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I am having a hard time while trying to fully understand Hadamard differentiability.

I use the following definition taken from a German source ( Martin Brokate, "Konvexe Analysis und Evolutionsprobleme", Lecture notes, Technische Universität München, 2014):

$F: V \rightarrow W$ is Hadamard differentiable in $u\in V$ iff

$$\lim_{t\searrow 0} \frac{F(u+th+r(t))-F(u)}{t}$$ exists for all $h\in V$ and every $r:(0,\infty)\rightarrow V$ with $\lim_{t\searrow 0}\frac{r(t)}{t}=0$.

My aim is to prove the chain rule formula for Hadamard derivable functions and highlight the contrast with respect to non-Hadamard-differentiable functions, for which the chain rule does not hold in general. To do so I am looking for an example of a simple function, for which the directional derivative exists but Hadamard derivative does not exist. I couldn't find any such example: can someone of you help or maybe provide a good source?

Thank you so much!

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    $\begingroup$ Consider $F:\mathbb{R}^2\to \mathbb{R}$ with $F(u_1,u_2) = u_1$, if $u_2=0$, and $F$ is $0$ everywhere else. At $u=0$, the directional derivative in direction $h=(1,0)$ exists, but not the Hadamard derivative. $\endgroup$
    – Fabian Wirth
    Commented Sep 5 at 7:43
  • $\begingroup$ I'm not sure why you would add the $r(t)$ term in the above definition. Maybe you could imagine some weird counterexamples, where differentiation in this sense is more general than the more simple directional derivative. The Wiki definition of Hadamard derivative does not include the higher order remainder $r$. $\endgroup$
    – Beni Bogosel
    Commented Sep 5 at 8:32
  • $\begingroup$ I think the sense of $r(t)$ in the definition is to make the chain rule automatic, since if $r(t)=o(t)$, $F(u+th+r(t))$ is then itself of the form $U+tH+R(t)$ for $U:=F(u)$, $H:=DF(u)h$ and $R(t)=o(t)$ $\endgroup$
    – Pietro Majer
    Commented Sep 5 at 9:26
  • $\begingroup$ The definition of the Wiki article seems equivalent, since $t_nh_n=th+(t_n-t)h+t_n(h_n-h)=th+o(t-t_n)$ . $\endgroup$
    – Pietro Majer
    Commented Sep 5 at 9:26

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Any $1$-homogeneous function $F:\mathbb R^2\to\mathbb R$ has all directional derivatives at $u:=0$, since it is linear along rays. But such a function, for e.g. $h,k\in\mathbb R^2$ and $r(t):=t^2k$, has $\displaystyle \frac{F(u+th+r(t))-F(u)}t=F(h+tk)$, that may fail to have a limit for $t\searrow 0$ if $F$ is discontinuous at $h$.

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  • $\begingroup$ Okay, so I tried to show this for the example of @FabianWirth where $F$ is discontinuous at $h=(1,0)$ using $r(t)=(0,t^2)$, but I get $\frac{F(u+th+r(t))-F(u)}{t}=\frac{0}{t}$ for $t>0$, which has a limit, right? What do I do wrong? $\endgroup$
    – Matthis
    Commented Sep 5 at 10:13
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    $\begingroup$ I'm not saying that if F is discontinuous, then that limit does not exist, I'm saying that the limit may fail to exist, that is, there are examples where the limit does not exist . For F 1-homogeneous such that e.g. F(1,t) =sin(1/t) the definition is not satisfied for h:=(1,0) and r(t):=(0,t²), producing a counterexample. $\endgroup$
    – Pietro Majer
    Commented Sep 5 at 11:06
  • $\begingroup$ @Matthis: The point is that in Hadamard differentiability, you need the limit to exist for all $r$. And if you have $r_1$, $r_2$ with different limits, then you can construct one which alternates between $r_1$ and $r_2$ and does not have a limit. E.g. in the example I gave, let $r(t) = (0,0)$ for rational $t$ and $r(t) = (0,t^2)$ for irrational $t$. $\endgroup$
    – Fabian Wirth
    Commented Sep 5 at 22:00

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