Use the implicit function theorem to prove smoothness

Quick description

Suppose you have a function ${\mathbb R}^n\rightarrow {\mathbb R}^m$ and you would like to see that $f$ is $C^r$ . One way to show this is to find a $C^r$ function ${\mathbb R}^n \times {\mathbb R}^m \rightarrow {\mathbb R}^m$ such that $\det(D_y F) \neq 0$ and

$F(x,f(x)) = c.$

The implicit function theorem will imply that $f$ is $C^r$ .

This article is incomplete. This article is missing large parts of what is required of a Tricki article. This is only one way I know the implicit function theorem is used to show regularity. This may not be the best way to state this idea.

Prerequisites

Calculus

Example 1

Deterministic and Stochastic Optimal Control, Wendell H. Fleming, Raymond W. Rishel, page 8. Take a $C^r$ cost function ${\mathbb R}^3 \rightarrow {\mathbb R}$ and consider the calculus of variation problem of minimizing

$J(x) = \int_{t_0}^{t_1} L(t,x(t),\dot{x}(t))dt,$ (1)

over piecewise $C^1$ functions $x$ with the given fixed end points $x(t_0)= x_0$ and $x(t_1) = x_1$ . A piecewise $C^1$ function $x$ is called an extremal of (1) if it satisfies

$-\int_{t_0}^t L_x(s,x(s), \dot{x}^*(s)) ds + L_{y}(t,x(t),\dot{x}(t)) = c$ (2)

for $t_0 \le t \le t_1$ where $c$ is a constant and where $y$ denotes the variable of $L$ for which $\dot{x}$ is substituted in (1) (the third variable).

Now let us assume that $L_{yy } > 0$ and consider an extremal $x$ of (1) that is also $C^1$ , i.e., $\dot{x}$ is continuous. We will show using the implicit function theorem that $x$ must be $C^r$ , i.e., if the extremal is $C^1$ then it has to be as smooth as the cost function $L$ .

Remark Before we proceed further a quick remark about the assumption $L_{yy} > 0$ . This assumption precludes $\dot{x}$ from sudden changes and forces it to be $C^1$ , even if this were not assumed. In order not to detract from the argument based on the implicit function theorem we assumed that $x$ is $C^1$ . The assumption $L_{yy} >0$ will also come into play in the invocation of the implicit function theorem.

Now let us continue with our argument that $x$ must be as smooth as $L$ . The argument proceeds by induction. We already have the base case that $x$ is $C^1$ . Now let us suppose that it is $C^k$ for a $k \le r-1$ . Then

$P(t) = \int_{t_0}^t L_x(s,x(s),\dot{x}(s) ) ds$

is $C^k$ as well. By (2)

$-P(t) + L_y(t,x(t),\dot{x}(t)) =c$ (3)

for some constant $c$ . Define

$\Phi(t,y) = -P(t) + L_y(t,x(t),y).$

$L$ is $C^r$ and $k \le r-1$ and $x$ is $C^k$ , it follows that $\Phi$ is $C^k$ . We can rewrite (3) as $\Phi(t,\dot{x}) = c.$ $\Phi_y = L_{yy}$ and this is strictly positive by assumption. These imply that $\dot{x}$ is at least as smooth as $\Phi$ , thus $\dot{x}$ is $C^k$ . It follows that $x$ is $C^{k+1}$ .

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Comments

are there other examples of the argument in example one?

Sat, 25/04/2009 - 07:11 — devin

The above example has a more sophisticated method than the one I tried to describe in the quick description section. In the generalization of the method of example 1, there would be a function $f$ and the goal would be to improve our understanding of its smoothness. To that end, we use $f$ itself to define another function

${\mathbb R}^n \times {\mathbb R}^{m\times n} \rightarrow {\mathbb R}^{m\times n}$

such that $\Phi(x,Df(x) ) = c$ , $\det(D_y\Phi) \neq 0$ and $\Phi$ is as smooth as $f$ . Now the implicit function theorem says $Df$ is as smooth as $\Phi$ and hence $f$ . The argument is continued as many steps as possible. Does anyone know of another application of this argument, or an argument similar to this?

Notation

Thu, 07/05/2009 - 00:41 — Brendan Murphy

The notation " $D_y\,F$ " isn't very clear. My first thought was: "partial derivative with respect to $y$ ". After this, I figured you meant the operator on ${\mathbb R}^m$ given by the decomposition of $D\, F$ (the Jacobian of $F$ ), since this is the condition required by the theorem. Still, the notation gives rise the possibility $D_y\,\Phi=\Phi_y$ .

What does the notation $x^*$ mean in equation 2?

Instead of using $L_y$ , why not use $L_{\dot x}$ ?

Notation

Sat, 09/05/2009 - 15:18 — devin

Thanks for the comment. I like the $D_y$ notation and frequently use it because it consisely states everything related to the operation (we are taking a derivative with respect to a variable.) The one you suggest ( $\Phi_y$ ) is also good I think. As for $L_{\dot{x}}$ . This is common notation in many books, including Fleming and Rishel. I don't prefer it because I think it is confusing to use the same symbol to mean two entirely different things on the same page. In this case, if we use the $L_{\dot{x}}$ notation, $\dot{x}$ will mean the derivative of the function $x$ with respect to time and also the name of a free variable.

$x^*$ is a is a free variable representing a function satisfying the Euler-Lagrange equation. It probably is not a good choice of notation because as you point out upon reading it one thinks it must be related to the $x$ in its context. I changed it to $x$ . This I think causes an abuse of notation, but hopefully not a confusing one.

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