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?

Question

Given that the Laplace transforms of \( J_0(x), J_0'(x), \) and \( J_0''(x) \) exist, where \( J_0(x) \) is the Bessel function. Let \( Y = Y(s) \) be the Laplace transform of the Bessel function \( J_0(x) \). Then, which one of the following is TRUE?

Answer 1

To determine the correct statement about the linear span \( W \) of all Legendre polynomials in the spaces \( X \) and \( Y \), we need to understand the concept of denseness in these spaces.

Let's break down the given information:

\( X = (C[-1, 1], \| \cdot \|_\infty) \): The space of all real-valued continuous functions defined on \([-1, 1]\) with the supremum norm \( \| f \|_\infty = \sup_{t \in [-1, 1]} |f(t)| \).
\( Y = (C[-1, 1], \| \cdot \|_2) \): The space of all real-valued continuous functions defined on \([-1, 1]\) with the L2 norm \( \| f \|_2 = \left( \int_{-1}^1 |f(t)|^2 \, dt \right)^{1/2} \).

Legendre polynomials form a complete orthogonal system on the interval \([-1, 1]\) with respect to the L2 norm. This means that any function \( f \) in \( Y \) can be approximated arbitrarily closely by a linear combination of Legendre polynomials, which implies that \( W \) is dense in \( Y \).

Moreover, by the Stone-Weierstrass theorem, which states that any subalgebra containing a non-zero constant and separating points of a compact space can uniformly approximate any continuous function on that space, \( W \) is also dense in \( X \). Since Legendre polynomials include the constant function and are continuous, they fulfill the conditions, making \( W \) dense in \( X \) as well.

Hence, the correct statement is:

\( W \) is dense in both \( X \) and \( Y \)

Answer 2

To determine the correct statement about the linear span \( W \) of all Legendre polynomials in the spaces \( X \) and \( Y \), we need to understand the concept of denseness in these spaces.

Let's break down the given information:

\( X = (C[-1, 1], \| \cdot \|_\infty) \): The space of all real-valued continuous functions defined on \([-1, 1]\) with the supremum norm \( \| f \|_\infty = \sup_{t \in [-1, 1]} |f(t)| \).
\( Y = (C[-1, 1], \| \cdot \|_2) \): The space of all real-valued continuous functions defined on \([-1, 1]\) with the L2 norm \( \| f \|_2 = \left( \int_{-1}^1 |f(t)|^2 \, dt \right)^{1/2} \).

Legendre polynomials form a complete orthogonal system on the interval \([-1, 1]\) with respect to the L2 norm. This means that any function \( f \) in \( Y \) can be approximated arbitrarily closely by a linear combination of Legendre polynomials, which implies that \( W \) is dense in \( Y \).

Moreover, by the Stone-Weierstrass theorem, which states that any subalgebra containing a non-zero constant and separating points of a compact space can uniformly approximate any continuous function on that space, \( W \) is also dense in \( X \). Since Legendre polynomials include the constant function and are continuous, they fulfill the conditions, making \( W \) dense in \( X \) as well.

Hence, the correct statement is:

\( W \) is dense in both \( X \) and \( Y \)

Show Hint

The Correct Option is C

Solution and Explanation

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