Show $\int\frac{f(u+\varepsilon v)-f(u)}\varepsilon\:{\rm d}\mu\xrightarrow{\varepsilon\to0}\int f'(u)v\:{\rm d}\mu$ for a large class of $f,u,v$ Let $(\Omega,\mathcal A,\mu)$ be a measure space. I want to show that $$\int\frac{g(u+tv)-g(u)}t\:{\rm d}\mu\xrightarrow{t\to0}\int g'(u)v\:{\rm d}\mu\tag1$$ for a preferably large class of differentiable $g:\mathbb R\to\mathbb R$ and $\mathcal E$-measurable $u,v:\Omega\to\mathbb R$.

Without any further assumption, we see that $$g(u(\omega)+tv(\omega))-f(u(\omega))=\int_0^tf'(u(\omega)+sv(\omega))v(\omega)\:{\rm d}s\tag2$$ for all $\omega\in\Omega$ and $t\in\mathbb R$ and $$\frac{f(u(\omega)+tv(\omega))-f(u(\omega))}t\xrightarrow{t\to0}f'(u(\omega))v(\omega)\tag3$$ for all $\omega\in\Omega$.

So, all we need to conclude is to find conditions under which the integrals in $(1)$ are well-defined and Lebesgue's dominated convergence theorem (or a theorem of that kind) is applicable.

We may note that if $(t_n)_{n\in\mathbb N}\subseteq\mathbb R\setminus\{0\}$ with $t_n\xrightarrow{n\to\infty}0$, then (by the mean value theorem) there is a $x_n(\omega)$ between $u(\omega)$ and $u(\omega)+t_nv(\omega)$ with $$\frac{g(u(\omega)+t_nv(\omega))-g(u(\omega))}{t_n}=g'(x_n(\omega))v(\omega)\tag4$$ for all $\omega\in\Omega$ and $n\in\mathbb N$.
So, some kind of linear growth condition on $g'$ (i.e. $|g'(x)|\le c(1+|x|)$ for all $x\in\mathbb R$ for some $c\ge0$) might be feasible.
 A: 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 ${\rm D}f(x)\in\mathfrak L(E_1,E_2)$.

