Trying to show that $z \mapsto f_z : \mathbb{C} \to L^1(\mathbb{R})$ is complex differentiable where $f_z(x) = e^{-(x+z)^2}$ Let $g$ be the entire function $g(z) = e^{-z^2}$. Note $g$ is integrable along every horizontal line. For each complex number $z \in \mathbb{C}$, define $f_z : \mathbb{R} \to \mathbb{C}$ by $f_z(x) = g(z+x)$.  I am trying to show that the map $z \mapsto f_z : \mathbb{C} \to L^1(\mathbb{R})$ is complex differentiable in the sense that, for all $z \in \mathbb{C}$,
$$\frac{f_{z +h} - f_z}{h} \to (f_z)'$$
in $L^1$ as the complex variable $h \to 0$. In other words, I want to show that, for all $z \in \mathbb{C}$,
$$ \int_{-\infty}^\infty \left| \frac{g(z+h+x) -g(z+x)}{h} - g'(z+x) \right| \ dx \to 0$$
as $h \to 0$. There is no harm in assuming $z$ is pure imaginary above since the integral is translation invariant in the real direction. 
So far I have had the idea to write
$$g(z+h+x) -g(z+x) = \int_0^1 g'(z+th +x) \ d(th) = h \int_0^1 g'(z+th+x) \ dt$$ and
$$ g'(x+x) = \int_0^1 g'(z+x) \ dt$$
so that
$$\frac{g(z+h+x) -g(z+x)}{h} - g'(z+x) = \int_0^1 \left( g'(z+th + x) - g'(z+x) \right) \ dt,$$
which is bounded in magnitude by the largest variation in $g'$ over the line from $z+x$ to $z+h+x$. This, would seem to mean, perhaps, that there is an estimate of
$$\int_{-\infty}^\infty \left| \frac{g(z+h+x) -g(z+x)}{h} - g'(z+x) \right| $$
related to an area integral of $|g''|$ over on the strip bounded between the lines $\{z+x:x + \mathbb{R}\}$ and $\{z+x+h:x \in \mathbb{R}\}$? 
Anyhow, I'm beginning to get muddled. Help would be appreciated.
 A: Differentiation under integral sign is painful, it's better to integrate. 
Accordingly, I would begin with $\psi(z)=-2ze^{-z^2}$, $\psi_z(x)=\psi(x+z)$ and consider 
the map $z\mapsto \psi_z $ from $\mathbb C$ to $L^1(\mathbb R)$. This map is continuous: 
as $h\to 0$, the functions $\psi_{z+h}$ converge pointwise to $\psi_{z}$ and they 
are all dominated by some Gaussian. 
Fix $z$.  Given $\epsilon>0$, we can find
$\delta$ such that $\|\psi_{z+h}-\psi_z\|_{L^1}<\epsilon$ whenever $|h|<\delta$. 
Since $$\frac{f_{z+h}-f_z}{h} = \int_0^1 \psi_{z+th} \,dt\tag{1}$$ 
it follows that
$$\left\|\frac{f_{z+h}-f_z}{h} - \psi_z\right\|_{L^1} 
\le  \int_0^1 \|\psi_{z+th}-\psi_z\|_{L^1} \,dt <\epsilon$$
whenever $|h|<\delta$. 
Concerning (1): the integral on the right can be understood as the limit of Riemann sums (say, over 
uniform partitions): the sums converge pointwise to the left hand side, and the dominated convergence theorem ensures
they converge in $L^1$
A: Given any two parabolas $p_1,p_2$, it is intuitively obvious that there is a third parabola $p$ such that $p \leq p_1,p_2$.  This is pretty well the idea behind the following:

Fact: Fix $M > 0$. Then, the parabola $x^2/2 - M^2$ is less than or equal to the parabola $(x+a)^2$ for all $a \in [-M,M]$. 

Since $x \mapsto e^{-x}$ is order-reversing, we get:

Corollary: Fix $M > 0$. Then each of the Gaussians $e^{-(x+a)^2}$, $a \in [-M,M]$, is bounded above by the Gaussian $e^{M^2} e^{-x^2/2}$.

OK, now let $f(z) = e^{-z^2}$ which is entire with complex derivative $\psi(z) = -2z e^{-z^2}$. Correspondingly, we get the families of functions $\mathbb{R} \to \mathbb{C}$
\begin{align*} f_z(x) = f(x+z) && \psi_z(x) = \psi(x+z) \end{align*}
with $z$ ranging over $\mathbb{C}$. 

Claim: If $z = a+ib$ where $|a|,|b| \leq M$, then we have the bounds:
  \begin{align*} | f_z(x) | \leq e^{2M^2} e^{-x^2/2} && |\psi_z(x)| \leq  2e^{2M^2} (|x| +M )e^{-x^2/2}. \end{align*}
  This shows that the families $\{f_z\}$, $\{\psi_z\}$, $z \in \mathbb{C}$ are in $L^1(\mathbb{R})$ and, moreover,  that, when $z$ is constrained to some bounded region, there will be an $L^1$ function dominating the lot.
Proof: $$f_z(x) = e^{-(x+a)^2} e^{-2i(x+a)b } e^{b^2}$$
  whence
  $$|f_z(x)| \leq e^{-(x+a)^2} e^{b^2} \leq \left( e^{M^2} e^{-x^2/2}  \right) \left(  e^{M^2} \right) = e^{2M^2} e^{-x^2/2}.$$
  Also $$|\psi_z(x)| = |-2(x+z) f_z(x)| \leq 2( |x| + M) e^{2M^2} e^{-x^2/2}$$

Now, we want to prove that, for any fixed $z \in \mathbb{C}$, 
$$ \left\| \frac{f_{z+h} - f_z}{h} - \psi_z \right\|_1 \to 0$$
as the complex variable $h$ tends to $0$. Obviously this holds pointwise over $\mathbb{R}$ since $\psi_z$ is indeed the derivative of $f_z$. We just want a dominating function so that dominated convergence will apply. For each $x \in \mathbb{R}$, we write
$$\frac{ f_{z+h}(x) - f_z(x)}{h} - \psi_z(x) = \frac{ f(x + z+h) - f(x+z)}{h} - \psi(x+z) = \int_0^1 \left( \psi(x+z+th) - \psi(x+z) \right) \ dt = \int_0^1 \left( \psi_{z+th}(x) - \psi_z(x) \right) \ dt$$ 
so that
$$ \left| \frac{ f_{z+h}(x) - f_z(x)}{h} \right| \leq  \int_0^1 \left| \psi_{z+th}(x) - \psi_z(x) \right| \ dt   \leq \max_{ 0 \leq t \leq 1} |\psi_{z+th}(x)| + | \psi_z(x)|$$
But, since $\{ z+th: 0 \leq t \leq 1\}$ is a bounded set of complex numbers (assuming, with no harm done, that $|h| \leq 1$), we will have our dominating function by the claim.
