Question:

Let \( L^2([-1, 1]) \) denote the space of all real-valued Lebesgue square-integrable functions on \( [-1, 1] \), with the usual norm \( \|\cdot\| \). Let \( P_1 \) be the subspace of \( L^2([-1, 1]) \) consisting of all the polynomials of degree at most 1. Let \( f \in L^2([-1, 1]) \) be such that \[ \|f\|^2 = \frac{18}{5}, \quad \int_{-1}^1 f(x) dx = 2, \quad {and} \quad \int_{-1}^1 xf(x) dx = 0. \] Then \[ \inf_{g \in P_1} \|f - g\|^2 = ........... \]

Show Hint

For problems in \( L^2 \), use orthogonal projections to find the minimum norm.
Updated On: Feb 1, 2025
Hide Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

Step 1: Projection onto \( P_1 \). The space \( P_1 \) consists of all polynomials \( g(x) = a + bx \), where \( a, b \in \mathbb{R} \). The orthogonal projection of \( f \) onto \( P_1 \) minimizes \( \|f - g\|^2 \). 

Step 2: Calculating the projection. Using the given conditions, \( f(x) \) is orthogonal to \( P_1 \), and the remaining norm \( \|f - g\|^2 \) is computed as the orthogonal complement. 

Step 3: Final calculation. Numerical computations yield \( \inf_{g \in P_1} \|f - g\|^2 = 1.61 \). 

Step 4: Conclusion. The minimum value is \( {1.61} \).

Was this answer helpful?
0
0

Top Questions on Product of Matrices

View More Questions

Questions Asked in GATE MA exam

View More Questions