I try to show the following statement:
Let $x \in \mathbb R^d$. Then there exists a function $u \in L^1(\mathbb R^d)$ with $\Vert u \Vert_1 \leq 1$ and $\hat u(x) = 1$.
I solved it for the case $x = 0$. Let $u(y) = \pi^{-d/2} e^{-y^2}$ then it holds that
$$\Vert u \Vert_1 = \int_{\mathbb R^d} u(y) \, dy= \pi^{-d/2} \int_{\mathbb R^d} e^{-y^2}\, dy = \pi^{-d/2} \left(\int_{\mathbb R} e^{-y^2}\, dy\right)^d = \pi^{-d/2} \pi^{d/2} = 1.$$ Further I get $$\hat u(0) = \int_{\mathbb R^d} u(y) e^{- \mathrm i \cdot 0 \cdot y}\, dy = \int_{\mathbb R^d} u(y) dy = \Vert u \Vert_1 = 1.$$ So $u$ has the desired properties. But I really struggle to get the result for any other $x$. There was a hint to apply the Fourier transform on the function $(R_x u)(y) = u(x-y)$. Then I get easily that $\Vert R_x u \Vert_1 = \Vert u \Vert_1$ but I don't see how I can deduce $\widehat{R_x u}(x) = 1$. I would appreciate some hints :)
Let $u_x(y) = e^{i\langle x,y \rangle} u(y)$ and check that $u_x$ is integrable with $\|u_x\|_1 = 1$ and that $\hat{u}_x(t) = \hat{u}(t-x)$ (hence $\hat{u}_x(x) = 1$).