Proving $\|(1+\xi^2)^{-1/2} \partial^2 \hat{f}\|_2 \le C \|\|(1+\xi^2)^{-1/2} \hat{f}\|_2$ for $\mathrm{supp}(f) \subset [-R,R]$? For $f \in L^2(\mathbb{R})$ with $\mathrm{supp}(f) \subset [-R,R]$, one has trivially
$$
\|\partial^2 \hat{f}\|_2 = \|\mathcal{F}(x^2 f)\|_2 =  \|x^2 f\|_2 \le R^2 \|f\|_2 = R^2 \|\hat{f}\|_2.
$$
My question is, whether one finds a $C > 0$ (depending on $R$ but not on $f$) such that for all such $f$:
$$
\|(1+\xi^2)^{-1/2} \partial^2 \hat{f}\|_2 \le C \|\|(1+\xi^2)^{-1/2} \hat{f}\|_2.
$$
Alternatively, what would be a counter-example?
I have a proof that works for $f \ge 0$ and $f \le 0$. In these cases $C = R^2$. 
For arbitrary $f$ however, the inequality does not hold with $C = R^2$, as can be seen with $\frac{1}{2}\delta(x-R) + \frac{1}{2} \delta(x+R) - a \delta(x)$. In this case $C = R^2 \sqrt{\coth{R}} > R^2$ is enough for all $a >0$.
I suspect that $C$ does not need to be much bigger for arbitrary $f$.
Proof for $f\ge 0$:
$$
\int_\mathbb{R} (1+\xi^2)^{-1} |\partial_\xi^2 \hat{f}|^2 \mathrm{d}{\xi}
= \int_\mathbb{R} (1+\xi^2)^{-1} \int_\mathbb{R} e^{i \xi x} x^2 f(x) \mathrm{d}{x} \int_\mathbb{R} e^{-i \xi x'} x'^2 f(x') \mathrm{d}{x'} \mathrm{d}{\xi} \\
 = \int_\mathbb{R} \int_\mathbb{R} x^2 f(x) x'^2 f'(x) \mathcal{F}((1+\xi^2)^{-1})(x'-x) \mathrm{d}{x} \mathrm{d}{x'} \\
 \le R^4 \int_\mathbb{R} \int_\mathbb{R} f(x) f'(x) \mathcal{F}((1+\xi^2)^{-1})(x'-x) \mathrm{d}{x} \mathrm{d}{x'}
= \int_\mathbb{R} (1+\xi^2)^{-1} |\hat{f}|^2 \mathrm{d}{\xi}.
$$
Here we used that $\mathcal{F}((1+\xi^2)^{-1}))(x) = c \exp(-|x|) > 0$.
 A: The inequality holds for $C = R^2 + 2R$.
Proof:
The Sobolev spaces $H^s(\mathbb{R})$ can be defined as preimages of the Bessel potential $H^s(\mathbb{R}) = J^{-s}(L^2(\mathbb{R}))$, where $J^s = \mathcal{F}^{-1} (1+\xi^2)^\frac{s}{2} \mathcal{F}$.
The inequality in question is thus equivalent to (using $\partial^2 \hat{f} = \mathcal{F}(x^2 f)$)
$$
\| x^2 f \|_{H^{-1}} \le C \|f \|_{H^{-1}}
$$
Now, since $H^{-1}(\mathbb{R})$ is an isometric realisation of $[H^1(\mathbb{R})]'$, we have the identity
$$
\| f \|_{H^{-1}} = \sup \{ |\langle f, \phi \rangle| : \phi \in H^1(\mathbb{R}), \| \phi \|_{H^1} =1 \}.
$$
Here, since $f \in L^2$ and $\phi \in H^1 \subset L^2$, the dual pairing is simply the inner product on $L^2$.
For every $\epsilon > 0$, we find a $\phi_\epsilon \in H^1$ with $\| \phi_\epsilon \|_{H^1} \le 1$, such that $| \langle x^2 f, \phi_\epsilon \rangle| + \epsilon \ge \|x^2 f \|_{H^{-1}}$.
Now one can easily construct, for every $\epsilon' >0$ a function $\zeta \in C^1_c(\mathbb{R})$, such that $\| \zeta \|_{C^1(\mathbb{R})} \le (1 + \epsilon') \| x^2 \|_{C^1([-R,R])}$ and $\zeta|_{[-R,R]} = x^2$.
This gives
$$
|\langle x^2 f, \phi_\epsilon \rangle| = |\langle \zeta f, \phi_\epsilon \rangle|=|\langle f, \zeta \phi_\epsilon \rangle| = \| \zeta \phi_\epsilon \|_{H^1} | \langle f, \zeta \phi_\epsilon / \|\zeta \phi_\epsilon \|_{H^1} \rangle|
$$
The inner product is bounded by the $H^{-1}$-norm,
$$
| \langle f, \zeta \phi_\epsilon / \|\zeta \phi_\epsilon \|_{H^1} \rangle| \le \| f \|_{H^{-1}},
$$
and
$$
\| \zeta \phi_\epsilon \|_{H^1} \le \| \zeta \|_{C^1} \| \phi_\epsilon \|_{H^1} \le \| \zeta \|_{C^1} \| \le (1+\epsilon') \|x^2 \|_{C^1([-R,R])}.
$$
Note that $\|x^2 \|_{C^1([-R,R])} = R^2 + 2R$.
Assembling all the pieces, we obtain
$$
\| x^2 f\|_{H^{-1}} \le (1+\epsilon') (R^2+2R) \|f\|_{H^{-1}} + \epsilon.
$$
Since $\epsilon$ and $\epsilon >0$ were chosen arbitrarily, the claim follows by $\epsilon, \epsilon' \to 0$.
