Understanding a step in the proof of Fourier Inversion Theorem by Stein I'm reading Stein and Shakarchi's Fourier Analysis, and have a question about the proof of Fourier inversion for Schwartz class functions. 
In the proof below, I have two questions. Here, $K_\delta(x)=\delta^{-1/2}e^{-\pi x^2/\delta}$.

 First, why does the integral $\int f(x)K_\delta (x)dx$ goes to $f(0)$ as $\delta$ tends to $0$ since $K_\delta$ is a good kernel? I've attached all the previous propositions in the text as well, but I can't see why it must follow because $K_\delta$ is a good kernel. My guess is, using Corollary 1.7, we have $(f*K_\delta) (0)=\int f(x)K_\delta (0-x) dx \to f(0)$ uniformly as $\delta \to 0$ and $K_\delta$ is even so $K_\delta (-x)=K_\delta (x)$ and we can get the result. But this requires $K_\delta$ being even and not just a good kernel. 
Finally, how does the second integral, $\int \hat{f}(\xi)G_\delta (\xi)d\xi$ converges to $\int \hat{f}(\xi)d\xi$ as $\delta \to 0$? I think we are supposed to change the order of $\lim_{\delta \to 0}$ and the integration, but how are we guaranteed to do this? 


 A: For the good kernel part you don't need symmetry of the kernel. As it integrates to one the claim follows from the more general result that for $f$ continuous and uniformly bounded we have
  $$ \lim_{\delta\rightarrow 0} \frac{1}{\sqrt{\delta}} \int_{\Bbb R} \left(f(x)-f(0)\right)\  e^{-\pi x^2/\delta} dx = 0 $$
Given $\epsilon>0$ find $\eta$ so that $|x|<\eta \Rightarrow |f(x)-f(0)|<\epsilon/2$ and then $\delta$ so that the contribution from the  integral over $|x|\geq \epsilon$  (the condition (iii)) is smaller than $\epsilon/2$
For the second part the easiest is to note that $\hat{f}$ is $L^1$ and use Dominated convergence (since $G_\delta(x)$ goes pointwise to 1). But you may also give an $\epsilon,\delta$ - proof by hand:
The function $\hat{f}$ is integrable, say $I=\int|\hat{f}|<+\infty$ so given $\epsilon>0$ first find $M$ so that 
$$\int_{|\xi|>M} |\hat{f}(\xi)| d\xi < \epsilon/2$$
Now find $\delta>0$ so that 
$$\sup_{|\xi|\leq M} (1-G_\delta(\xi)) < \frac{\epsilon}{2 I}$$
Then $$ \left|\int_{|\xi|\leq M}\ \hat{f}(\xi)(1-G_\delta(\xi)) d\xi \right|<\epsilon/2 $$ Combining with the above, noting that $0< G_\delta\leq 1$  we obtain the result.
