Frechet derivative I want to find the Fréchet derivative of the following functional:
$$
\begin{align}
F : C[-1,1] &\rightarrow \mathbb{R}\\
x &\mapsto x(0)\int_0^1 \sin\ x(t) \, dt.
\end{align}
$$
How can I do it?
 A: Let
$$
G:C([-1,1])\to\mathbb{R}:x\mapsto x(0)\\
H:C([-1,1])\to\mathbb{R}:x\mapsto \int_0^1\sin(x(t))dt
$$
then $F(x)=G(x)\cdot H(x)$ and $F'(x)=G(x)\cdot H'(x)+H(x)\cdot G'(x)$. So we need to find Frechet derivatives of $G$ and $H$. Since $G$ is linear, then $G'(x)(h)=G(h)$. As for the $H$ we had to compute its derivative directly from definition. For any $x,h\in C([-1,1])$ we have
$$
\begin{align}
H(x+h)-H(x) &=\int_0^1(\sin(x(t)+h(t))-\sin(x(t)))dt=
\int_0^1(\cos(x(t))h(t)+o(h(t)))dt \\
&=\int_0^1\cos(x(t))h(t)+\int_0^1o(h(t))dt=
\int_0^1\cos(x(t))h(t)dt+o(h)
\end{align}
$$
Thus we see
$$
H'(x)(h)=\int_0^1\cos(x(t))h(t)dt
$$
And the final result is
$$
F'(x)(h)=h(0)\int_0^1\sin(x(t))dt+x(0)\int_0^1\cos(x(t))h(t)dt
$$
A: Since for every $a,\delta \in \mathbb{R}$ we have
$$
\sin(a+\delta)=\sin a+\delta\cos a+o(\delta),
$$
taking $x, h \in C([-1,1])$ we get
\begin{eqnarray}
F(x+h)&=&[x(0)+h(0)]\int_0^1\sin(x(t)+h(t))dt\cr
&=&[x(0)+h(0)]\int_0^1[\sin(x(t))+h(t)\cos(x(t))+o(h(t))]dt\cr
&=&x(0)\int_0^1\sin(x(t))dt+x(0)\int_0^1h(t)\cos(x(t))dt+h(0)\int_0^1\sin(x(t))dt+R(x,h)\cr
&=&F(x)+L_x[h]+R(x,h),
\end{eqnarray}
where
\begin{eqnarray}
L_x[h]&=&x(0)\int_0^1h(t)\cos(x(t))dt+h(0)\int_0^1\sin(x(t))dt\cr
R(x,h)&=&x(0)\int_0^1o(h(t))dt+h(0)\int_0^1[h(t)\cos(x(t))+o(h(t))]dt.
\end{eqnarray}
Clearly $L_x: C([-1,1]) \to \mathbb{R},\ h\mapsto L_x[h]$ is a linear continuous map, and
$$
|R(x,h)| \le |x(0)|o(|h|_\infty)+|h|_\infty[|h|_\infty+o(|h|_\infty)]=o(|h|_\infty).
$$
Hence the Frechet derivative $DF$ of $F$ is gievn by $DF(x)[h]=L_x[h]$ for every $x, h \in C([-1,1])$.
A: A super easy way to solve this is using Leibniz's rules:
Differentiation under integral sign. If $f:[a, b] \times \mathrm{O} \to \mathrm{F}$ (where $\mathrm{O}$ is an open set in a normed and complete linear space $\mathrm{E}$ and $\mathrm{F}$ is a complete normed linear space) is such that it is continuous and $\mathbf{D}_2f:\mathrm{O} \to \mathscr{L}_\mathrm{E}(\mathrm{F})$ (the latter is the space of continuous linear maps $\mathrm{E} \to \mathrm{F}$) is also continuous then the function $\displaystyle g(x) = \int\limits_a^b dt\ f(t, x)$ is differentiable with continuity and its derivative is given in accordance with the rule
$$\mathbf{D}g(x) = \int\limits_a^b dt\ \mathbf{D}_2f(t,x);$$
notice both sides belong to $\mathscr{L}_\mathrm{E}(\mathrm{F}).$
Product rule. Let $\mathrm{E}, \mathrm{F}, \mathrm{G}$ and $\mathrm{H}$ be four vector spaces, normed and assume they are complete with their corresponding norms. If $\mathrm{B}$ is the restriction of a bilinear function $\mathrm{E} \times \mathrm{F} \to \mathrm{G}$ to an open set, then for any pair of differentiable functions $f$ with values in $\mathrm{E}$ and $g$ with values in $\mathrm{F},$ both of them defined on an open set of $\mathrm{H},$ then their product relative to $\mathrm{B},$ defined to be $\mathrm{B}(f, g)$ is differentiable and its derivative is the linear function $$\mathbf{D}\mathrm{B}(f,g)(x) h = \mathrm{B}(\mathbf{D}f(x) h, g(x)) + \mathrm{B}(f(x), \mathbf{D}g(x)h).$$
Knowing the previous two results, the desired derivative of the proposed exercise is readily seen to be:
$$\mathbf{D}\mathrm{F}(x) h = h(0) \int\limits_0^1 dt\ \sin(x(t)) + x(0) \int\limits_0^1 dt\ \cos(x(t)) h(t)$$
