If $u\in L^1(\mathbb{R}^n)$ then its Furier transform is $ \hat{u}(y):=\frac{1}{(2\pi)^{n/2})}\int_{\mathbb{R}^n} e^{-ix\cdot y}u(x) d x $ and its inverse Fourier transform $ \check{u}:=\frac{1}{(2\pi)^{n/2})}\int_{\mathbb{R}^n} e^{ix\cdot y}u(x) d x. $ Plancherel's theorem, as stated in the book by Evans, tells us that if $u\in L^1(\mathbb{R}^n)\cap L^2(\mathbb{R}^n)$ then $\hat{u},\check{u}\in L^2(\mathbb{R}^n)$ and $ \| \hat{u}\|_{L^2(\mathbb{R}^n)}= \| \check{u}\|_{L^2(\mathbb{R}^n)}=\|u\|_{L^2(\mathbb{R}^n)}. $
First question. In the first part of the demonstration (see figure below) why establish identity $$ \frac{1}{(2\pi)^{n/2})}\int_{\mathbb{R}^n} \hat{w}(y)e^{-\epsilon |y|^2}u(x) d y = \frac{1}{(2\pi)^{n/2})}\int_{\mathbb{R}^n} w(y)e^{\frac{-|y|^2}{4\epsilon}}u(x) d x \quad ? $$
Second question. I suspect that in the second part of the proof (see figure below) can be concluded that $ \hat{u} \in L^2 (\mathbb{R}^n) $. How can I get this conclusion from the second part of the proof?
Third question. In the second part of the demonstration it is concluded that $ \| \hat{u} \|_{L^2(\mathbb{R}^n)} = \| u \|_{L^2 (\mathbb{R}^n)} $. How can I get this other conclusion from the second part of the demonstration?