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Pretty simple question.

This definition of the Gateaux Derivative presents it as: \begin{equation} df(u,\psi)=\dfrac{d}{d\epsilon}f(u+\epsilon\psi)\Bigg{|}_{\epsilon=0} \end{equation} Why is the derivative being evaluated at $\epsilon=0$ in that definition? The fact that I'm interpreting it in a rather similar way as the directional derivative is causing some confusion.

Johann Wagner
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    Where else would you evaluate it, if you want the derivative at the point $u$? – Hans Lundmark Mar 31 '22 at 20:57
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    "Interpreting it in a rather similar way as the directional derivative" is exactly what you should be doing though, since it's a generalization of the concept of directional derivative. As for why everything is evaluated at $\epsilon=0$, perhaps imagine $\epsilon$ as representing a time parameter, $u$ an initial position, and $\psi$ a direction. The path $\gamma(\epsilon) = u+\epsilon\psi$ is motion through the point $u$. If I want to track how quickly $f(\gamma(\epsilon))$ changes in $\epsilon$ at the point $u$, then I want to check things at $u = \gamma(0)$, i.e., at $\epsilon=0$. – Glare Mar 31 '22 at 20:59
  • So, that's like setting an origin, a "start", possibly, for the operation we're applying to the smooth curve? Like, evaluating your derivative at a time parameter being equal to zero in order to see how your system evolves from that point? That would make it quite intuitive actually. Note that I'm already making a parallel with classical mechanics here, that's why I stated this last phrase in such a way. Just to clear, since we're on MSE. – Johann Wagner Mar 31 '22 at 21:11

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