Question:
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?
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?
Show Hint
In functional analysis, the density of a set in a normed space means that any element of the space can be approximated arbitrarily closely by elements of the set. Legendre polynomials are dense in spaces of continuous functions under both \( L^\infty \) and \( L^2 \) norms.
Updated On: Feb 2, 2026
- \( W \) is dense in \( X \) but not in \( Y \)
- \( W \) is dense in \( Y \) but not in \( X \)
- \( W \) is dense in both \( X \) and \( Y \)
- \( W \) is dense neither in \( X \) nor in \( Y \)
Show Solution
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The Correct Option is C
Solution and Explanation
We are given two spaces \( X \) and \( Y \) of continuous functions on \( [-1, 1] \), equipped with different norms. The Legendre polynomials form an orthogonal basis in the space of square-integrable functions over \( [-1, 1] \) under the \( L^2 \) norm (the norm \( \| \cdot \|_2 \)).
Step 1: Density of \( W \) in \( Y \)
Since the Legendre polynomials are orthogonal with respect to the \( L^2 \) inner product, they form a complete orthonormal system in the Hilbert space \( Y \). Therefore, the span of the Legendre polynomials is dense in \( Y \).
Step 2: Density of \( W \) in \( X \)
The Legendre polynomials are also dense in the space \( X \) equipped with the sup norm \( \| \cdot \|_\infty \). In fact, any continuous function on \( [-1, 1] \) can be approximated uniformly (in the sup norm) by polynomials, and the Legendre polynomials are a subset of the polynomial functions. Therefore, the span of the Legendre polynomials is dense in \( X \) as well.
Final Answer
\[ \boxed{(C) \quad W \text{ is dense in both } X \text{ and } Y.} \]
Step 1: Density of \( W \) in \( Y \)
Since the Legendre polynomials are orthogonal with respect to the \( L^2 \) inner product, they form a complete orthonormal system in the Hilbert space \( Y \). Therefore, the span of the Legendre polynomials is dense in \( Y \).
Step 2: Density of \( W \) in \( X \)
The Legendre polynomials are also dense in the space \( X \) equipped with the sup norm \( \| \cdot \|_\infty \). In fact, any continuous function on \( [-1, 1] \) can be approximated uniformly (in the sup norm) by polynomials, and the Legendre polynomials are a subset of the polynomial functions. Therefore, the span of the Legendre polynomials is dense in \( X \) as well.
Final Answer
\[ \boxed{(C) \quad W \text{ is dense in both } X \text{ and } Y.} \]
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