For $f\in L^{1}(\mathbb R),$ we define its Fourier transform as follows: $\hat{f}(t)=\int_{\mathbb R} f(x) e^{-ix\cdot t} dx ,(t\in \mathbb R).$
Suppose that $f\in L^{1}(\mathbb R)$ with $\hat{f}(0)=1.$
Let $\phi\in \mathcal{S}(\mathbb R),$ (Schwartz space). For $\lambda >0,$ we define, $\phi_{\lambda}(x)=\lambda^{-1} \phi(x/\lambda).$ Then $\|\phi_{\lambda}- \phi_{\lambda} \ast f \|_{L^{1}(\mathbb R)}\to 0$ as $\lambda \to \infty.$
[Since $\hat{f}(0)=1,$ for $x\in \mathbb R,$ we have, $\phi_{\lambda}(x)-\phi_{\lambda}\ast f(x) = \int_{\mathbb R} f(y) (\phi_{\lambda}(x)-\phi_{\lambda}(x-y)) dy;$ so $\|\phi_{\lambda}-\phi_{\lambda}\ast f\|_{L^{1}(\mathbb R)} \leq \int_{\mathbb R} |f(y)|(\int_{\mathbb R}\lambda^{-1}|(\phi(x/\lambda)- \phi ((x-y)/\lambda)| dx) dy= \int_{\mathbb R}|f(y)|( \int_{\mathbb R} (\phi(z)-\phi(z-\lambda^{-1}z)| dz)dy$
;and the inner integral is at most $2\|\phi\|_{L^{1}},$ and it tends to zero for every $y\in \mathbb R,$ as $\lambda \to \infty.$ Hence by dominated convergence theorem, we get, $\|\phi_{\lambda}- \phi_{\lambda} \ast f \|_{L^{1}(\mathbb R)}\to 0$ as $\lambda \to \infty.$]
My Question is: (1) Suppose $f\in L^{2}(\mathbb R)$ with $\hat{f}(0)=1.$ Then can we expect, $\|\phi_{\lambda}- \phi_{\lambda} \ast f \|_{L^{2}(\mathbb R)}\to 0$ as $\lambda \to \infty ?$ (2) If answer is yes, can we expect similar result for other $L^{p}$ ?
Edit: It is well-known that, for $f\in L^{p}, (1\leq p <\infty),$ we have $\|f-\phi_{\lambda}\ast f\|_{L^{p}} \to 0$ as $\lambda \to 0.$ (Bit roughly speaking, approximate identity )