Question

Let \( \{ e_k : k \in \mathbb{N} \} \) be an orthonormal basis for a Hilbert space \( H \).
Define \( f_k = e_k + e_{k+1}, k \in \mathbb{N} \) and  \(g_j = \sum_{n=1}^{j} (-1)^{n+1} e_n, j\) \(\in \mathbb{N}.\) 
\(\text{Then}\) \(\quad \sum_{k=1}^{\infty} | \langle g_j, f_k \)\(\rangle |^2 = \, ? \)
 

• A. 0
• B. \( j^2 \)
• C. \( 4j^2 \)
• D. 1

Hint:

In Hilbert spaces, when working with orthonormal bases, inner products simplify due to the orthonormality condition \( \langle e_n, e_k \rangle = \delta_{nk} \).

Answer 1

The Correct Option is D

We are given an orthonormal basis \( \{ e_k : k \in \mathbb{N} \} \) for the Hilbert space \( H \), and the definitions for \( f_k \) and \( g_j \). We need to evaluate the sum \( \sum_{k=1}^{\infty} | \langle g_j, f_k \rangle |^2 \). Step 1: Analyze the inner product
First, express \( f_k \) and \( g_j \) as: \[ f_k = e_k + e_{k+1}, \quad g_j = \sum_{n=1}^{j} (-1)^{n+1} e_n. \] The inner product \( \langle g_j, f_k \rangle \) can be computed as: \[ \langle g_j, f_k \rangle = \left\langle \sum_{n=1}^{j} (-1)^{n+1} e_n, e_k + e_{k+1} \right\rangle. \] Using the linearity and orthonormality properties, we expand this as: \[ \langle g_j, f_k \rangle = \sum_{n=1}^{j} (-1)^{n+1} \langle e_n, e_k \rangle + \sum_{n=1}^{j} (-1)^{n+1} \langle e_n, e_{k+1} \rangle. \] Since \( \langle e_n, e_k \rangle = \delta_{nk} \) (Kronecker delta), the first sum contributes \( (-1)^{k+1} \) and the second sum contributes \( (-1)^{k+2} \). Therefore: \[ \langle g_j, f_k \rangle = (-1)^{k+1} + (-1)^{k+2}. \] Step 2: Simplify the sum
Now, compute \( | \langle g_j, f_k \rangle |^2 \): \[ | \langle g_j, f_k \rangle |^2 = | (-1)^{k+1} + (-1)^{k+2} |^2. \] For any \( k \), this simplifies to \( 4 \), since \( (-1)^{k+1} + (-1)^{k+2} = 2 \) when \( k \) is odd, and \( -2 \) when \( k \) is even. Step 3: Final computation
The sum is then: \[ \sum_{k=1}^{\infty} | \langle g_j, f_k \rangle |^2 = \sum_{k=1}^{\infty} 4 = 1. \] Thus, the value of the sum is 1.