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I'm reading the book of Drabek, Milota - Methods of Nonlinear Analysis, and at page 121, they state:

extract of Mil Dra

but I can't manage to find such counterexample. For clarity the Gateaux derivative is defined in this way:

definition

I need some kind of hints about how to build such counterexample because I'm like going nowhere with my trials. According to me $f$ and $g$ can't be continuous, otherwise G-derivative would be Frechét-derivative and for this kind of derivative the chain rule holds. It is sufficient requiring that only one function is non-continuos?

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1 Answer 1

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Hint: define $f: {\mathbb R}^2 \to {\mathbb R}$ such that $f(x,y) = 0$ unless $ x^2 < y < 2 x^2$. Note that the intersection of any line through the origin with the exceptional set $A = \{(x,y): x^2 < y < 2 x^2\}$ misses some interval around the origin, so what $f$ does in $A$ does not affect the Gâteaux derivative at the origin.

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  • $\begingroup$ Thanks for the answer! sorry for the late comment, but I tried to work out your hint. I don' understand the passage "misses some interval around the origin, […] at the origin". Can you please expand it a little? A is the set between the two parabolas, without the origin, so for example the vertical line (y-axis) don't miss anything. Maybe I didn't understand what do you mean by "misses some interval" $\endgroup$
    – Riccardo
    Commented Mar 9, 2014 at 22:20
  • $\begingroup$ definitely I can't go any further with your hint. I started trying with an $f$ 0 everywhere but for $x \in A$. but for $g$ I don't have any idea $\endgroup$
    – Riccardo
    Commented Mar 9, 2014 at 23:14
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    $\begingroup$ The $y$ axis misses the whole set $A$, as does the $x$ axis. The line $x=y$ intersects $A$ for $1/2 < x < 1$ and so misses the interval $-1/2 < x < 1/2$. $\endgroup$ Commented Mar 10, 2014 at 0:23
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    $\begingroup$ Try defining $g$ so that $g(t)$ is in $A$ for all $t \ne 0$. $\endgroup$ Commented Mar 10, 2014 at 0:25
  • $\begingroup$ I've found another counterexample without following your hints, $h: \mathbb{R}^2 \to \mathbb{R}^2 , \ \ h(x,y) = (x,y^2)$ and $g(x,y) = \frac{y(x^2+y^2)^{3/2}}{(x^2+y^2)^2+y^2} $ if $(x,y) \neq 0$ and $g(0,0)=0$. In $(0,0)$ the chain rule doesn't hold. What was your functions instead? $\endgroup$
    – Riccardo
    Commented Mar 10, 2014 at 17:49

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