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Let $\cal H$ be a Hilbert space with inner product $(\cdot, \cdot)$ and norm $\|\cdot\|:=\sqrt{(\cdot,\cdot)}$. The gradient of a real-valued functional $E:\cal H\to\mathbb R$ at a point $u\in\cal H$, denoted as $\nabla E[u]$, is defined as $$ (\nabla E[u] , v) = \lim\limits_{\varepsilon \to 0}\frac{E[u+\varepsilon v] - E[u]}{\varepsilon}.$$ If $\cal H = L^2$, this definition corespondents to the functional derivative

My question is whether there exists a functional $E:\cal H\to \mathbb R$ suth that its gradient satisfies $$ \nabla E[u] = \nabla f\circ \nabla g[u],$$ where $f,g:\cal H \to\mathbb R$ are continuously differential and ‘$\circ$’ denotes the composition of two operators. Moreover, g can be viewed as a conjugate functional of a strongly convex functional and thus $\nabla g$ is Lipschitz continuous. Precisely, the conjugate of $g$ defined as $$ g^*[v] = \sup_{\text{dom} g}\{(u,v)-g[u]\}$$

Similar problem:

Inverse functional derivatives: Find a functional whose functional derivative is a given function

Exterior derivative of a form and $d(d\omega)=0$?

Thanks

ChangChen
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    This fails already in two dimensions. Let $f(x_1,x_2)=x_1^2$ and $g(x_1,x_2)= x_1x_2$ then $\nabla f(\nabla g(x)) = (2x_2,0)$, which is not the derivative of any function. – s.harp May 05 '20 at 10:45
  • Thanks! It’s a good counter-example. Do you know any sufficient condition to make this conclusion true? – ChangChen May 06 '20 at 03:11
  • In finite dimensions you want $$\sum_{k=1}^n(\partial_k \partial_i f[\nabla g(x)]) \cdot \partial_j \partial_k g(x) )= \sum_{k=1}^n (\partial_k \partial_j f[\nabla g(x)]) \cdot \partial_i \partial_k g(x) ) $$ to hold for all $i,j$ from $1,...,n$ (and for $f, g$ to be defined on all of $\Bbb R^n$). This condition is necessary and sufficient in that case. A more succint form of writing it down would be $d [\nabla f \circ \nabla g[u]]^T = 0$ where $d$ is the exterior derivative. I think this condition can be adapted to infinite dimensions, but I don't know for sure (keywords: deRahm cohomology) – s.harp May 06 '20 at 11:09
  • Appreciate very much! I learned a lot from your answers. – ChangChen May 07 '20 at 09:12

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