If $1<p<2$, then it is not possible to have the inequality $$ \|f\|_p \lesssim \|\widehat{f}\|_{p'} \quad\quad\quad\quad\quad (1) $$ for all $f\ge 0$. This follows from the existence of (positive) purely singular measures $\mu$ with $\widehat{\mu}\in L^{p'}$ (in fact, $\widehat{\mu}$ can have power decay). (I used this fact also in my answer to this question; here, however, decay of $\widehat{\mu}$ in an average sense is enough, which is much easier to obtain.)

I think it is intuitively clear that such a $\mu$ refutes (1), but to elaborate some more on this, consider $f_t=\varphi_t*\mu$, with $\varphi_t(x)=(1/t)\varphi(x/t)$ and $\varphi\ge 0$, $\varphi\in C_0^{\infty}$. Then, if we had (1), it would follow that $f_t$ is a Cauchy sequence in $L^p$; however, this sequence converges to $\mu$ in the sense of distributions, and as $\mu$ is singular, clearly $\mu\notin L^p$.

If $1<p<2$, then it is not possible to have the inequality $$ \|f\|_p \lesssim \|\widehat{f}\|_{p'} \quad\quad\quad\quad\quad (1) $$ for all $f\ge 0$. This follows from the existence of (positive) purely singular measures $\mu$ with $\widehat{\mu}\in L^{p'}$ (in fact, $\widehat{\mu}$ can have power decay). (I used this fact also in my answer to this question; here, however, decay of $\widehat{\mu}$ in an average sense is enough, which is much easier to obtain.)

I think it is intuitively clear that such a $\mu$ refutes (1), but to elaborate some more on this, consider $f_t=\varphi_t*\mu$, with $\varphi_t(x)=(1/t)\varphi(x/t)$ and $\varphi\ge 0$, $\varphi\in C_0^{\infty}$. Then, if we had (1), it would follow that $f_t$ is a Cauchy sequence in $L^p$; however, this sequence converges to $\mu$ in the sense of distributions, and as $\mu$ is singular, clearly $\mu\notin L^p$.

If $1<p<2$, then it is not possible to have the inequality $$ \|f\|_p \lesssim \|\widehat{f}\|_{p'} \quad\quad\quad\quad\quad (1) $$ for all $f\ge 0$. This follows from the existence of (positive) purely singular measures $\mu$ with $\widehat{\mu}\in L^{p'}$ (in fact, $\widehat{\mu}$ can have power decay). (See perhaps my answer to this question for some background.)

I think it is intuitively clear that such a $\mu$ refutes (1), but to elaborate some more on this, consider $f_t=\varphi_t*\mu$, with $\varphi_t(x)=(1/t)\varphi(x/t)$ and $\varphi\ge 0$, $\varphi\in C_0^{\infty}$. Then, if we had (1), it would follow that $f_t$ is a Cauchy sequence in $L^p$; however, this sequence converges to $\mu$ in the sense of distributions, and as $\mu$ is singular, clearly $\mu\notin L^p$.

If $1<p<2$, then it is not possible to have the inequality $$ \|f\|_p \lesssim \|\widehat{f}\|_{p'} \quad\quad\quad\quad\quad (1) $$ for all $f\ge 0$. This follows from the existence of (positive) purely singular measures $\mu$ with $\widehat{\mu}\in L^{p'}$ (in fact, $\widehat{\mu}$ can have power decay). (I used this fact also in my answer to this question; here, however, decay of $\widehat{\mu}$ in an average sense is enough, which is much easier to obtain.)

I think it is intuitively clear that such a $\mu$ refutes (1), but to elaborate some more on this, consider $f_t=\varphi_t*\mu$, with $\varphi_t(x)=(1/t)\varphi(x/t)$ and $\varphi\ge 0$, $\varphi\in C_0^{\infty}$. Then, if we had (1), it would follow that $f_t$ is a Cauchy sequence in $L^p$; however, this sequence converges to $\mu$ in the sense of distributions, and as $\mu$ is singular, clearly $\mu\notin L^p$.