Question

Let \( a>0 \). Define \( D_a : L^2(\mathbb{R}) \to L^2(\mathbb{R}) \) by \[ (D_a f)(x) = \frac{1}{\sqrt{a}} f\left( \frac{x}{a} \right), \text{ almost everywhere, for } f \in L^2(\mathbb{R}). \] Then, which of the following statements are TRUE?

• A. \( D_a \) is a linear isometry
• B. \( D_a \) is a bijection
• C. \( D_a \circ D_b = D_{a+b}, \, b>0 \)
• D. \( D_a \) is bounded from below

Hint:

When working with linear operators, checking if they preserve norms (isometries) and verifying injectivity and surjectivity can help determine if the operator is a bijection.

Answer 1

The Correct Option is A, B, D

We are given the operator \( D_a \) defined on the space \( L^2(\mathbb{R}) \), and we need to determine which of the provided statements are true. Step 1: Checking if \( D_a \) is a linear isometry
The operator \( D_a \) is defined by: \[ (D_a f)(x) = \frac{1}{\sqrt{a}} f\left( \frac{x}{a} \right). \] To verify if \( D_a \) is a linear isometry, we need to check if it satisfies linearity and if it preserves the \( L^2 \)-norm. - Linearity: Since \( D_a \) involves a linear transformation (scaling and shifting) of the function \( f \), it is linear. - Isometry: To check if \( D_a \) is an isometry, we compute: \[ \| D_a f \|_{L^2}^2 = \int_{-\infty}^{\infty} \left| (D_a f)(x) \right|^2 dx = \int_{-\infty}^{\infty} \left| \frac{1}{\sqrt{a}} f\left( \frac{x}{a} \right) \right|^2 dx = \frac{1}{a} \int_{-\infty}^{\infty} |f(x)|^2 dx = \| f \|_{L^2}^2. \] Thus, \( D_a \) preserves the \( L^2 \)-norm, meaning \( D_a \) is an isometry. Step 2: Checking if \( D_a \) is a bijection
To check if \( D_a \) is a bijection, we need to verify if \( D_a \) is both injective (one-to-one) and surjective (onto). - Injectivity: If \( D_a f = 0 \), then \( f(x) = 0 \) almost everywhere (since the scaling and shifting do not affect the function being zero). Therefore, \( D_a \) is injective. - Surjectivity: For any function \( g \in L^2(\mathbb{R}) \), we can find a function \( f \in L^2(\mathbb{R}) \) such that \( D_a f = g \). Specifically, \( f(x) = \sqrt{a} g(ax) \) satisfies \( D_a f = g \). Therefore, \( D_a \) is surjective. Since \( D_a \) is both injective and surjective, it is a bijection. Step 3: Checking if \( D_a \circ D_b = D_{a+b} \)
The statement \( D_a \circ D_b = D_{a+b} \) is incorrect. The correct relation is: \[ D_a \circ D_b = D_{ab}. \] This is because the scaling of \( x \) by \( a \) and \( b \) results in the scaling by their product. Step 4: Checking if \( D_a \) is bounded from below
We need to check if there exists a constant \( C>0 \) such that: \[ \| D_a f \|_{L^2} \geq C \| f \|_{L^2} \quad \forall f \in L^2(\mathbb{R}). \] Since \( D_a \) is an isometry, we know that \( \| D_a f \|_{L^2} = \| f \|_{L^2} \). Thus, \( D_a \) is bounded from below with \( C = 1 \). Step 5: Conclusion
The correct answers are: - (A) \( D_a \) is a linear isometry.
- (B) \( D_a \) is a bijection.
- (D) \( D_a \) is bounded from below.
Thus, the correct answer is (A), (B), (D).