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

  • 1
    $\begingroup$ What do you mean by $\langle f(u), u\rangle$ when $u$ is only an $L^1$ function? One needs $L^2$ to speak of inner products. $\endgroup$ – nullUser Mar 26 '16 at 23:52
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    $\begingroup$ you are right, fixed $\endgroup$ – Nightwing Mar 26 '16 at 23:55
  • $\begingroup$ And what is meant by $\nabla F$ when the domain of $F$ is $L^2$ functions? $\endgroup$ – nullUser Mar 26 '16 at 23:59
  • $\begingroup$ Do you have assumptions like $\langle df(x) u, v\rangle = \langle df(x) v, u\rangle$? $\endgroup$ – user99914 Mar 27 '16 at 5:59
  • $\begingroup$ no, only what comes out there.I thought I would write the integral as $ \int_{0}^u f (v) \cdot dv$ making $ v = tu; dv = udt $ then derived. $\endgroup$ – Nightwing Mar 27 '16 at 6:10

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.


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