Frechet Derivative
Last updated
Nov 1, 2022
# Definition
Suppose V,W are Normed Vector Space
Normed Vector Space
Definition
A is a X equipped with a ∣∣⋅∣∣....
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s and f:V→W. Suppose for v∈V there exists T∈BL(V,W) s.t.
$$\lim\limits_{h \to 0} \frac{|f(v + h) - f(v) - Th|{W}}{|h|{V}} = 0$$
Then we say f has Frechet Derivative
Frechet Derivative
Definition
Suppose V,W are s and f:V→W. Suppose for v∈V there exists $T \in BL(V,...
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Dvf=T.
# Properties
# Continuity of f
# Statement
If Dvf exists for v∈V, then f is Continuous Function
Continuous Function
Definition 1
Let (X,τ),(Y,ρ) be s and let f:X→Y be a . Then f is...
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at v. TODO
# Proof
Existence of the Frechet Derivative
Frechet Derivative
Definition
Suppose V,W are s and f:V→W. Suppose for v∈V there exists $T \in BL(V,...
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means
$$\begin{align*}
&&\lim\limits_{h \to 0} \frac{|f(v + h) - f(v) - Th|{W}}{|h|{V}} = 0\\ &\Rightarrow& 0 = \lim\limits_{h \to 0} \Big| \big( f(v + h) - f(v) - Th\big) - \mathbf{0}{W} \Big|{W}\\ \end{align*}$$
By definition of Function Limit
Function Limit
Definition
Let (M,d1) and (N,d2) be two s. Let f:M→N be a . Suppose there exists...
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, we have that
$$\begin{align*}
&\lim\limits_{h \to 0} \big( f(v + h) - f(v) - Th\big) = \mathbf{0}{W}\\ &\Rightarrow \lim\limits{h \to 0} \big( f(v + h) - f(v) \big) - \lim\limits_{h \to 0} Th = \mathbf{0}{W}\\ &\Rightarrow \lim\limits{h \to 0} \big( f(v + h) - f(v) \big) = \mathbf{0}_{W}
\end{align*}$$
Where the last line follows because Bounded Linear Maps are Continuous (how to prove this?). The last line is the definition of continuity
Continuous Function
Definition 1
Let (X,τ),(Y,ρ) be s and let f:X→Y be a . Then f is...
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and f is Continuous Function
Continuous Function
Definition 1
Let (X,τ),(Y,ρ) be s and let f:X→Y be a . Then f is...
11/7/2022
.