Understanding a step in Stein's proof that the Schwartz space is closed under convolution. I am looking at Stein's proof that if $f,g \in \mathcal{S}(\mathbb{R})$, then so is $f*g$. The definition of the Schwartz space on $\mathbb{R}$ given is, the set of all indefinitely differentiable functions $f$ so that $f$ and all its derivatives are rapidly decreasing, in the sense that 
$$\sup_{x\in \mathbb{R}}|x|^k |f^{(l)}(x)| &lt; \infty \; \text{for every}\; k,l\ge 0.$$
 In the proof below, I have four questions.

First, how can I check the assertion that we have $\sup_x |x|^l |g(x-y)|\le A_l (1+|y|)^l$? I tried considering the two cases, but I cannot figure out how to get this bound. In fact, I don't really understand why we need to consider two separate cases. Shouldn't we be able to get $|x|^l |g(x-y)|\le |x-y|^l|g(x-y)|+|y|^l|g(x-y)|\le (1+|y|^l)A_l\le(1+|y|)^l A_l$, where $A_l=\sup_x |x|^l |g(x)|$?

Next, how can we guarantee that the integral $\int |f(y)|(1+|y|)^l dy$ is bounded for every $l\ge 0$? 

Finally, in the last paragraph, how is the interchange of differentation and integration justified in the case by the rapid decrease of $dg/dx$? Also, why do we only take the differentiation on $g$ and not $f$? I am not aware of this kind of  theorem. 

I would greatly appreciate any hints or solutions.
For your proof
$$|x|^l |g(x-y)|≤|x-y|^l |g(x-y)| + |y|^l |g(x-y)|$$
is incorrect, the triangle inequality doesn't hold for $|\cdot|^l$. Consider ie $x=2,y=1$, $l=2$ you have $|2|^2=4 \not≤|2-1|^2+|1|^2=2$.

I'm not sure how the hint should be used, but you can also do it like this:
$$|x-y|^l|g(x)|≤\sum_kC_k |x|^{l-k}|y|^k|g(x)|≤A_l(1+|y|)^l$$
Where the Schwartz property of $g$ was used to bound $|x|^{l-k}|g(x)|$ in each summand. Then
$$\sup_{x}|x|^l |g(x-y)|=\sup_{x} |x+y|^l|g(x)|$$
and the result follows.
The integral is bounded because
$$|f(y)|(1+|y|^l)(1+|y|^{2})≤C$$
follows from the Schwartz condition. Then you have since $|f|(1+|y|^l)$ is measurable and pointwise dominated by $\frac C{1+|y|^2}$ it is integrable.
$$\frac{d}{dx}\int f(y)g(x-y)dy=\frac{d}{dx}\lim_{n\to\infty}\int_{-n}^n f(y)g(x-y)dy$$
We want to swap the limit with the derivative, this is possible if $\int_{-n}^n f(y)g(x-y) dy$ and $\int_{-n}^n f(y) g'(x-y) dy$ converge uniformly on compacta (wrt $x$). Here:
$$\left|\int_{\mathbb R} f(y)g(x-y) dy -\int_{-n}^n f(y) g(x-y)dy\right|≤\left(\int_{-\infty}^{-n}+\int_n^\infty\right)|f(y)g(x-y)|dy$$
From the Schwartz condition bound $g$ by some constant and bound $f(y)$ by $\frac c{1+y^2}$ to see that the expression goes to zero uniformly. One does essentially the same for the derivative integral.
This gives that
$$\frac{d}{dx}\int f(y) g(x-y) dy = \lim_{n\to\infty}\frac{d}{dx}\int_{-n}^n f(y) g(x-y) dy = \lim_{n\to\infty}\int_{-n}^n f(y) g'(x-y) dy\\
=\int f(y) g'(x-y)dy$$
From the rule of derivation under the integral. Proceed via induction to get it for $\frac{d^k}{dx^k}$.
The expression $\frac{d^k}{dx^k}(f*g)=(f*g^{(k)})$ is useful because we know already that for any Schwartz $f,g$:
$$\sup_{x}|x|^l |f*g|(x)&lt;\infty$$
Thus, since $g^{(k)}$ is Schwartz, you have:
$$\sup_{x}|x|^l|(f*g)^{(k)}|(x)=\sup_{x}|x|^l |f*g^{(k)}|(x)&lt;\infty$$
Schwartz space is invariant under Fourier transform, and it is closed under pointwise multiplication. And Fourier transform takes convolution to pointwise multiplication.