Sobolev inequalities and Wiener algebra

Question

It follows from the Gagliardo-Nirenberg inequality that for a locally integrable function $f$ defined on $\mathbb R^d$ (we assume $d\ge 3$) such that $\nabla f$ belongs to $L^2(\mathbb R^d)$ and $$ \lim_{\vert x\vert\rightarrow \infty} f(x)=0, \tag{0}$$ we have $$ f\in L^{\frac {2d}{d-2}}(\mathbb R^d). \tag{1}$$ Now, the hypothesis means also that $\mathbb R^d\ni\xi\mapsto \vert \xi\vert\hat f(\xi)$ belongs to $L^2(\mathbb R^d)$ and since $$ \frac{2d}{d+2}<2<\frac{2d}{d-2}, $$ we may wonder, thinking about the Hausdorff-Young inequality, whether $$ \hat f \in L^{\frac {2d}{d+2}}(\mathbb R^d), \tag{2}$$ which would imply (1).

Question In fact, I do not believe that (2) follows from (0) and $\nabla f\in L^2$, but although it is easy to prove that (2) does not follow from $\nabla f\in L^2$, the standard counterexample given by a logarithmic perturbation of a critical power does not obviously satisfy (0). So my question is related to finding $f$ satisfying (0), $\nabla f\in L^2$ but not (2).

Answer 1

It is indeed false, (2) does not follow from (0) and $\nabla f\in L^2$. The idea is that since you consider the critical scaling, the potential inequality (on the Fourier side) fails both at $0$ and at $\infty$, so we can find a counterexample which is supported only in the vicinity of $0$. Next, since $\frac{2d}{d+2} > 1$, we can also additionally guarantee that $\hat{f}\in L^1$. But then $f(x)\to 0, |x|\to\infty$ follows simply by the Riemann–Lebesgue lemma.

Explicitly, take $$\hat{f}(x) =\begin{cases}|x|^{-\frac{d+2}{2}}\log(\frac{1}{x})^{-\frac{d+2}{2d}}, |x| \le \frac{1}{2},\\ 0, |x| > \frac{1}{2},\end{cases}$$ Then $x\hat{f}(x)\in L^2$, $\hat{f}(x)\notin L^{\frac{2d}{d+2}}$ and $\hat{f}(x)\in L^1$ (the last one is true with a big margin of error since $\frac{d+2}{2} < d$). So, $\nabla f(x)\in L^2$, $\hat{f}(x)\notin L^{\frac{2d}{d+2}}$ and $f(x)\to 0, |x| \to \infty$, as required.