The desired claim will follow from the following more general result:
Let
- $E_i$ be a normed $\mathbb R$-vector space
- $x_0\in E_1$
- $\varepsilon>0$
- $U:=B_\varepsilon(x_0)$
- $h\in E_1\setminus\{0\}$ and $$\tilde\varepsilon:=\frac\varepsilon{\left\|h\right\|_{E_1}}$$
We first remember the following mean value theorem: If $f:U\to E_2$ is differentiable in direction $h$, then $$\frac{\left\|f(x_0+th)-f(x_0)\right\|_{E_2}}{|t|}\le\sup_{0<|\tau|<|t|}\left\|{\rm D}_hf(x_0+\tau h)\right\|_{E_2}\;\;\;\text{for all }0<|t|<\tilde\varepsilon\tag5.$$
The proof of $(5)$ is a straightforward application of the ordinary mean value theorem to $$g(t):=f(x_0+th)\;\;\;\text{for }|t|<\tilde\varepsilon.$$
Now we can state and prove the following:
Let
- $(\Omega,\mathcal A,\mu)$ be a measure space;
- $E_2$ be a $\mathbb R$-Banach space; and
- $f:\Omega\times U\to E_2$.
If
- $f(\;\cdot\;,x)\in\mathcal L^1(\mu;E_2)$ for all $x\in U$;
- $f(\omega,\;\cdot\;)$ is differentiable in direction $h$ for all $\omega\in\Omega$; and
- There is a $\psi\in\mathcal L^1(\mu)$ with $$\sup_{x\in U}\left\|{\rm D}_hf(\omega,x)\right\|_{E_2}\le\psi(\omega)\;\;\;\text{for }\mu\text{-almost all }\omega\in\Omega,$$ then $$F(x):=\int f(\omega,x)\:\mu({\rm d}x)\;\;\;\text{for }x\in U$$ is well-defined and differentiable at $x_0$ in direction $h$ with $${\rm D}_hF(x_0)=\int{\rm D}_hf(\omega,x_0)\:\mu({\rm d}\omega)\tag6.$$
We prove it in the following way: Let $(t_n)_{n\in\mathbb N}\subseteq(-\tilde\varepsilon,\tilde\varepsilon)$ with $t_n\xrightarrow{n\to\infty}0$. Then $$x_n:=x_0+t_nh\in U\;\;\;\text{for }n\in\mathbb N.$$ Let $$g_n(\omega):=\frac{f(\omega,x_n)-f(\omega,x_0)}{t_n}\;\;\;\text{for }\omega\in\Omega\text{ and }n\in\mathbb N.$$ By 2., $$g_n(\omega)\xrightarrow{n\to\infty}g(\omega):={\rm D}_hf(\omega,x_0)\;\;\;\text{for all }\omega\in\Omega$$ and hence (by 1.) $g$ is strongly $\mathcal A$-measurable. By the mean value theorem and 3., $$\left\|g_n\right\|_{E_2}\le\psi\;\;\;\mu\text{-almost everywhere}.$$ Thus, by Lebesgue's dominated convergence theorem, $g\in\mathcal L^1(\mu)$ with $$\left\|g_n-g\right\|_{L^1(\mu)}\xrightarrow{n\to\infty}0$$ which yields the claim.
We immediately obtain the following corollary: If additionally to 1.,
- $f(\omega,\;\cdot\;)$ is differentiable in each direction for all $\omega\in\Omega$; and
- For all $h\in E_1\setminus\{0\}$, there is a $\psi\in\mathcal L^1(\mu)$ with $$\sup_{0<|\tau|<\frac\varepsilon{\left\|h\right\|_{E_1}}}\left\|{\rm D}_hf(\omega,x_0+\tau h)\right\|_{E_2}\le\psi(\omega)\;\;\;\text{for }\mu\text{-almost all }\omega\in\Omega,\tag7$$
then $F$ is differentiable at $x_0$ in each direction $h\in E_1$ with the directional derivative ${\rm D}_hF(x_0)$ given by $(6)$. If additionally to 1., 4. and 5.,
- $f(\omega,\;\cdot\;)$ is Gâteaux differentiable at $x_0$ for all $\omega\in\Omega$; and
$\exists\phi\in\mathcal L^1(\mu)$ with $$\left\|{\rm d}f(\omega,x_0)\right\|_{\mathfrak L(E_1,\:E_2)}\le\phi(\omega)\;\;\;\text{for }\mu\text{-almost all }\omega\in\Omega\tag8,$$ then $F$ is Gâteaux differentiable at $x_0$ with $$\left\|{\rm d}F(x_0)\right\|_{\mathfrak L(E_1,\:E_2)}\le\int\psi\:{\rm d}\mu\tag9.$$ If additionally to 1., 4. and 5.,
$f(\omega,\;\cdot\;)$ is Fréchet differentiable at $x_0$ for all $\omega\in\Omega$, and
- $\exists\varphi\in\mathcal L^1(\mu)$ with $$\sup_{x\in U}\left\|{\rm D}f(\omega,x)\right\|_{\mathfrak L(E_1,\:E_2)}\le\varphi(\omega)\;\;\;\text{for }\mu\text{-almost all }\omega\in\Omega,\tag{10}$$ then $F$ is Fréchet differentiable at $x_0$.
As for the question: We need to assume that $$\exists\varepsilon>0:\forall|t|<\varepsilon:\int|g(u+tv)|\:{\rm d}\mu<\infty\tag{11}$$ and $$\exists\phi\in\mathcal L^1(\mu):\sup_{|t|<\varepsilon}|g'(u+tv)v|\le\phi\;\mu\text{-almost everywhere}\tag{12}.$$ Now we apply the corollary above to $$f:\Omega\times U\to\mathbb R\;,\;\;\;(\omega,t)\mapsto g(u(\omega)+tv(\omega)),$$ where $U:=(-\varepsilon,\varepsilon)$. Then it hols
- $f(\;\cdot\;,t)\in\mathcal L^1(\mu)$ for all $t\in U$;
- $f(\omega,\;\cdot\;)$ is (Fréchet) differentiable at $0$ for all $\omega\in\Omega$; and
- $\sup_{t\in U}\left\|{\rm D}f(\omega,t)\right\|_{\mathfrak L(\mathbb R)}=\sup_{t\in U}\left|g'(u(\omega)+tv(\omega))v(\omega)\right|\le\phi(\omega)$ for $\mu$-almost all $\omega\in\Omega$
and hence $$F(t):=\int f(\omega,t)\:\mu({\rm d}\omega)=\int g(u+tv)\:{\rm d}\mu\;\;\;\text{for }t\in U$$ is well-defined and (Fréchet) differentiable at $0$ with $$F'(0)=\int g'(u)v\:{\rm d}\mu\tag{13}.$$
This is precisely the desired claim from the question. I think the assumptions I imposed here cannot be relaxed, but feel free to correct me, if I'm wrong.
BTW: We've basically shown that $$u\mapsto\int g(u)\:{\rm d}\mu\tag{14}$$ is differentiable at $u$ in direction $v$.
Since the terminology is not consistent in the literature, let me briefly define the following: Let $E_i$ be a normed $\mathbb R$-vector space, $U\subseteq E_1$, $f:U\to E_1$ and $x\in U^\circ$. Then $f$ is called
- differentiable at $x$ in direction $h\in E_1$ if $$\frac{f(x+th)-f(x)}t\xrightarrow{t\to0}{\rm D}_hf(x)\tag1$$ for some ${\rm D}_hf(x)\in E_2$;
- Gâteaux differentble at $x$ if $f$ is differentiable at $x$ in each direction and $${\rm D}_hf(x)={\rm d}f(x)h\;\;\;\text{for all }h\in E_1$$ for some ${\rm d}f(x)\in\mathfrak L(E_1,E_2)$;
- Fréchet differentiable at $x$ if $$\frac{\left\|f(x+h)-f(x)-{\rm D}f(x)h\right\|_{E_2}}{\left\|h\right\|_{E_1}}\xrightarrow{h\to0}\tag2$$ for some ${\rm D}f(x)\in\mathfrak L(E_1,E_2)$.
$$\frac{f(u(x)+\varepsilon v(x))-f(u(x))}\varepsilon=v(x)\cdot\frac{f(u(x)+[\varepsilon v(x)])-f(u(x))}{[\varepsilon v(x)]}\to v(x)f'(u(x)) \quad ?$$
The way the differential quotient is written just doesn't seem right (also in $(1)$).
– Mars Plastic Aug 05 '19 at 18:30