Inequalities in proof of Bernstein-type lemmas I'm working through the proof of Lemma 2.1 in Bahouri-Chemin-Danchin's Fourier Analysis and Nonlinear PDE. The place where I'm stuck boils down to (I think) proving the following inequality:
$$\|f\|_{L^1(\mathbb R^d)}\leq\||\cdot|^{2d}f\|_{L^\infty(\mathbb R^d)}$$
where we have that $f$ is not compactly supported. I think I've proved that there is some $n$ such that $\|f\|_{L^1}\leq\||\cdot|^nf\|_{L^\infty}$, but not the stronger result that $n=2d$ works. Am I on the right track?
 A: The desired inequality is false, even with constants, since it does not have the right sort of homogeneity. Take a function $f$ for which the inequality does hold, and define
$$g_{\lambda}(x) = \lambda^{d} f(\lambda x)$$
This is the $L^1$ dilation of $f$, and $$\|g_{\lambda} \|_1 = \|f\|_1$$ for all $\lambda > 0$. On the other hand,
\begin{align*}
\big| |x|^{2d} g_{\lambda}(x) \big| = \lambda^{-d} \big| |\lambda x|^{2d} f(\lambda x)\big| = \lambda^{-d} |y|^{2d} |f(y)|
\end{align*}
It follows that 
$$\big\| |.|^{2d} g_{\lambda} \big\| = \lambda^{-d} \big\||.|^{2d} f\big\|$$
But then we have
$$\|g_{\lambda}\|_1 = \|f\|_1 \le \big\||.|^{2d} f\big\|_{\infty} = \lambda^d \big\||.|^{2d} g_{\lambda}\big\|_{\infty}$$
Now send $\lambda$ to zero to find a counterexample.
A: The inequality you are trying to prove is indeed false, but the following is true : A positive constant C exists for which 
$∥f∥_{L^1(R^d)}≤ C ∥ (1 + |⋅|^{2})^d f∥_{L^∞((R^d))}$
And it comes from the fact that the function $\frac{1}{(1 + |⋅|^{2})^d}$ is integrable, thus one may write : 
$∥f∥_{L^1(R^d)} =  ∥(1 + |⋅|^{2})^df \frac{1}{(1 + |⋅|^{2})^d}∥ _{L^1((R^d))}$
And the results comes from the application of Hölder inequality with p=1 and q = $\infty$
I think this is the argument that is intended in Bahouri-Chemin-Danchin's book. 
