Frechet Derivatives and the Implicit function theorem

In this section $X_{1},\cdots$ etc and are Banach spaces over $\Bbbk ,$ is an open neighbourhood of .

m-linear Bounded Operators

An operator

is called -linear and bounded if it is linear in each of its arguments and there exists a such that

The norm of is defined by
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Then

Proposition

Every -linear bounded operator is continuous.

In the sequel we will write for

Frechet Derivatives

Definition

Suppose $f:U\rightarrow Y.$

The differential of at exists if there exists a bounded linear operator such that

with
The second differential of at exists if there exists a bounded bilinear operator such that

with

Inductively we define
The $\left( n+1\right)$ th differential of at exists if there exists a bounded $\left( n+1\right)$ -linear operator such that

with

The differentials are also dentoed by We also write = and so on. The diferentials are called the Frechet derivatives of at To formally find the Frechet derivatives we use the formula

Once we have a formula for the derivative we proceed to prove that it is the desired one. This involves establishing two things:

the resulting operator is multilinear,bounded, and
the remainder is

Definition

is continuous at if for each $\varepsilon >0$ there exists a such that
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This means that for all and all

$\$

Definition

is called a $C^{k}$ function if is continuous on

Example

If has continuous paritial derivatives of order then is $C^{k}$ and
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where

Proof

( $N=2,\ k=1$ ) write Then

for some Hence, where
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so that
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and the continuity of the partial derivatives give that

Example

(Bilinear Operators)

Suppose is a bilinear bounded operator. Then is $C^{\infty }.$

Proof

Set Let's illustrate first the formal calculation of $dB\left( u\right)$ and
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Next, we verify that these are the required derivatives. We begin with $dB\left( u\right)$

Then i.e., $dB\left( u\right)$ is a bounded linear operator on $X_{1}\times X_{2}$ into .
where Hence,

Similarly, for we have:

Then i.e., is a bounded bilinear operator on into .
where

Next we establish the continuity of
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Hence, i.e., $dB\left( u\right)$ is actually Lipschitz continuous. The continuity of is easier to see since it is independent of Finally,

Proposition

Suppose $f,g:U\rightarrow Y$ have Frechet derivatives of order Then
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Partial Derivatives

Suppose Let and define the function by

If $dg\left( u\right)$ exists, we define the partial Frechet derivative of at $\left( u,v\right)$ by

The partial derivative of at $\left( u,v\right)$ is defined in a similar manner. Partial derivatives are also denoted by

Proposition

If exists then and both exist and
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Proof

With the above definition of we have
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where i.e., $dg\left( u\right)$ exists and Hence,
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Similarly,
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Using the linearity of we obtain the assertion of the proposition.

Application to Analytic Operators

Power Operators

Let be a -linear bounded operator which is symmetric with respect ot all its arguments. This means that is invariant under any permutation of its argument. A power operator generated by is defined by

and
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where

Properties of Power Operators

(e.g., for since is symmetric.)
when

To see this, we write
$Mu^{k}$ has Lipschitz continuous derivatives of all orders and

For example, when

where with

Also, which shows the continuity of the operator $kMu^{k-1}.$ Hence, The Lipschitz continuity of can be shown as follows:

for

Power Series

Suppose is a power operator for $k=0,1,\cdots .$ Furthermore, suppose that the series
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for all $\rho >0.$ The operator defined by
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is called a power series. Note that () gurantees that $A\left( u\right)$ is a well defined element of

Definition

An operator is called analytic at the point $u_{0}\in U$ if there exists a $\rho >0$ such that $A\left( u\right)$ admits a power series representation () that is absolutely convergent (i.e., () converges) for every

Theorem

If is analytic at the point $u_{0}\in U$ then $A\in C^{\infty }$ on for any $\varepsilon >0,$ where $\rho$ is as in Definition . Moreover, $\cdots$ can be found by term by term differentiation of ().

Proof

We will show that
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For this we have
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and
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where
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Now, for
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Furthermore,
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where Since as then Hence,
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Therefore,
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where It remains to show that This follows from
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The statement about higher derivatives can be shown in the same manner.

A Useful Formula for Calculating Higher Derivatives

Proposition

Suppose is a bounded bilinear operator, such that exist at some point $t\in U.$ Then is differentiable and
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Proof

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where Hence,
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Example

Denote the subset of of operators with bounded inverse by Define
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by
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Then $\Phi$ is analytic at , and
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for all and all

Proof

For let . Then
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Since the series
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converges, it follows that the series for is absolutely convergent. Hence $\Phi$ is analytic at To find we use the formal calculation
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Now noting that $E\mapsto EBA$ is a bilinear operator, we use formula to compute higher derivatives of $\Phi$ . For the second derivative we get
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This formula can, in turn, be used to obtain higher derivatives of $\Phi .$

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