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I have read (In French)that the differential of a function depends on the topology and not the norm, the latter is rather easy to grasp, the first is hard for me to construct.

Norms being equivalent in finite dimensional spaces, I am looking for an example (in a functional space maybe) to showcase to what extent differentials may be just so different depending on the initial topology (implied by two different non-equivalent norms). Could someone suggest a classic example thereof?

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  • $\begingroup$ If I understand you correctly, you want an example of a vector space $V$, a function $f:V\to\mathbb C$, a topology $\tau_1$ on $V$ such that $f$ is differentiable when viewed as a function $(V,\tau_1)\to \mathbb C$, another topology $\tau_2$ such that $f$ is differentiable when viewed as a function $(V,\tau_2)\to \mathbb C$. But you want the two differentials to be different. Is that indeed what you want? $\endgroup$
    – André Henriques
    Commented Dec 7, 2017 at 23:00
  • $\begingroup$ @AndréHenriques Exactly. I could not find one. $\endgroup$
    – Averroes
    Commented Dec 7, 2017 at 23:01
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    $\begingroup$ The differential of $f$ at zero, if it exists, is an element of the dual vector space. The dual of $(V,\tau_1)$ is not the same as the dual of $(V,\tau_2)$. So the two differentials live in different vector spaces. You can of course map the dual of $(V,\tau_1)$ and the dual of $(V,\tau_2)$ into the algebraic dual of $V$. Viewed as elements of the algebraic dual of $V$, the two differentials always agree. $\endgroup$
    – André Henriques
    Commented Dec 7, 2017 at 23:20
  • $\begingroup$ @AndréHenriques Thank you for your insights and Sorry for the confusion.My question is rather general. instead of $\mathbb{C}$ the function is from $V \rightarrow W$ with both $V$ and $W$ being metric spaces but $\mathbb{C}$ does the job because the essential property lies in $V$ . I also am looking for concrete examples if you kindly have any. $V$ in this case must be of infinite dimension. $\endgroup$
    – Averroes
    Commented Dec 7, 2017 at 23:26
  • $\begingroup$ If you look at maps $f:\mathbb R\to V$ instead of maps $V\to \mathbb C$, then you can cook up an example of a vector space $V$ equipped with two distinct topologies $\tau_1$ and $\tau_2$ such that the derivative of $f$ at $0$ is a certain element of $V$ when $V$ isa equipped with a certain topology and is another element of $V$ when $V$ is equipped with another topology. $\endgroup$
    – André Henriques
    Commented Dec 7, 2017 at 23:43

1 Answer 1

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Let $V:=\mathcal S(\mathbb R)$, the space of smooth functions with rapid decay.
Let $\gamma:\mathbb R\to V$ be the map which sends $t$ to the function $x\mapsto e^{-(x-t)^2}$.
Then $\gamma$ is differentiable.

Pick a Hamel basis of $V$ which includes all the elements $\gamma(t)$, and which also includes all the elements $\gamma'(t)$.

Consider a linear automorphism $a:V\to V$ which is the identity on all the elements of the Hamel basis, except that it exchanges $\gamma'(0)$ and $\gamma'(1)$.

Then you can pull back the topology of $V$ by the map $a$. This gives a new topology on $V$.

When computed w.r.t that new topology, the derivative of $\gamma$ at $0$ is the element $\gamma'(1)\in V$.

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  • $\begingroup$ Is it obvious that there is such a Hamel basis? That seems Schanuel's-conjecture-flavoured. $\endgroup$
    – LSpice
    Commented Dec 8, 2017 at 17:22
  • $\begingroup$ I haven't thought very deeply about whether such a Hamel basis exists. But I think that it shouldn't bee too hard... If you have a finite linear combination of functions $e^{-(x-t)^2}$ and $xe^{-(x-t)^2}$, you look at the rate of decay as $x\to \infty$, and that tells you the leading term in the linear combination. Subtract that term, and you're done by induction. $\endgroup$
    – André Henriques
    Commented Dec 8, 2017 at 17:32
  • $\begingroup$ Oh, sorry, I was thinking of independence of the values, rather than of the functions. Yes, I agree that this is very plausible. $\endgroup$
    – LSpice
    Commented Dec 8, 2017 at 18:00

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