What is the inverse operation of the Gateaux derivative?

Question

Context

This is purely a question of curiosity, I can't provide much context.

I am taking a course in computational fluid dynamics to say that a class of problems can be classified as "saddle point problems".

We introduced the Gateaux derivative and made some application examples just to see how to use it.

---

Let $V$ be a Hilbert space and let $F$ be a generic functional from $V$ to $\mathbb{R}$.

As a definition in our space we said that if

$${}_{V'}\langle d,w\rangle_{V}:=\lim_{\lambda\to 0^{+}}\frac{F(v+\lambda w)-F(v)}{\lambda}$$

then we define the Gateaux derivative as

$$d=F'(v)$$

---

We have given two examples of Gateaux derivative, one for a linear functional and one for a quadratic functional.

$$\text{If }F(v)={}_{V'}\langle f,v\rangle_V\quad\text{then}\quad F'(v)=f\quad \forall v\in V$$

$$\text{If }F(v)=\frac{1}{2}a(v,v)\quad\text{then}\quad F'(v)=a(v,\cdot)\quad\forall v\in V$$

Where $a$ is a symmetric bilinear form.

---

Question

Since the analogy with derivatives is obvious, I was wondering what the inverse operation was.

Because

• in the case of a linear operator the inverse operation is the duality with a test function essentially (and it makes sense because duality is an integral anyway)
• In the case of the symmetric bilinear operator it is essentially evaluating the function on a test function, but then in the case of it being symmetric you also have to divide by $2$

Clarification, if $a$ was a non-symmetric bilinear operator I would have had

$$\text{if }F(v)=\frac{1}{2}a(v,v)\text{ then }F'(v)=\frac{1}{2}(a(v,\cdot)+a(\cdot,v))\quad\forall v\in V$$

Answer 1

I think what you have in mind is the potential operator. An operator $A \colon X \to X^*$ between Banach spaces is called a potential operator if there exists a functional $f \colon X \to \mathbb R$, called the potential of $A$, which is Gateaux differentiable and $f'=A$.

You can find many more details in Chapter 5 of the book Basic Monotonicity Methods with Some Applications

Answer 2

Not an answer, just some thought. The inverse operation is the same as finding the anti-gradient (potential function) of a vector field in Calculus 3. And I don't see much trouble caused by infinite dimensions.

But anti-gradient does not necessarily exist for a given vector field, so the inverse operation is not always well-defined. Assuming good smoothness, one necessary condition of existence is that the second Gateaux derivative (as a bilinear form) is symmetric.

You may think of the Hessian matrix of a smooth scalar function defined on $\mathbb{R}^d$. Or you can calculate the further derivative of your two examples to check this.