Let $\phi:\mathbb R \to \mathbb C$ be a function; and define $$\phi_{t}(x):=\frac{1}{t}\phi (x/t), (t>0).$$
We let, $\phi\in L^{1}(\mathbb R),$ then $\phi_{t}\in L^{1}(\mathbb R);$ in fact, we have, $\|\phi \|_{L^{1}}= \|\phi_{t}\|_{L^{1}(\mathbb R)}.$
We also assume that, $\hat{\phi}(0)=1= \int_{\mathbb R} \phi(x) dx;$ then $\{\phi_{t}\}_{t>0}$ forms a bounded approximate identity in $L^{1}(\mathbb R);$ that is, if $f\in L^{1}(\mathbb R)$, then $\|f\ast \phi_{t}- f\|_{L^{1}(\mathbb R)} \to 0$ as $t\to 0.$
My Question is: Fix $\xi_{0}\in \mathbb R.$ Then can we expect to find a sequence $\{k_{\lambda}\}_{\lambda >0} \subset L^{1}(\mathbb R);$ with the following properties: (1) $\hat{k_{\lambda}} (\xi_{0}) =0,$ for all $\lambda $, (2) $\|k_{\lambda}\|_{L^{1}(\mathbb R)} \leq C$ for all $\lambda$ and $C$ is some constant, (3) If $f\in L^{1}(\mathbb R),$ then $\|f\ast k_{\lambda} - f\|_{L^{1}(\mathbb R)} \to 0$ as $\lambda \to 0$ (or $\lambda \to \infty$ , depending on the choice of $k_{\lambda}$).
(Bit Roughly, problem states that, can we expect a bounded approximate identity in $L^{1}(\mathbb R)$, whose Fourier transform at particular point is zero. )
Edit: If I assume $\hat{f}(\xi_{0})=0$, then what can we say in in the above problem.( In view of the following remark)
Thanks,