Find the Gateaux differential

Question

Let $E = C[0,1]= \{x:[0,1]\to \mathbb{R},\hbox{ } x \hbox{ continium} \}$ and $g:\mathbb{R} \times [0,1]\to \mathbb{R}$, a $C^1$ function with partial derivative $g_x$ bounded. Let be the functional

\begin{align} f: E & \to \mathbb{R} \\ x & \to f(x) = \int_{0}^{1} g[x(t),t]dt \\ \label{eq:1} \end{align}

We know that the Gâteaux differential $\delta f(x; h)$ of $f$ at $x\in E$ in the direction $h \in E$ is defined as:

$$ \delta f(x; h) = \lim_{\alpha \to 0} \frac{f(x + \alpha h) - f(x) }{\alpha} $$

So I want to find the Gâteaux differential $ \delta f(x; h)$ whith $f$ defined above.

With some simple calculus using the definition, we have

$$ \delta f(x; h) = \lim_{\alpha \to 0} \int_{0}^{1} \frac{g[x(t) + \alpha h(t),t ] - g[x(t),t]}{\alpha} dt$$

The problem now is the interchange between the limit and the integral. My question is:

Can I use the:

Dominated convergence theorem: Let ${f_n}$ be a sequence of real-valued measurable functions on a measure space $(S, \Sigma, \mu)$. Suppose that the sequence converges almost avery where to a function $f:S\to \mathbb{R}$ and that $$|f_n(x)| \leq K, \quad \forall n \in \mathbb{N}$$
Then $$\lim_{n \to \infty}\int f_n d\mu = \int f d\mu$$ ? How?

If not, how I find the Gateaux differential?

I would like to remark that $g_x$ is bounded, so by the mean value theorem, we have for all $t \in [0,1]$, there is some $\alpha^{'}$ that

$$| \frac{g[x(t) + \alpha h(t),t ] - g[x(t),t]}{\alpha}| = |g_x[x(t) + \alpha^{'} h(t),t]| \leq K$$

Answer 1

For all $t \in \left[ {0,1} \right]$, we have $$\mathop {\lim }\limits_{\alpha \to 0} \frac{{g\left( {x\left( t \right) + \alpha h\left( t \right),t} \right) - g\left( {x\left( t \right),t} \right)}}{\alpha } = {D_1}g\left( {x\left( x \right),t} \right) \cdot h\left( t \right).$$ Further more, we have $$\left| {\frac{{g\left( {x\left( t \right) + \alpha h\left( t \right),t} \right) - g\left( {x\left( t \right),t} \right)}}{\alpha }} \right| = {\left\| g \right\|_{{C^1}}}{\left\| h \right\|_{{C^0}\left( {\left[ {0,1} \right]} \right)}} < + \infty .$$ By Dominated convergence theorem, we have $$\mathop {\lim }\limits_{\alpha \to 0} \int_0^1 {\frac{{g\left( {x\left( t \right) + \alpha h\left( t \right),t} \right) - g\left( {x\left( t \right),t} \right)}}{\alpha }dt} = \int_0^1 {{D_1}g\left( {x\left( x \right),t} \right) \cdot h\left( t \right)dt} .$$ Finally, we have $$\delta \left( {f;h} \right) = \int_0^1 {{D_1}g\left( {x\left( x \right),t} \right) \cdot h\left( t \right)dt} .$$