Derivative

Note to self: don't forget to mention the [ilmath]h[/ilmath] or [ilmath]x-x_0[/ilmath] thing doesn't matter

Definition

Note: there are 2 definitions of differentiability, I will state them both here, then prove them equivalent.

Let [ilmath]U[/ilmath] be an open set of a Banach space [ilmath]X[/ilmath], let [ilmath]Y[/ilmath] be another Banach space.

• Let [ilmath]f:X\rightarrow Y[/ilmath] be a given map
• Let [ilmath]x_0\in X[/ilmath] be a point.

Definition 1

We say that [ilmath]f[/ilmath] is differentiable at a point [ilmath]x_0\in X[/ilmath] if:

• there exists a continuous linear map, [ilmath]L_{x_0}\in L(X,Y)[/ilmath] such that:
  • [ilmath]T(x_0+h)-T(x_0)=L_{x_0}(h)+r(x_0,h)[/ilmath] where [math]\ \lim_{h\rightarrow 0}\left(\frac{\Vert r(x_0,h)\Vert}{\Vert h\Vert}\right)=0[/math]

Definition 2

We say that [ilmath]f[/ilmath] is differentiable at a point [ilmath]x_0\in X[/ilmath] if:

• there exists a continuous linear map, [ilmath]L_{x_0}\in L(X,Y)[/ilmath] such that:
  • [math]\lim_{h\rightarrow 0}\left(\frac{f(x_0+h)-f(x_0)-L_{x_0}(h)}{\Vert h\Vert}\right)=0[/math]

Hybrid definition

These naturally lead to: We say that [ilmath]f[/ilmath] is differentiable at a point [ilmath]x_0\in X[/ilmath] if:

• there exists a continuous linear map, [ilmath]L_{x_0}\in L(X,Y)[/ilmath] such that:
  • [math]\lim_{h\rightarrow 0}\left(\frac{\Vert f(x_0+h)-f(x_0)-L_{x_0}(h)\Vert}{\Vert h\Vert}\right)=0[/math]

Extra workings for proof

• there exists a continuous linear map, [ilmath]L_{x_0}\in L(X,Y)[/ilmath] such that:
  • [ilmath]T(x_0+h)-T(x_0)=L_{x_0}(h)+r(x_0,h)[/ilmath] where [math]\ \lim_{h\rightarrow 0}\left(\frac{\Vert r(x_0,h)\Vert}{\Vert h\Vert}\right)=0[/math]
  • This can be interpreted as [math]\ \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][/math]
    • Which is
      1. [math]\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][/math]
         • Does this make sense though? We need [ilmath]r(x_0,\cdot)[/ilmath] to be given, surely a form with [ilmath]r(x_0,h)=T(x_0+h)-T(x_0)-L_{x_0}(h)[/ilmath] in the numerator would make more sense? No of course not.

this sort of outlines the proof I'll need for definitions 1 and 2