Derivative (analysis)
From Maths
Contents
[hide]Definition
There are 2 kinds of derivative, a strong derivative (AKA the Fréchet derivative, total derivative[reqref 1]) one and a directional derivative one (AKA the Gateaux derivative, weak derivative)
Strong derivative
TODO: List of conditions
TODO: Weak derivative
Required references
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References
OLD PAGE
Note to self: don't forget to mention the h or x−x0 thing doesn't matter
Definition
- Note: there are 2 definitions of differentiability, I will state them both here, then prove them equivalent.
Let U be an open set of a Banach space X, let Y be another Banach space.
- Let f:X→Y be a given map
- Let x0∈X be a point.
Definition 1
We say that f is differentiable at a point x0∈X if[1][2]:
- there exists a continuous linear map, Lx0∈L(X,Y) such that:
- T(x0+h)−T(x0)=Lx0(h)+r(x0,h) where lim
Definition 2
We say that f is differentiable at a point x_0\in X if[2]:
- there exists a continuous linear map, L_{x_0}\in L(X,Y) such that:
- \lim_{h\rightarrow 0}\left(\frac{f(x_0+h)-f(x_0)-L_{x_0}(h)}{\Vert h\Vert}\right)=0
Hybrid definition
These naturally lead to: We say that f is differentiable at a point x_0\in X if:
- there exists a continuous linear map, L_{x_0}\in L(X,Y) such that:
- \lim_{h\rightarrow 0}\left(\frac{\Vert f(x_0+h)-f(x_0)-L_{x_0}(h)\Vert}{\Vert h\Vert}\right)=0
Extra workings for proof
- there exists a continuous linear map, L_{x_0}\in L(X,Y) such that:
- T(x_0+h)-T(x_0)=L_{x_0}(h)+r(x_0,h) where \ \lim_{h\rightarrow 0}\left(\frac{\Vert r(x_0,h)\Vert}{\Vert h\Vert}\right)=0
- This can be interpreted as \ \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]
- Which is
- \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]
- Does this make sense though? We need r(x_0,\cdot) to be given, surely a form with 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.
- \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]
- Which is