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Let $E$ be a real (or complex) Banach space. By $C^\infty(E) $ we mean the space of all functions $f:E\to \mathbb{R}(f:E\to \mathbb{C})$ which are smooth in the sense of Frechet diffetentiability. A derivation is a linear map $D:C^\infty (E)\to C^\infty(E)$ which satisfies $D(fg)=D(f)g+fD(g)$. As an example of a derivation we assume that $H:E\to E$ is a smooth map then we consider vector field $Z'=H(z)$ on $E$. So we have a derivation $D$ with $D(f)(x)=df(x).H(x)$ where $df$ is the Frechet differential of $f$.

Is there a Banach space $E$ with a derivation $D$ which does not arrive from a vector field as in the above example?

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    $\begingroup$ Any linear form $\phi$ on $E^*$ produces a derivation via $D(f)(x):=\langle \phi, df(x)\rangle$ in your sense, and it arrives from a vector field iff it is an evaluation on a constant vector. So, if you allow non-continuous $\phi$, any infinite dimensional Banach space has one not coming from a vector field; if you want it continuous, any non-reflexive Banach has it. $\endgroup$
    – Pietro Majer
    Commented Dec 6, 2019 at 11:06
  • $\begingroup$ @PietroMajer Thanks for your answer. To be honnest the motivation for this post is the following: How can one define the Lie bracket of two vector fields in infinit dimensional space. I would like to apply it to the following case: We consider the vector field$ X:\; Z'=Z^2-Z$ on a $C^*$ algebra $A$ then we try to find a vector field $Y:\; Z'=g(Z)$ with $[X,Y]=0$ then idempotents of $A$ are invariant under flow of $Y$. $\endgroup$
    – Ali Taghavi
    Commented Dec 6, 2019 at 18:59
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    $\begingroup$ The most natural way I think is to define $[X,Y]$ as the Lie derivative $L_XY$ (and the Lie derivative can be defined via the flow of $X$). $\endgroup$
    – Pietro Majer
    Commented Dec 6, 2019 at 19:39
  • $\begingroup$ @PietroMajer You mean $L_X Y -L_Y X$? $\endgroup$
    – Ali Taghavi
    Commented Dec 6, 2019 at 20:00
  • $\begingroup$ @PietroMajer my apology for this second message. May you read my previous comment? $\endgroup$
    – Ali Taghavi
    Commented Dec 31, 2019 at 15:34

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