Questions on the proof of $f*g\in C^\infty(\mathbb R)$ when $f\in L^2(\mathbb R)$ and $g\in C_c^\infty(\mathbb R)$ I am working through the proof that the convolution of a square integrable function with a compactly supported continuously differentiable function is itself continuously differentiable:
"Let $f\in L^2(\mathbb R)$ and $g\in C_c^\infty(\mathbb R)$. Show that $f*g\in C^\infty(\mathbb R)$ and that $(f*g)^{(k)}=f*g^{(k)}$ for $k\in\mathbb N$."
To this end I have been making use of the following questions and the particularly linked answers:
Convolution of locally integrable and compactly supported infinitely differentiable function
Derivative of convolution
Differentiating under integral for convolution
I am confident that I understand the idea of the proof, however, there are some points which are common throughout each of the attempts which I am not entirely certain on and would appreciate to have better explained.


*

*How, exactly, is the Lebesgue Dominated Convergence Theorem being applied? In taking the difference quotient (in working with the derivative) we obtain something like,
$$\lim_{h\to0}\int_\mathbb{R}f(z)\frac{g(x+h-z)-g(x-z)}h\text{d}z,$$
for which we want to find some integrable function $\psi$ so that for all $z\in\mathbb R$,
$$\left|\frac{g(x+h-z)-g(x-z)}h\right|<\psi(z).$$
That is to say, we want to find a function, $\psi$, which dominates the above difference quotient. I see that that the hypotheses of the LDCT are satisfied (since $g\in C_c^\infty$ it is Borel measurable), but how do we rectify the fact that our sequence (the difference quotient) isn't indexed by the natural numbers? How do we apply LDCT when we have $0<h<1$, which is uncountable.

*In order to find the dominating $\psi$, as mentioned above, we make use of the Mean Value Theorem (Rather than just assume that such a dominating function exists, we should move to draw out the specific existence of such a function). In the first of the linked questions, this is done as follows,
\begin{eqnarray*}
 \left|g(x + h - z) - g(x - z)\right|
 & = & 
 \left|
 \int_0^1\frac{\rm d}{{\rm d}s} g(x - z + sh)\; {\rm d} s
 \right|
 \\
 &\leq & 
 |h|\max_{x\in \mathbb R} |g'(x)|. 
\end{eqnarray*}
How is it that $g'(x-z+sh)$ for $s\in(0,1)$ is bounded by $g'(x)$ as a function of $x$ alone? I am thinking that one defines $g'_s(x):=g'(x - z + sh)$ for $s\in(0,1)$ and then argues that for all $s\in(0,1)$, $|g'_s(x)|<|g'(x)|$ for all $x\in\mathbb R$ so that the sequence of functions $(g_s)_{s\in(0,1)}$ is uniformly bounded. But how does one transition from dealing with $g'(x-z+sh)$ to $g'(x)$? And how does this affect our considerations of the support we are on?
 A: Nice questions!
For 1. There is a version of the dominated convergence theorem that works for general limits (not only sequences) but you can also derive it as follows:
Pick any sequence $h_n\rightarrow 0$. Look at the term $\psi(h):=\frac{g(x+h-z)-g(x-z)}{h}$. The sequence $\psi(h_n)$ satisfies the condition of the dominated convergence theorem  and so $\lim_{n\rightarrow\infty} \int \psi(h_n) d\mu$ convergence. It is not hard to see that the limit for every $h_n$ is the same (it's always the integral of the derivative of $g$). Therefore by Heine's theorem this means that $\lim_{h\rightarrow 0} \int \psi(h_n) d\mu$ exists and equal to the same limit.
For 2. You're just confused because they use the same element to denote two different things. $g'$ is a function. Let $M=\max_{y\in\mathbb{R}} |g'(y)|$.
Then if you pick $t=x-z+sh$, this is clearly a real number and so $|g'(t)|\leq M$. 
A: *

*The DCT works not only with sequences, also in cases like this one, in which $h\to0$. You can convince yourself by considering sequences $h_n\to0$.

*It is not using $|g'_s(x)|\le|g'(x)|$, but $|g'_s(x)|\le\max_{x\in\Bbb R}|g'(x)|$.
