} \leq c ||\hat f||_{p'}$ for $f \in \mathscr{S}$ with the constant only depending on $p$.

This is easy: The function $f_\lambda=e^{-\pi i \lambda x^2-\pi x^2}$ satisfies $||f_\lambda||_p=c$ independent of $\lambda$, whereas $||\hat f_\lambda||_{p'} \leq c \lambda^{1/p'-1/2}$. But there is no constant such that $c \leq \lambda^{1/p'-1/2} $ for all $\lambda >0$. Therefore the fourier transform is not surjective from $L^p \to L^{p'}$ for $1\leq p<2$

If $1<p<2$ then $\mathscr{F}: L^p \to L^{p'}$ is not surjective. I had this as a homework problem a week back.

The reason is the bounded inverse theorem: $\mathscr{F}: L^p \to L^{p'}$ is injective, (by fourier inversion on the dense subspace of schwarz functions). If the map were surjective then there would be an inverse that would be continuous, since $\mathscr{F}$ is an open map under this assumption.

Thus we just need to prove that there is no bounded inverse: for $f \in \mathscr{S}$, there is no $c$ such that $||\hat f||_{p'} \leq c ||f||_{p}$ with the constant only depending on $p$. Equivalently we can show that there is no constant $c$ such that $||\hat f||_{p'} \leq c ||f||_p$ for $f \in \mathscr{S}$.

This is easy: The function $f_\lambda=e^{-\pi i \lambda x^2-\pi x^2}$ satisfies $||f_\lambda||_p=c$ independent of $\lambda$, whereas $||\hat f_\lambda||_{p'} \leq c \lambda^{1/p'-1/2}$. But there is no constant such that $\lambda^{1/p'-1/2} \geq c$ for all $\lambda >0$. Therefore the fourier transform is not surjective from $L^p \to L^{p'}$ for $1<p<2$

If $1<p<2$ then $\mathscr{F}: L^p \to L^{p'}$ is not surjective. I had this as a homework problem a week back.

The reason is the bounded inverse theorem: $\mathscr{F}: L^p \to L^{p'}$ is injective, (by fourier inversion on the dense subspace of schwarz functions). If the map were surjective then there would be an inverse that would be continuous, since $\mathscr{F}$ is an open map under this assumption.

Thus we just need to prove that there is no bounded inverse: for $f \in \mathscr{S}$, there is no $c$ such that $||\hat f||_{p} \leq c ||f||_{p'}$ with the constant only depending on $p$. Equivalently we can show that there is no constant $c$ such that $||\hat f||_{p'} \geq c ||f||_p$ for $f \in \mathscr{S}$.

This is easy: The function $f_\lambda=e^{-\pi i \lambda x^2-\pi x^2}$ satisfies $||f_\lambda||_p=c$ independent of $\lambda$, whereas $||\hat f_\lambda||_{p'} \leq c \lambda^{1/p'-1/2}$. But there is no constant such that $\lambda^{1/p'-1/2} \geq c$ for all $\lambda >0$. Therefore the fourier transform is not surjective from $L^p \to L^{p'}$ for $1<p<2$