I am trying to understand the proof of Tomas's theorem:

[![enter image description here][1]][1]


The proof reads

[![enter image description here][2]][2]


  [1]: https://i.sstatic.net/n5FQd.png
  [2]: https://i.sstatic.net/x8Ssi.png

My question:

How do we get the estimates 

$$\|T_k\ast f \|_{\infty}\lesssim 2^{-(n-1)k/2}\|f\|_{1},\qquad\qquad (1)$$
$$\|T_k\ast f \|_{2}\lesssim 2^{k}\|f\|_{2}.\qquad\qquad (2)$$


We can prove (1) by showing that 

$$\|T_k \|_{\infty}\lesssim 2^{-(n-1)k/2}\qquad \qquad (3)$$

And we can get (2) by proving that

$$\|T_k \|_{1}\lesssim 2^{k}\qquad\qquad (4)$$.

I am stuck with (3) and (4). 

By definition of $\widehat{d\theta}$ and $K$, we have that, for large enough $k$, 
 
$$T_k (x)=\int_{2^{k-1}\leq |x|\leq 2^k} \left(K\left(\frac{|x|}{2^k}\right)-K\left(\frac{|x|}{2^{k-1}}\right) \right)\int_{\mathbb{S^{n-1}}}e^{\dot{\imath}x\cdot \theta}d\theta.$$


I am not sure of the latter claim of mine. We have
$$k(x)=g(\frac{|x|}{2^k})$$ where $g\in \mathcal{S}(\mathbb{R})$. So, given, $\epsilon>0$, we can find $k_{\epsilon}>1$ such that $\int_{2^{k_{\epsilon}}}^{\infty} |g(r)|dr<\epsilon$. 


A second question:

The proof implicitly uses the fact that 

$$\sum_{k\geq 1} \|T_k\ast f\|_{p^{\prime}}\lesssim \sum_{k\geq 1} \|f\|_{p^{\prime}}$$. 

I think a hint of how to deal with (3) or (4) would help me figure this out.