13
$\begingroup$

If you have a function $V: L \rightarrow \mathbb{R}$, where $L$ is an infinite dimensional topological vector space, there are multiple notions of differentiability. For $x,u \in L$, $V$ is Gateaux differentiable at $x$ in the direction $u$ if the limit $\underset{t \rightarrow 0}{\lim} \frac{V(x + tu) - V(x)}{t}$ exists. Supposing that you are in a Banach space, $V$ is Frechet differentiable if the above limit exists for all $u$ in a ball around $x$, and importantly, with the convergence being uniform over this neighborhood.

The question is, what difference does it make for a function to be Frechet differentiable versus Gateaux differentiable, maybe with respect to proving theorems that generalize the finite-dimensional setting, where the two notions of differentiability more or less agree. What kind of pathological behavior can functions exhibit that are merely Gateaux differentiable in every direction? There are also intermediate forms of differentiability between Frechet and Gateaux, defined in terms of uniform convergence of the difference quotients over some preferred family of sets (a bornology). Are there any intermediate kinds of differentiability that are important?

$\endgroup$
1

2 Answers 2

15
$\begingroup$

For Lipschitz functions in finite dimensional spaces, Gateaux and Frechet differentiability are the same, but there are huge differences when the domain is infinite dimensional. Lipschitz functions are Gateaux differentiable off a null set when the domain is a separable Banach space and the range has the Radon Nikodym property (e.g., is reflexive), where "null set" can have any of several different meanings. On the other hand, it is rare for a Lipschitz function on an infinite dimensional space to have a Frechet derivative anywhere. A great theorem of David Preiss says that a real valued Lipschitz function on an Asplund space has a point of Frechet differentiability, but it is not known whether a complex valued Lipschitz function on an Asplund space has a point of Frechet differentiability (although some wonderful progress was made recently by Lindenstrauss and Preiss).

If a bi-Lipschitz equivalence from a Banach space X to a Banach space Y has a Frechet derivative at some point, then the derivative is an isomorphism from X onto Y. If it is only Gateaux differentiable, then the derivative is only an into isomorphism. It gets worse when you look at Lipschitz quotients (introduced in my paper with Bates, Lindenstrauss, Preiss, and Schechtman): The Frechet derivative of a Lipschitz quotient is a surjective linear operator, while the Gateaux derivative can be anything!

Look at the book by Benyamini and Lindenstrauss [BL], Geometric nonlinear functional analysis, to learn more.

For a useful concept that is weaker than Frechet differentiability, take a look at the sections in [BL] that treat $\epsilon$-Frechet differentiability. There are much better existence theorems about Lipschitz functions having for all $\epsilon > 0$ points of $\epsilon$-Frechet differentiability than having points of Frechet differentiability, and for many applications it is just as good to have such points as to have points of Frechet differentiability.

$\endgroup$
3
$\begingroup$

I know this question has been inactive for such a long time, but I will try to add something.

Dieudonne in Treatise on analysis defines an "intermediate" concept of differentiability between Fréchet and Gâteaux ones called quasi-differentiability. Using your notation, $V$ is quasi-differentiable at $x$ if $V \circ g$ is differentiable at 0, for all $g : [0,1] \to L$ which are continuous, right-differentiable in $0$, and $g(0) = x$.

(I am not sure about it, but i think you could caracterize quasi-differentiability in terms of uniform convergence on compact sets, i.e. the limit $ \lim_{t\to 0} \frac{V(x+tu) - V(x)}{t}$ ` is uniform on compact sets.)

If $L$ is finite dimensional, then quasi-differentiability implies Fréchet differentiability. If $V$ is Lipschitz, then Gâteaux differentiability implies quasi-differentiability.

Nice thing about quasi-differentiability is that the theorem of differentiation of composite functions holds (wich is not the case with Gâteaux differentiability). Moreover, if $F$ is quasi-differentiable and $G$ is Gâteaux differentiable, then $F \circ G$ is G-differentiable.

Unfortunately, it's possible to construct a function $F : L^2 \to L^2$ which is everywhere quasi-differentiable, its differential is a right inverse (thus injective), but is nowhere locally injective. This is not possible if $F$ is Fréchet differentiable.

$\endgroup$
2
  • $\begingroup$ Hi! May I ask you about the explicit example on the "unfortunately" fact? Or some references? Thank you! $\endgroup$
    – matgaio
    Nov 11, 2015 at 6:19
  • $\begingroup$ It was part of my master thesis, which is available online but unfortunately is in Italian. poisson.phc.unipi.it/~lucia//tesi/tesi-magistrale.pdf But you can try to read Esempio 3.2. on page 49. The function is $Fu(x) = (cos u(x), sin u(x))$ from $L^2(R)$ to $L^2(R^2)$. $\endgroup$ Feb 17, 2016 at 17:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.