EDB — 1GZ

Exercises

1. [1GZ] In the same assumptions as the previous theorem [1GD]↺↻, show that there exist \(\varepsilon {\gt}0\) and a continuous function \(\tilde g:V→ ℝ\) where \(I=(\overline a-\varepsilon ,\overline a+\varepsilon )\) and \(V=U'× I\) is open in \(ℝ^ n\), such that

   \begin{equation} ∀ (x',a)∈ V \quad ,\quad (x',\tilde g(x',a))∈ U \quad \text{e} \quad f\big(x',\tilde g(x',a)\big)=a \quad .\label{eq:tilde_ g_ teor_ funz_ inv} \end{equation}

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   Vice versa if \(x∈ U\) and \(a=f(x)\) and \(a∈ I\) then \(x_ n=\tilde g(x',a)\).

   Note that the previous relation means that, for each fixed \(x'∈ U'\), the function \(\tilde g(x',⋅)\) is the inverse of the function \(f(x',⋅)\) (when defined on appropriate open intervals).

   So, moreover, the function \(\tilde g\) is always differentiable with respect to \(a\), and the partial derivative is

   \[ \frac{\partial ~ }{\partial {a}} \tilde g(x',a)=\frac{1}{\frac{\partial ~ }{\partial {x_ n}} f(x',\tilde g(x',a))}\quad . \]

   The other derivatives instead (obviously) are as in the theorem [1GD]↺↻.

   The regularity of \(\tilde g\) is the same as \(g\): if \(f\) is Lipschitz then \(\tilde g\) is Lipschitz; if \(f∈ C^ k(U)\) then \(\tilde g∈ C^ k(V)\).

   Solution 1

   [1H0]↺↻