gradient and inner product $f(·)=\nabla F(·)$ I have the following:
If $f:L^2(\mathbb{R})\to H$ is locally lipschitz with $f(0)=0$ and    $\langle f(u),u\rangle \leq 0$. Define 
$$F(u)=\int_{0}^1\langle f(tu),u\rangle dt$$ 
with $u\in L^1(\mathbb{R})$ and $H \subset L^2(\mathbb{R})$.
Show that $F(0)=0 , F(u)\leq 0$ and $f(·)=\nabla F(·)$ on $L^2(\mathbb{R})$.
I need to check the calculations of $f(·)=\nabla F(·)$, the theoretical part not interest me much.
I tried derivative $F$ with respect to $u$, but did not get anything.
Any help is appreciated
 A: I will prove your claim under the extra assumption that 
$$\tag{1} \langle df(x) u, v\rangle = \langle df(x)v, u\rangle,\ \ \ \forall x, u, v\in L^2(\mathbb R).$$
By definition and $(1)$, we have 
$$\begin{split}
dF(u)(v) &= \int_0^1 \langle df(tu)(tv), u\rangle+ \langle f(tu), v\rangle dt \\
&= \int_0^1 t\langle df(tu)v, u\rangle+ \langle f(tu), v\rangle dt\\
&=  \int_0^1 t\langle df(tu)u, v\rangle+ \langle f(tu), v\rangle dt
\end{split}$$
Since 
$$\frac{d}{dt} \langle f(tu), v\rangle = \langle df(tu) u, v\rangle,$$
using integration by part we have 
$$\begin{split}
 \int_0^1 t\langle df(tu)u, v\rangle dt &= t\langle f(tu), v\rangle\bigg|_0^1 - \int_0^1 \langle f(tu), v\rangle dt\\
&= \langle f(u), v\rangle  - \int_0^1 \langle f(tu), v\rangle dt.
\end{split} $$
Thus
$$ dF (u) v =\langle f(u), v\rangle,\ \ \ \forall u, v$$
and so $f(\cdot) = \nabla F (\cdot)$. 
On the other hand, the following shows that $(1)$ is almost necessary: 

If there is a $C^2$ function $G : L^2(\mathbb R) \to \mathbb R$ so that $\nabla G = f$, then $(1)$ holds. 

Proof: Let $x, u, v \in L^2(\mathbb R)$ consider
$$A(s, t) = G(x + su + tv)$$
Since $G$ is $C^2$, then so is $A$. Thus partial derivatives commute and 
$$ \langle df(x) u, v \rangle = \partial_s \partial_t A|_{s=t=0} = \partial_t \partial_s A|_{s=t=0} = \langle df(x) v, u \rangle.$$

Aside: In finite dimensional cases, Your $F$ is a standard way to show that all closed (that is, $(1)$ holds) one form $f$ on $\mathbb R^N$ is exact. 

