Question

Let M be a positive real number and let u, v: \(\R^2\rightarrow\R\) be continuous functions satisfying \(\sqrt{u(x,y)^2+v(x,y)^2}\ge M\sqrt{x^2+y^2}\ for\ all\ (x,y)\isin\R^2.\)
Let F: \(\R^2\rightarrow\R^2\) be given by
F(x, y) = (u(x, y), v(x, y)) for (x, y)\(\isin\R^2\).
Then which of the following is/are true?

• A. F is injective.
• B. If K is open in \(\R^2\), then F(K) is open in \(\R^2\).
• C. If K is closed in \(\R^2\), then F(K) is closed in \(\R^2\).
• D. If E is closed and bounded in \(\R^2\), then F^-1(E) is closed and bounded in \(\R^2\).

Answer 1

The Correct Option is C, D

To solve this problem, we need to analyze the behavior of the function \( F: \mathbb{R}^2 \rightarrow \mathbb{R}^2 \) defined by \( F(x, y) = (u(x, y), v(x, y)) \), where the functions \( u \) and \( v \) satisfy \( \sqrt{u(x, y)^2 + v(x, y)^2} \ge M\sqrt{x^2+y^2} \) for all \( (x, y) \in \mathbb{R}^2 \).

1. Analysis of injectivity: To determine if \( F \) is injective, we need to check if \( F(x_1, y_1) = F(x_2, y_2) \) implies \( (x_1, y_1) = (x_2, y_2) \). The given inequality does not necessarily enforce that the map is one-to-one, as the inequality gives a bound on the magnitude of \( F(x, y) \), not distinctness. Hence, F is not guaranteed to be injective.
2. Openness of F(K) when K is open: If \( K \) is open in \( \mathbb{R}^2 \), openness of \( F(K) \) would require \( F \) to be an open map. Without additional properties like differentiability everywhere or conditions on derivatives, we cannot conclude that \( F(K) \) is open.
3. Closedness of F(K) when K is closed: To verify if \( F(K) \) is closed, consider that the image of a closed set under a continuous function is closed only under certain conditions such as being a proper map or compact sets in certain contexts. However, since the inequality ensures \( F(x, y) \) is "controlled" with respect to \( (x, y) \), it is reasonable to suspect that this strong form of magnitude control pushes us toward closure properties being preserved. Therefore, if \( K \) is closed in \( \mathbb{R}^2 \), then \( F(K) \) is closed.
4. Closedness and boundedness of the inverse image: The function \( F^{-1}(E) \) under the conditions of closed and bounded \( E \) implies continuity or "properness" due to the scaling inequality suggesting a factor of proportional location preservation. Thus, like the above analysis, as a possible major consequence of the inequality, if \( E \) is closed and bounded in \( \mathbb{R}^2 \), then \( F^{-1}(E) \) is closed and bounded.

Conclusion: The correct options are:

• If \( K \) is closed in \( \mathbb{R}^2 \), then \( F(K) \) is closed in \( \mathbb{R}^2 \).
• If \( E \) is closed and bounded in \( \mathbb{R}^2 \), then \( F^{-1}(E) \) is closed and bounded in \( \mathbb{R}^2 \).