Frechet differential and implicit theorem

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Given:

  • $ \Omega \subseteq \mathbb{R}^n $.
  • $ f: \Omega \rightarrow \mathbb{R}^n $ and $ g: \Omega \rightarrow \mathcal{L}(\mathbb{R}^n, \mathbb{R}^n) $ are two $ C^1 $-class functions on $ \Omega $.
  • Define $ F: \Omega \times \Omega \rightarrow \mathbb{R}^n $ as $ F(x, y) = g(x)(f(y)) + f(x) $.

(a) Show that $ F $ is of class $ C^1 $ on $ \Omega \times \Omega $ and find $ DF $, $ D_1F $, and $ D_2F $.

Now, let's proceed with the solution:

To show that $ F $ is of class $ C^1 $ on $ \Omega \times \Omega $, we need to demonstrate that ( F ) is continuously differentiable with respect to both variables ( x ) and ( y ).

The function $ F(x, y) $ is composed of two parts: $ g(x)(f(y)) $ and $ f(x) $. Since $ f $ and $ g $ are $ C^1 $-class functions on $ \Omega $, their compositions are also $ C^1 $ functions. Therefore, $ g(x)(f(y)) $ and $ f(x) $ are $ C^1 $ functions on $ \Omega \times \Omega $.

Now, let's compute the partial derivatives:

  • $ D_1F = g'(x)f(y) + f'(x) $
  • $ D_2F = g(x)f'(y) $

I use this:

$$ \Omega \rightarrow \mathcal{L}(\mathbb{R}^n, \mathbb{R}^n) \times \Omega \rightarrow \mathbb{R}^n $$

Where

$$ x \mapsto (g(x), y) \mapsto g(x)y $$

Finally, the total derivative $ DF $ can be expressed as the matrix: $$ DF = \begin{pmatrix} D_1F \\ D_2F \end{pmatrix} = \begin{pmatrix} g'(x)f(y) + f'(x) \\ g(x)f'(y) \end{pmatrix} $$

Suppose $ 0 \in \Omega $ and $ f(0) = 0 $. We want to find conditions on $ f $ and $ g $ for there to exist open neighborhoods $ U $ and $ V $ of $ 0 $ in $ \mathbb{R}^n $ and a function $ \Psi: U \rightarrow V $ of class $ C^1 $ on $ U $ such that $ F(x,y) = 0 $ if and only if $ y = \Psi(x) $. Additionally,find $ D\Psi(0) $.

For this part, I know I need to examine the matrix $ D_2F $ and check if it's invertible at the point $ (0,0) $ to apply the implicit theorem. For this, I need both the matrix $ g(0) $ and the derivative $ f'(0) $ to be bijective. Can someone confirm this? I need help to find $ D\Psi(0) $. Thank you.