It is easy to see that $$f\in L^\infty(\mathbb R^d), \ f\in L^1(\mathbb R^d) \implies f\in L^p(\mathbb R^d) \ \text{for all}\ p\in[1,\infty] \label{1}\tag{A}$$
Indeed, $ \int|f|^p \le \|f\|_{L^\infty}^{p-1}\|f\|_{L^1}. $
Let $\hat f$ denotes the Fourier transform. Another well-known easy result is $$ \|\hat f\|_{L^\infty} \le \|f\|_{L^1}.$$
It seems natural then to ask if the following generalisation of \eqref{1} could hold? $$f\in L^\infty(\mathbb R^d), \ \hat f\in L^\infty(\mathbb R^d) \implies f\in L^p(\mathbb R^d) \ \text{for some}\ p\in(1,\infty] \label{2}\tag{B}$$ I know that \eqref{2} for $p=1$ is false in general by taking $d=1,f(x)=\frac{\sin x}x\notin L^1$ since $f\in L^p$ for all $p>1$ and $\hat f = C \mathbf 1_{[-1,1]} \in L^\infty$. Could it be true for any other values of $p$? Dare I hope for all $p\in(1,\infty]$?
What you expect is not true. Below is an example.
Let $$\Lambda:=\{\lambda\in\mathbb C: |\operatorname{Im}\lambda|<|\operatorname{Re}\lambda|\}.$$ Let $$f_\lambda(x)=e^{-\pi\lambda^2|x|^2}, \qquad \lambda\in\mathbb C,\ x\in\mathbb R^d,$$ then we have $$\widehat {f_\lambda}(\xi)=\lambda^{-d}e^{-\pi\frac{|\xi|^2}{\lambda^2}}, \qquad \lambda\in\Lambda, \ \xi\in\mathbb R^d,\tag{$*$}$$ where the Fourier transform of $f$ is defined by $$\hat f(\xi)=\int_{\mathbb R^d}f(x)e^{-2\pi ix\cdot\xi}\,dx,\qquad \xi\in\mathbb R^d.$$
Now we take a sequence $\{\lambda_n\}_{n=1}^\infty\subset\Lambda$ such that $\lim_{n\to\infty}\lambda_n=\lambda_0=e^{\frac{i\pi}4}\in\bar\Lambda$. It can be checked that $(*)$ holds for $\lambda_0$ in the sense of distribution, with $$f_{\lambda_0}(x)=e^{-i\pi|x|^2}, \qquad \widehat{f_{\lambda_0}}(\xi)=e^{-\frac{i\pi}4d}e^{i\pi|\xi|^2}.$$ We have $f_{\lambda_0}\in L^\infty(\mathbb R^d)$ and $\widehat{f_{\lambda_0}}\in L^\infty(\mathbb R^d)$, but clearly $$f_{\lambda_0}\notin L^p(\mathbb R^d), \qquad \widehat{f_{\lambda_0}}\notin L^p(\mathbb R^d),\qquad \forall p\in(0,\infty).$$