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| − | Note to self: don't forget to mention the {{M|h}} or {{M|x-x_0}} thing doesn't matter
| + | {{Disambiguation}} |
| | + | * [[Derivative (analysis)]] - the derivative of a [[mapping]] between two [[normed]] spaces (this is probably the one you want) |
| | + | * [[Derivative (measure theory)]] - the Raydon-Nikon (spelling?) derivative and stuff |
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| − | ==Definition==
| + | {{Todo|Flesh out}} |
| − | : '''Note:''' there are 2 definitions of differentiability, I will state them both here, then prove them equivalent.
| + | {{Definition|Analysis|Measure Theory}} |
| − | Let {{M|U}} be an [[open set]] of a [[Banach space]] {{M|X}}, let {{M|Y}} be another Banach space.
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| − | * Let {{M|f:X\rightarrow Y}} be a given [[map]]
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| − | * Let {{M|x_0\in X}} be a point.
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| − | ===Definition 1===
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| − | We say that ''{{M|f}} is differentiable at a point {{M|x_0\in X}}'' if{{rAPIKM}}{{rITTGG}}:
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| − | * there exists a continuous linear map, {{M|L_{x_0}\in L(X,Y)}} such that:
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| − | ** {{M|1=T(x_0+h)-T(x_0)=L_{x_0}(h)+r(x_0,h)}} where {{MM|1=\ \lim_{h\rightarrow 0}\left(\frac{\Vert r(x_0,h)\Vert}{\Vert h\Vert}\right)=0}}
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| − | ===Definition 2===
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| − | We say that ''{{M|f}} is differentiable at a point {{M|x_0\in X}}'' if<ref name="ITTGG"/>:
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| − | * there exists a continuous linear map, {{M|L_{x_0}\in L(X,Y)}} such that:
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| − | ** {{MM|1=\lim_{h\rightarrow 0}\left(\frac{f(x_0+h)-f(x_0)-L_{x_0}(h)}{\Vert h\Vert}\right)=0}}
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| − | ===Hybrid definition===
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| − | These naturally lead to:
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| − | We say that ''{{M|f}} is differentiable at a point {{M|x_0\in X}}'' if:
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| − | * there exists a continuous linear map, {{M|L_{x_0}\in L(X,Y)}} such that:
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| − | ** {{MM|1=\lim_{h\rightarrow 0}\left(\frac{\Vert f(x_0+h)-f(x_0)-L_{x_0}(h)\Vert}{\Vert h\Vert}\right)=0}}
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| − | ==Extra workings for proof==
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| − | * there exists a continuous linear map, {{M|L_{x_0}\in L(X,Y)}} such that:
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| − | ** {{M|1=T(x_0+h)-T(x_0)=L_{x_0}(h)+r(x_0,h)}} where {{MM|1=\ \lim_{h\rightarrow 0}\left(\frac{\Vert r(x_0,h)\Vert}{\Vert h\Vert}\right)=0}}
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| − | ** This can be interpreted as {{MM|1=\ \exists L_{x_0}\in L(X,Y)\left[\lim_{h\rightarrow 0}\left(\frac{\Vert r(x_0,h)\Vert}{\Vert h\Vert}\right)=0\implies T(x_0+h)-T(x_0)=L_{x_0}(h)+r(x_0,h)\right]}}
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| − | *** Which is
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| − | ***# {{MM|1=\exists L_{x_0}\in L(X,Y)\forall\epsilon>0\exists\delta>0\forall h\in X\left[0<\Vert h-x\Vert<\delta\implies\frac{\Vert r(x_0,h)\Vert}{\Vert h\Vert}<\epsilon\implies T(x_0+h)-T(x_0)=L_{x_0}(h)+r(x_0,h)\right]}}
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| − | ***#* Does this make sense though? We need {{M|r(x_0,\cdot)}} to be given, surely a form with {{M|1=r(x_0,h)=T(x_0+h)-T(x_0)-L_{x_0}(h)}} in the numerator would make more sense? No of course not.
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| − | '''this sort of outlines the proof I'll need for definitions 1 and 2'''
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