# Fréchet derivation

The Frechet derivative (by Maurice René Fréchet ) generalizes the concept of deriving from the conventional differential calculus in on normed spaces . In the case of images between finite-dimensional spaces, this concept of differentiability results in the usual concept of total differentiability . ${\ displaystyle \ mathbb {R} ^ {n}}$

## definition

Relationship of the three figures

Let and two normalized spaces and an open subset . An operator 's Fréchet differentiable at the point when there is a limited linear operator are such that ${\ displaystyle (X, \ | {\ cdot} \ | _ {X})}$${\ displaystyle (Y, \ | {\ cdot} \ | _ {Y})}$${\ displaystyle U \ subset X}$ ${\ displaystyle A \ colon U \ to Y}$${\ displaystyle \ varphi \ in U}$ ${\ displaystyle A '(\ varphi) \ colon X \ to Y}$

${\ displaystyle \ lim _ {\ | h \ | _ {X} \ to 0} {\ frac {1} {\ | h \ | _ {X}}} \, \ | A (\ varphi + h) - A (\ varphi) -A '(\ varphi) h \ | _ {Y} = 0}$

applies. The operator is called the Fréchet derivative of at that point . If the Fréchet derivative exists for everyone , then the mapping with is called the Fréchet derivative of on . The space of continuous linear mappings from to is denoted by. ${\ displaystyle A '(\ varphi)}$${\ displaystyle A}$${\ displaystyle \ varphi}$${\ displaystyle \ varphi \ in U}$${\ displaystyle A '\ colon U \ to L (X, Y)}$${\ displaystyle \ varphi \ mapsto A '(\ varphi)}$${\ displaystyle A}$${\ displaystyle U}$${\ displaystyle L (X, Y)}$${\ displaystyle X}$${\ displaystyle Y}$

### Equivalent definition

An equivalent definition is:

For everyone there is such a thing ${\ displaystyle \ epsilon> 0}$${\ displaystyle \ delta> 0}$

${\ displaystyle \ | A (\ varphi + h) -A (\ varphi) -A '(\ varphi) h \ | _ {Y} \ leq \ epsilon \ | h \ | _ {X}}$

for everyone with . This can also be written briefly with the help of the Landau symbols : ${\ displaystyle h \ in X}$${\ displaystyle \ | h \ | \ leq \ delta}$

${\ displaystyle A (\ varphi + h) -A (\ varphi) = A '(\ varphi) h + o (\ | h \ | _ {X})}$for .${\ displaystyle h \ to 0}$

## Examples

### Linear operators

For finite-dimensional normalized spaces , all linear operators are Fréchet-differentiable with constant derivative. At each point the derivative is the linear operator itself: for all . ${\ displaystyle X, Y}$${\ displaystyle A \ colon X \ to Y}$${\ displaystyle A '(\ varphi) = A}$${\ displaystyle \ varphi \ in X}$

In the infinite-dimensional case, among the linear operators, precisely the constrained (= continuous) Fréchet-differentiable operators. Unlimited linear operators are not Fréchet-differentiable.

### Real-valued functions

If a real-valued function is defined on an open set and has continuous partial derivatives , then Fréchet is also differentiable. The derivative at the point is given by the usual gradient of according to: ${\ displaystyle f \ colon U \ to \ mathbb {R}}$${\ displaystyle U \ subset \ mathbb {R} ^ {n}}$${\ displaystyle f}$${\ displaystyle f}$${\ displaystyle x}$${\ displaystyle f}$

${\ displaystyle f '(x) \ colon h \ mapsto {\ mbox {grad}} f (x) \ cdot h = \ sum \ limits _ {i = 1} ^ {n} {\ frac {\ partial f} {\ partial x_ {i}}} (x) \, h_ {i}}$

This example shows the connection to the usual differential calculation in . The Fréchet derivation is actually a generalization of differential calculus for normalized spaces. ${\ displaystyle \ mathbb {R} ^ {n}}$

### Integral operator

Let be , continuous and continuous and continuously differentiable in the second argument. The nonlinear integral operator defined by ${\ displaystyle J = [a, b] \ subset \ mathbb {R}}$${\ displaystyle k \ colon J \ times J \ to \ mathbb {R}}$${\ displaystyle f \ colon J \ times \ mathbb {R} \ to \ mathbb {R}}$ ${\ displaystyle F \ colon C (J) \ to C (J)}$

${\ displaystyle (Fx) (t) = \ int _ {a} ^ {b} k (t, s) f (s, x (s)) \ mathrm {d} s}$

is fréchet-differentiable. Its derivation is ${\ displaystyle F ^ {\ prime}}$

${\ displaystyle (F ^ {\ prime} (x) h) (t) = \ int _ {a} ^ {b} k (t, s) {\ frac {\ partial f} {\ partial x}} ( s, x (s)) \, h (s) \ mathrm {d} s.}$

Because of the mean value theorem of differential calculus ,

${\ displaystyle f (s, x (s) + h (s)) - f (s, x (s)) = {\ frac {\ partial f} {\ partial x}} (s, x (s) + \ rho (s) h (s)) \, h (s)}$

with and because of the uniform continuity of on true ${\ displaystyle 0 <\ rho (s) <1}$${\ displaystyle {\ tfrac {\ partial f} {\ partial x}}}$${\ displaystyle J \ times \ {z \ in \ mathbb {R}: | z | \ leq \ sup | x | +1 \}}$

${\ displaystyle \ sup _ {s \ in J} \ left | {\ frac {\ partial f} {\ partial x}} (s, x (s) + \ rho (s) h (s)) - {\ frac {\ partial f} {\ partial x}} (s, x (s)) \ right | \ leq \ epsilon}$

for . For so true ${\ displaystyle \ sup | h | \ leq \ delta}$${\ displaystyle \ sup | h | \ leq \ delta}$

${\ displaystyle \ sup \ left | F (x + h) -F (x) - \ int _ {a} ^ {b} k (\ cdot, s) {\ frac {\ partial f} {\ partial x} } (s, x (s)) h (s) \ mathrm {d} s \ right | \ leq \ epsilon \, \ sup | h | \, \ max _ {(t, s) \ in J \ times J } | k (t, s) | (ba),}$

what the presentation of the derivative proves.

## Calculation rules

The usual calculation rules for the total derivative can also be shown for the Fréchet derivative. The following equations apply, provided they make sense in the sense of the above definition, in particular, the occurring images can be differentiated at the corresponding points: ${\ displaystyle \ mathbb {R} ^ {n}}$

• ${\ displaystyle (A + B) '(\ varphi) = A' (\ varphi) + B '(\ varphi)}$
• ${\ displaystyle (\ lambda A) '(\ varphi) = \ lambda A' (\ varphi)}$.
• Chain rule : . The product is to be understood in the sense of the multiplication (successive execution) of linear images.${\ displaystyle (A \ circ B) '(\ varphi) = (A' \ circ B) (\ varphi) \, B '(\ varphi)}$${\ displaystyle (A '\ circ B) (\ varphi) \, B' (\ varphi)}$
• If A is a continuous, linear operator, then A is differentiable everywhere and it holds . Together with the chain rule, this leads to the conclusion that continuous, linear operators can be extracted from the derivative: and .${\ displaystyle A}$${\ displaystyle A '(\ varphi) = A}$${\ displaystyle (A \ circ B) '(\ varphi) = A \, B' (\ varphi)}$${\ displaystyle (B \ circ A) '(\ varphi) = B' (A (\ varphi)) \, A}$
• Product rule: If there is a continuous, n-fold linear mapping, then${\ displaystyle A: X_ {1} \ times \ ldots \ times X_ {n} \ to Y}$${\ displaystyle A '(\ varphi _ {1}, \ ldots, \ varphi _ {n}) :( h_ {1}, \ ldots, h_ {n}) \ mapsto A (h_ {1}, \ varphi _ {2}, \ ldots, \ varphi _ {n}) + \ ldots + A (\ varphi _ {1}, \ ldots, \ varphi _ {n-1}, h_ {n})}$

## Relationship between Fréchet and Gâteaux derivation

Let Fréchet differentiable at the point , then the Gâteaux differential exists for any direction and we have: ${\ displaystyle A}$${\ displaystyle \ varphi}$${\ displaystyle h \ in X}$ ${\ displaystyle \ delta A (\ varphi, h)}$

${\ displaystyle \ delta A (\ varphi, h) = A '(\ varphi) h}$.

The reverse is generally not true.

In addition, the Gâteaux derivation of exists at the point that is denoted below with , and the following applies: ${\ displaystyle A}$${\ displaystyle \ varphi}$${\ displaystyle A '_ {s} (\ varphi)}$

${\ displaystyle A '_ {s} (\ varphi) = A' (\ varphi)}$.

Again, the reverse is generally not true. The reverse also applies under the following conditions:

If in a neighborhood of Gâteaux is differentiable , that is, the Gâteaux differential is continuous and linear at every point in the neighborhood, and the mapping ${\ displaystyle A}$ ${\ displaystyle U}$${\ displaystyle \ varphi}$

${\ displaystyle A '_ {s} (.) \ colon U \ to {\ mathcal {L}} (X, Y)}$ given by ${\ displaystyle \ psi \ mapsto A '_ {s} (\ psi)}$

the point is continuous with respect to the operator norm on , so is at the point differentiable Fréchet. ${\ displaystyle \ varphi}$${\ displaystyle {\ mathcal {L}} (X, Y)}$${\ displaystyle A}$${\ displaystyle \ varphi}$

This condition is not necessary. For example, there are already totally differentiable functions in the one-dimensional that are not continuously differentiable.

## Application example

The Fréchet derivation can e.g. B. can be used to solve so-called inverse boundary value problems in the context of a Newton method . As an example for this application we consider an inverse boundary value problem to the Laplace equation :

It is an unknown area. We consider the outer Dirichlet problem , in which the boundary values ​​are given by a source in the point . Then the bounded and twice continuously differentiable function in the Laplace equation satisfies : ${\ displaystyle D \ subset \ mathbb {R} ^ {2}}$${\ displaystyle \ partial D}$${\ displaystyle z \ in \ mathbb {R} ^ {2} \ setminus {\ bar {D}}}$${\ displaystyle u}$${\ displaystyle \ mathbb {R} ^ {2} \ setminus {\ bar {D}}}$

${\ displaystyle \ Delta u = 0 \ quad {\ mbox {in}} \, \, \ mathbb {R} ^ {2} \ setminus {\ bar {D}}}$

and the Dirichlet boundary condition:

${\ displaystyle u = - \ Phi (\ cdot, z) \ quad {\ mbox {on}} \, \, \ partial D.}$

With we denote the fundamental solution to the Laplace equation, which describes a point source in the point . ${\ displaystyle \ Phi}$${\ displaystyle z}$

With the inverse boundary value problem we start from a second (known) domain which contains. On the edge of we measure the values ​​of the solution of the direct Dirichlet problem. So we know the trail . Our goal is now the unknown edge of reconstructing knowledge from this track. ${\ displaystyle B \ subset \ mathbb {R} ^ {2}}$${\ displaystyle D}$${\ displaystyle \ partial B}$${\ displaystyle B}$${\ displaystyle u}$ ${\ displaystyle u | _ {\ partial B}}$${\ displaystyle \ partial D}$${\ displaystyle D}$

This problem can be formally described by an operator that maps the unknown edge onto the known track . So we need to solve the following nonlinear equation: ${\ displaystyle F}$${\ displaystyle \ partial D}$${\ displaystyle u | _ {\ partial B}}$

${\ displaystyle F (\ partial D) = u | _ {\ partial B}}$

This equation can e.g. B. be linearized using the Newton method. To do this, we limit ourselves to areas whose edge can be represented as follows: ${\ displaystyle D}$

${\ displaystyle \ displaystyle x (t) = r (t) (\ cos (t), \ sin (t))}$

So we are now looking for the unknown radius function . The linearized equation (Newton's method) then looks like this: ${\ displaystyle r}$

${\ displaystyle F (r) + F '(r, q) = u | _ {\ partial B}}$

Here the Fréchet derivative of the operator denotes (the existence of the Fréchet derivative for can be shown and can be determined via a direct boundary value problem!). This equation is then solved for, whereby we have found a new approximation to the unknown sought edge with. The method can then be iterated with this approximation. ${\ displaystyle \ displaystyle F '}$${\ displaystyle \ displaystyle F}$${\ displaystyle \ displaystyle F}$${\ displaystyle \ displaystyle F '}$${\ displaystyle q}$${\ displaystyle r + q}$