Question

Let \( H \) be a complex Hilbert space. Let \( u, v \in H \) be such that \( \langle u, v \rangle = 2 \). Then \[ \frac{1}{2\pi} \int_0^{2\pi} \| u + e^{it} v \|^2 e^{it} dt = \underline{\hspace{1cm}}. \]

Hint:

Use properties of inner products and expand the norm to solve integrals in Hilbert spaces.

Answer 1

Correct Answer: 2

Using the properties of the inner product and expanding the norm, we find: \[ \| u + e^{it} v \|^2 = \| u \|^2 + 2 \text{Re}( \langle u, e^{it} v \rangle ) + \| v \|^2. \] Integrating over \( t \) from \( 0 \) to \( 2\pi \), we get: \[ \frac{1}{2\pi} \int_0^{2\pi} \| u + e^{it} v \|^2 e^{it} dt = 2. \] Thus, the value of the integral is \( \boxed{2} \).