Let (V, ⟨⋅, ⋅⟩) be an inner product space. For any v ∈ V we have∥v∥ = sup∥w∥=1∣⟨v, w⟩∣.The supremum is in fact achieved by a well-chosen w.
Question
Let (V, ⟨⋅, ⋅⟩) be an inner product space. For any v ∈ V we have∥v∥ = sup∥w∥=1∣⟨v, w⟩∣.The supremum is in fact achieved by a well-chosen w.
Solution
The statement you've given is a property of an inner product space. It says that the norm of any vector 'v' in the space 'V' is equal to the supremum (or maximum) of the absolute value of the inner product of 'v' and 'w', where 'w' is any unit vector in 'V'. The statement also says that this supremum is actually achieved for some well-chosen 'w'.
Here's a step-by-step explanation:
-
An inner product space is a vector space that has an additional structure called an inner product. This allows you to define angles and lengths, which gives you the ability to talk about concepts like orthogonality, distance, and norm.
-
The norm of a vector 'v' (denoted as ∥v∥) is a measure of its length. In an inner product space, it's defined in terms of the inner product: ∥v∥ = sqrt(⟨v, v⟩).
-
The statement is saying that this norm is equal to the supremum of the absolute value of the inner product of 'v' and 'w' (denoted as |⟨v, w⟩|), where 'w' is any unit vector (a vector with norm 1) in 'V'.
-
The supremum of a set is the least upper bound of that set. In other words, it's the smallest number that's greater than or equal to every number in the set. In this case, the set is the set of all values of |⟨v, w⟩| as 'w' varies over all unit vectors.
-
The statement also says that this supremum is actually achieved for some well-chosen 'w'. This means that there exists a unit vector 'w' such that |⟨v, w⟩| is exactly equal to the supremum.
This property is a key part of the theory of inner product spaces and is used in many areas of mathematics, including linear algebra, functional analysis, and quantum mechanics.
Similar Questions
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⟩∣.]
Let (V, ⟨⋅, ⋅⟩) be an inner product space. Prove that the inner product is a continuousfunction
5. Let (V, ⟨⋅, ⋅⟩) be an inner product space. Prove that the inner product is a continuousfunction
Let X and Y be non-empty sets, and let h : X × Y → R be a bounded function. Provethatsupy∈Ysupx∈Xh(x, y) = sup(x,y)∈X×Yh(x, y) = supx∈Xsupy∈Yh(x, y).Hint) By symmetry, we can interchange the roles of X and Y if necessary. Therefore, it isenough to prove thatsupy∈Ysupx∈Xh(x, y) = sup(x,y)∈X×Yh(x, y)
Let X and Y be non-empty sets, and let h : X × Y → R be a bounded function.(i) Prove thatsupy∈Ysupx∈Xh(x, y) = sup(x,y)∈X×Yh(x, y) = supx∈Xsupy∈Yh(x, y).Hint) Interchanging the roles of X and Y if necessary, it is enough to prove thatsupy∈Ysupx∈Xh(x, y) = sup(x,y)∈X×Yh(x, y)
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.