Use the implicit function theorem to prove smoothness

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Use the implicit function theorem to prove smoothness

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[QUICK DESCRIPTION]
Suppose you have a function $f:{\mathbb R}^n\rightarrow {\mathbb R}^m$ and you would like to see that $f$ is $C^r$. One way to show this is to find a $C^r$ function $F:{\mathbb R}^n \times {\mathbb R}^m \rightarrow {\mathbb R}^m$ such that $\det(D_y F) \neq 0$ and 
[maths]
F(x,f(x)) = c.
[/maths]
The implicit function theorem will imply that $f$ is $C^r$.

[note article 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.
[/note]

[PREREQUISITES]
Calculus
[EXAMPLE]
Deterministic and Stochastic Optimal Control, Wendell H. Fleming, Raymond W. Rishel, page 8. Take a $C^r$ cost function $L :{\mathbb R}^3 \rightarrow {\mathbb R}$ and consider the calculus of variation problem of minimizing
[math cprob]
J(x) = \int_{t_0}^{t_1} L(t,x(t),\dot{x}(t))dt,
[/math] 
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 [eqref cprob] if it satisfies
[math euler]
-\int_{t_0}^t L_x(s,x(s), \dot{x}^*(s)) ds + L_{y}(t,x(t),\dot{x}(t)) = c
[/math]
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 [eqref cprob] (the third variable).

Now let us assume that $L_{yy } > 0$ and consider an extremal $x$ of [eqref cprob] 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.
[/remark]

Now let us continue with our argument that $x$ must be as smooth as $L$. The argument proceeds by [[induction front page| 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 
[maths]
P(t) = \int_{t_0}^t L_x(s,x(s),\dot{x}(s) ) ds
[/maths]
is $C^k$ as well. By [eqref euler] 
[math apeuler]
-P(t) + L_y(t,x(t),\dot{x}(t)) =c
[/math] 
for some constant $c$. Define
[maths]
\Phi(t,y) = -P(t) + L_y(t,x(t),y).
[/maths]
$L$ is $C^r$ and $k \le r-1$ and $x$ is $C^k$, it follows that
$\Phi$ is $C^k$. 
We can rewrite [eqref apeuler] 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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Use the implicit function theorem to prove smoothness

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