Higher order derivatives

  • Thread starter cliowa
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Let E, F be Banach spaces, and let [itex]L(E;F)[/itex] denote the space of linear, bounded maps between E and F. My goal is to understand better higher order derivatives.
Let's take [itex]E=\mathbb{R}^2, F=\mathbb{R}[/itex]. Consider a function [itex]f:U\subset\mathbb{R}^2\rightarrow\mathbb{R}[/itex], where U is an open subset of [itex]\mathbb{R}^2[/itex]. Then [itex]D^2 f:U\rightarrow L(\mathbb{R}^2;L(\mathbb{R}^2;\mathbb{R}))[/itex].
Now, I read that for [itex]u\in U, v,w\in\mathbb{R}^2[/itex] by definition [itex]D^2 f(u)\cdot (v,w):=D((Df)(.)\cdot w)\cdot v[/itex]. My question now is: Why was this defined precisely this way?
Does it have something to do with "using the product rule", which would amount to [itex]D((Df)(.)\cdot w)=D^2 f(.)\cdot w+Df(.)\cdot D(w)=D^2 f(.)\cdot w[/itex]?
Thanks for any help. Best regards...Cliowa
 

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  • #2
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It has to do with the definition of derivatives. ##D^2(f)= D(D(f))##. We get a bilinear functional from that. How to consider a derivative is a matter of purpose. E.g. look at the ten point list at the beginning of
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/where I listed a few of such perspectives. Your view is: ##D## is a linear functional, and ##D^2## a bilinear. Look up differential forms. E.g.
https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/
 

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