Question

Consider the following equation in a 2-D real-space: \[ |x_1|^p + |x_2|^p = 1 \text{for } p > 0 \] Which of the following statement(s) is/are true?

• A. When \(p=2\), the area enclosed by the curve is \(\pi\).
• B. When \(p \to \infty\), the area enclosed by the curve tends to 4.
• C. When \(p \to 0\), the area enclosed by the curve is 1.
• D. When \(p=1\), the area enclosed by the curve is 2.

Hint:

The equation \(|x_1|^p + |x_2|^p = 1\) defines the unit \(L^p\) norm ball in 2D. As \(p\) increases, the shape changes from diamond (\(p=1\)) to circle (\(p=2\)) to square (\(p \to \infty\)).

Answer 1

The Correct Option is A, B

Step 1: Recognize the curve.
The given equation represents the boundary of an \(L^p\)-norm unit ball in 2D.

Step 2: Case \(p=2\).
Equation becomes: \[ x_1^2 + x_2^2 = 1 \] This is a circle of radius 1. \[ \text{Area} = \pi (1^2) = \pi \] Thus, statement (A) is true.

Step 3: Case \(p \to \infty\).
Equation becomes: \[ \max(|x_1|, |x_2|) = 1 \] This describes a square with vertices \((\pm 1, \pm 1)\). \[ \text{Area} = 2 \times 2 = 4 \] Thus, statement (B) is true.

Step 4: Case \(p=1\).
Equation becomes: \[ |x_1| + |x_2| = 1 \] This is a diamond (square rotated 45°) with diagonals length 2. Area: \[ \frac{d_1 d_2}{2} = \frac{2 \times 2}{2} = 2 \] Thus, statement (D) is true.

Step 5: Case \(p \to 0\).
As \(p \to 0\), the set shrinks towards the coordinate axes and enclosed area tends to 0, not 1. Thus, statement (C) is false.

Final Answer:
\[ \boxed{(A), (B), (D)} \]