# Trying to understand Gateaux differential

1. Dec 15, 2013

### BruceW

Hi everyone!
I've been trying to get my head around the Gateaux differential. (or derivative, however it is called). So anyway, I've seen it defined like this:
$$\lim_{\tau \rightarrow 0} \frac{F(u + \tau \psi) - F(u)}{\tau}$$
where the function $F$ takes $X$ to $Y$ (which are both locally convex topological vector spaces). Also, $u$ is an element of some open subset of $X$ and $\psi$ is any element of $X$. i.e. the Gateaux differential only exists at $u$ if this limit is the same for any choice of $\psi$. Now, I'm not a mathematician. So I'm not used to this kind of stuff, so please say if I've done something stupid. But I decided to try a really simple example, just have $X$ and $Y$ the real numbers. And, to keep it super-simple, use the function $F(x)=x$. So, now check if the Gateaux differential exists:
$$\lim_{\tau \rightarrow 0} \frac{u + \tau \psi - u}{\tau}$$
Which is then equal to:
$$\lim_{\tau \rightarrow 0} \frac{\tau \psi}{\tau}$$
Which, after cancelling the $\tau$ simply gives:
$$\psi$$
But clearly, this is not the same for any choice of $\psi$. So does that mean the Gateaux differential does not exist for this simple example? Or is $\psi$ meant to have unit magnitude, in some sense? hmm. the fact that $X$ is a locally convex topological vector space means that it must be a vector space with a family of seminorms. So it would make sense if $\psi$ must be defined to have a set norm or seminorm. For my super-simple example, we could define the norm as the absolute value, and if we say that $\psi$ must have a set norm, then the Gateaux differential would exist.

So, is that the answer? $\psi$ is not any element of $X$, but its norm must be set, and then if the limit exists (for any $\psi$ with the same norm), then the Gateaux differential exists?

And so, in the context of Lagrangian mechanics, when we calculate the functional derivative of the Action, we vary the test function, but we cannot vary the norm of the test function. So, say if we chose $L^2$ as our space of test functions, then for our Gateaux differential to exist, we require that we get the same answer with any $L^2$ function that has the same value for its square integral. Also, I realize that to calculate the functional derivative of the action, we don't really need to care about keeping the norm of the test function the same, because we can just say that far away from the minimized path, the test function takes whatever values it needs so that the norm of the test function stays the same. And I think we can usually do this with nice enough functions, so I won't worry too much about this when I'm doing Lagrangian mechanics.

Anyway, I hope someone can verify some of this, or has some insight on this topic. It's pretty interesting, I think.

edit: in fact, the use of the Gateaux differential in extremizing the action is kinda weird, because when the Euler-Lagrange equations are not satisfied, the Gateaux differential does not exist. So it's true we are choosing the action to be stationary, but it is not like the functional derivative of the action is non-zero on nearby paths. The functional derivative does not exist on nearby paths! Although, I guess we don't care, since if we have found a stationary action, that is all we need.

p.s. I've been assuming that we want to use the Gateaux differential as the functional derivative simply because (I think) the Gateaux differential is the most general mathematical definition for the functional derivative.

Last edited: Dec 15, 2013
2. Dec 15, 2013

### R136a1

The Gateaux derivative is a generalization of the directional derivative, so it depends on the choice of $\psi$. A different $\psi$ gives a different derivative, in much the same way as $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ are different (where the former is the Gateaux derivative with $\psi = e_1$ and the latter is with $\psi = e_2$.

So, there is no unique Gateaux derivative, but rather only one depending on the vector $\psi$. Of course, we can always define a function which sends $\psi$ to its Gateaux derivative at $x$ at direction $\psi$. In this sense, it is unique.

3. Dec 15, 2013

### R136a1

You might also want to check out the frechet derivative then: http://en.wikipedia.org/wiki/Fréchet_derivative

4. Dec 15, 2013

### BruceW

I'm only interested in the situations where $F$ is Gateaux differentiable. So when I was saying "the Gateaux differential only exists at $u$ if this limit is the same for any choice of $\psi$" I mean where $F$ is Gateaux differentiable. And so the Gateaux differential is unique, where $F$ is Gateau differentiable.

edit: thanks for the reply though :)

5. Dec 16, 2013

### R136a1

You seem to have a very restrictive definition of Gateaux differentiable. Usually, the definition is that a function is Gateaux differentiable if the Gateaux differentials exist. Nothing says that they must equal.

What you ask is for a function $F:\mathbb{R}^2\rightarrow \mathbb{R}$ that all directional derivatives equal in $(0,0)$. This means that there is a unique $L$ such that for all $\psi = (\alpha, \beta)$ holds that

$$\alpha \frac{\partial F}{\partial x}(0,0) + \beta \frac{\partial F}{\partial y}(0,0) = L.$$

This is almost never satisfied, except for perhaps constant $F$.

6. Dec 16, 2013

### BruceW

not just constant $F$, but yeah, I agree that Gateaux differentiability is probably 'rare' in a certain mathematical sense. The reason I'm interested in Gateaux differentiability is because (as far as I can work out) the variational method in Lagrangian mechanics requires a map which is Gateaux differentiable at least for some elements of the domain. And the entire principle of Lagrangian mechanics (and therefore all of quantum field theory), relies on Gateaux differentiability.

p.s. you must define some norm for $\psi=(\alpha ,\beta )$ and the norm must be held constant when you take the Gateaux differential. This was my main question in my first post. Now I am almost certain that it is true.

7. Dec 16, 2013

### R136a1

Well, not even the identity $\mathbb{R}\rightarrow \mathbb{R}$ will be Gateaux differentiable when interpreted in your sense. Seems like a completely useless notion to me then.

The usual notion of Gateaux differentiable just asks for the limits to exists, but not to equal. In this sense, the identity is differentiable.

8. Dec 16, 2013

### BruceW

I am starting to realize that I was totally wrong about this. Thanks for helping me through it. OK, so we don't set the norm of $\psi$ to some value. The norm of $\psi$ can take whatever value it wants. And therefore, it is very rare indeed to get the kind of differentiability that I was looking for. And in the context of Lagrangian mechanics and variational methods, we do get this kind of differentiability at the points where the Euler-Lagrange equations are satisfied, but for most points, we will generally not get this kind of differentiability.

Also, OK, my version of Gateaux differentiable is not the usual definition. And my version occurs very rarely. This is my first time learning about the Gateaux differential. And I was thinking it was in some way analogous to the idea of complex differentiable, in the sense that if a function is complex differentiable, then the limit is unique, and does not depend on the direction. But now I see that the Gateaux differential is not like the complex differential in that way.