To determine where the inverse function theorem is applicable, we need to analyze the Jacobian matrix of the function \( F: \mathbb{R}^2 \to \mathbb{R}^2 \). The inverse function theorem can be applied at a point if the determinant of the Jacobian at that point is nonzero.
The function given is:
\( F(x, y) = (x^3 - 3xy^2 - 3x, 3x^2y - y^3 - 3y) \).
We will compute the Jacobian matrix \( J \) of \( F \), which is the matrix of first-order partial derivatives:
\[ J = \begin{bmatrix} \frac{\partial F_1}{\partial x} & \frac{\partial F_1}{\partial y} \\ \frac{\partial F_2}{\partial x} & \frac{\partial F_2}{\partial y} \end{bmatrix} \]where \( F_1(x, y) = x^3 - 3xy^2 - 3x \) and \( F_2(x, y) = 3x^2y - y^3 - 3y \).
Calculating the partial derivatives, we have:
Thus, the Jacobian matrix is:
\[ J = \begin{bmatrix} 3x^2 - 3y^2 - 3 & -6xy \\ 6xy & 3x^2 - 3y^2 - 3 \end{bmatrix} \]The determinant of the Jacobian matrix can be calculated as:
\[ \text{det}(J) = (3x^2 - 3y^2 - 3)(3x^2 - 3y^2 - 3) - (-6xy)(6xy) \] \[ = (3x^2 - 3y^2 - 3)^2 - (-6xy)^2 \] \[ = (3x^2 - 3y^2 - 3)^2 - 36x^2y^2 \]Setting \(\text{det}(J) = 0\) will give us the points where the inverse function theorem is not applicable:
\[ (3x^2 - 3y^2 - 3)^2 = 36x^2y^2 \]This implies:
\[ 3x^2 - 3y^2 - 3 = \pm 6xy \]Solving for these conditions involves finding points \((x, y)\) on \(\mathbb{R}^2\) where these equalities hold.
Analyzing these equations, you get two different solutions, leading to two specific points where \(\text{det}(J) = 0\), making the inverse function theorem not applicable. Hence the correct choice is:
Let \( 0<\alpha<1 \). Define \[ C^\alpha[0, 1] = \left\{ f : [0, 1] \to \mathbb{R} \ : \ \sup_{s \neq t, \, s,t \in [0, 1]} \frac{|f(t) - f(s)|}{|t - s|^\alpha}<\infty \right\}. \] It is given that \( C^\alpha[0, 1] \) is a Banach space with respect to the norm \( \| \cdot \|_\alpha \) given by \[ \| f \|_\alpha = |f(0)| + \sup_{s \neq t, \, s,t \in [0, 1]} \frac{|f(t) - f(s)|}{|t - s|^\alpha}. \] Let \( C[0, 1] \) be the space of all real-valued continuous functions on \( [0, 1] \) with the norm \( \| f \|_\infty = \sup_{0 \leq t \leq 1} |f(t)| \).
If \( T: C^\alpha[0, 1] \to C[0, 1] \) is the map \( T f = f \), where \( f \in C^\alpha[0, 1] \), then which one of the following is/are TRUE?
Consider the following two spaces:
\[ \begin{aligned} X &= (C[-1, 1], \| \cdot \|_\infty), \quad \text{the space of all real-valued continuous functions} \\ &\quad \text{defined on } [-1, 1] \text{ equipped with the norm } \| f \|_\infty = \sup_{t \in [-1, 1]} |f(t)|. \\ Y &= (C[-1, 1], \| \cdot \|_2), \quad \text{the space of all real-valued continuous functions} \\ &\quad \text{defined on } [-1, 1] \text{ equipped with the norm } \| f \|_2 = \left( \int_{-1}^1 |f(t)|^2 \, dt \right)^{1/2}. \end{aligned} \]
Let \( W \) be the linear span over \( \mathbb{R} \) of all the Legendre polynomials. Then, which one of the following is correct?