Let V, W be inner product spaces and let f ∈ B(V, W ). Prove that∥f ∥ = sup∥v∥V =∥w∥W =1∣⟨f (v), w⟩W ∣.[Hint: Use Exercise 4.2 which says that ∥v∥ = sup∥w∥=1 ∣⟨v, w⟩∣.]

This seems to be a mathematical proof question related to linear algebra, specifically about inner product spaces and bounded linear operators. Here's a step-by-step guide to prove the statement:

Step 1: Define the terms V and W are inner product spaces. f is a bounded linear operator from V to W. The norm of f is defined as the supremum of the norms of f(v) where v ranges over all vectors in V of norm 1.

Step 2: Use the hint According to Exercise 4.2, the norm of a vector v in an inner product space is equal to the supremum of the absolute value of the inner product of v and w, where w ranges over all vectors of norm 1.

Step 3: Apply the definition of the operator norm We have ∥f ∥ = sup∥v∥V =1 ∥f(v)∥W.

Step 4: Apply the result from Exercise 4.2 to ∥f(v)∥W We can replace ∥f(v)∥W with sup∥w∥W =1 ∣⟨f (v), w⟩W ∣.

Step 5: Combine the results So, we have ∥f ∥ = sup∥v∥V =1 sup∥w∥W =1 ∣⟨f (v), w⟩W ∣, which completes the proof.

Please note that this is a high-level explanation and the actual proof may require more detailed arguments depending on the level of rigor required.