Bounding $\Vert f''(x)\mathrm{sech}(x)\Vert_{L^2(\mathbb{R})}$ for functions $f$ supported on small frequencies.

Question

Consider a real-valued function $f\in L^2(\mathbb{R})$ such that its fourier transform is supported in some ball $B(0,\delta)$ for some $\delta>0$ small enough. Is it true that if $\delta$ is small enough then we have $$ \Vert f''(x) \mathrm{sech}(x)\Vert_{L^2}\leq \varepsilon(\delta) \Vert f(x)\mathrm{sech}(x)\Vert_{L^2} $$ for some $\varepsilon(\delta)<1$ satisfying that $\varepsilon(\delta)\to 0$ as $\delta \to0^+$? Intuitively I would like to say that if the support of the Fourier transform of $f$ is small enough (and centered around $0$), then some weighted $L^2$-norms of $f''$ should be way smaller than those of $f$. At least, if the sech(x) wasn't there, the property is trivially true in the $L^2$-norm, and the function $\varepsilon(\delta)$ is explicit, due to Plancharel. But once adding the sech(x), we have to deal with the convolution if we want to use Plancharel. Does anybody have an intuition why this should or should not be true? A counterexample maybe? Is there any additional hypothesis that could make this true?