Brezis Exercise 3.9 Solution And Explanation Functional Analysis

Hey guys! Today, we're diving deep into a fascinating problem from Brezis' Functional Analysis – Exercise 3.9. This exercise beautifully intertwines concepts from Banach spaces, linear subspaces, weak topology, and annihilators. So, buckle up, and let's unravel this problem together!

Problem Statement

Before we jump into the solution, let's clearly state the problem. We're given:

• A Banach space E.
• A linear subspace M of E.
• A functional f₀ belonging to the continuous dual space E^* (denoted as E^*).

Our mission, should we choose to accept it (and we do!), is to prove the existence of a g₀ in M^⊥ (the annihilator of M) that satisfies this condition:

inf{||f₀ - g|| : g ∈ M<sup>⊥</sup>} = sup{|<f₀, x>| : ||x|| ≤ 1, x ∈ M}

In simpler terms, we need to show that the infimum of the distances between f₀ and elements in M^⊥ is equal to the supremum of the absolute values of f₀ evaluated on elements of M with norm less than or equal to 1. Sounds like a mouthful, right? Let's break it down step by step.

Decoding the Problem

To truly conquer this problem, we need to understand the key players. Let's start by defining some crucial concepts:

Banach Space

A Banach space is a complete normed vector space. "Complete" means that every Cauchy sequence in the space converges to a limit within the space. This completeness property is super important for many theorems and results in functional analysis.

Linear Subspace

A linear subspace M of a vector space E is a subset of E that is itself a vector space under the same operations of addition and scalar multiplication. Think of it as a vector space within a vector space.

Continuous Dual Space (E^*)

The continuous dual space E^* consists of all bounded linear functionals from E to the scalar field (usually real or complex numbers). A bounded linear functional is a linear map that doesn't "blow up" – it maps bounded sets in E to bounded sets in the scalar field.

Annihilator (M^⊥)

The annihilator M^⊥ of a subset M of E is the set of all functionals in E that "annihilate" M. That is, M^⊥ contains all g in E such that g(x) = 0 for all x in M. It's like the orthogonal complement, but for functionals!

The Infimum and Supremum

The infimum (inf) is the greatest lower bound, and the supremum (sup) is the least upper bound. In our case, the infimum represents the smallest possible distance between f₀ and any functional in M^⊥, while the supremum represents the largest possible value of |f₀(x)| when x is in M and has a norm less than or equal to 1.

Solution Strategy

Now that we have a solid grasp of the definitions, let's formulate a plan of attack. Our goal is to prove the equality:

inf{||f₀ - g|| : g ∈ M<sup>⊥</sup>} = sup{|<f₀, x>| : ||x|| ≤ 1, x ∈ M}

A common strategy for proving such equalities is to show two inequalities:

1. inf||f₀ - g|| g ∈ M^⊥ ≥ sup|<f₀, x>| ||x|| ≤ 1, x ∈ M
2. inf||f₀ - g|| g ∈ M^⊥ ≤ sup|<f₀, x>| ||x|| ≤ 1, x ∈ M

If we can establish both inequalities, we've successfully proven the equality. Let's tackle them one by one.

Part 1: Proving inf||f₀ - g|| g ∈ M^⊥ ≥ sup|<f₀, x>| ||x|| ≤ 1, x ∈ M

Let's start by considering any g in M^⊥ and any x in M with ||x|| ≤ 1. Because g is in M^⊥, we know that g(x) = 0. Now, let's use the triangle inequality and the definition of the norm of a functional:

|<f₀, x>| = |<f₀, x> - <g, x>| = |<(f₀ - g), x>| ≤ ||f₀ - g|| ||x|| ≤ ||f₀ - g||

This inequality holds for any g in M^⊥ and any x in M with ||x|| ≤ 1. Therefore, ||f₀ - g|| is an upper bound for the set *|<f₀, x>| ||x|| ≤ 1, x ∈ M* for any g in M^⊥. Since the supremum is the least upper bound, we have:

sup{|<f₀, x>| : ||x|| ≤ 1, x ∈ M} ≤ ||f₀ - g||

This inequality holds for all g in M^⊥. Now, we take the infimum over all g in M^⊥ on the right-hand side. The left-hand side is a constant (it doesn't depend on g), so we get:

sup{|<f₀, x>| : ||x|| ≤ 1, x ∈ M} ≤ inf{||f₀ - g|| : g ∈ M<sup>⊥</sup>}

This is exactly the first inequality we wanted to prove! Awesome!

Part 2: Proving inf||f₀ - g|| g ∈ M^⊥ ≤ sup|<f₀, x>| ||x|| ≤ 1, x ∈ M

This part requires a bit more finesse. We'll use a clever application of the Hahn-Banach theorem. Let's denote the supremum on the right-hand side by:

α = sup{|<f₀, x>| : ||x|| ≤ 1, x ∈ M}

Our goal is to show that for any ε > 0, there exists a g in M^⊥ such that:

||f₀ - g|| ≤ α + ε

This will imply that inf||f₀ - g|| g ∈ M^⊥ ≤ α, which is the inequality we're aiming for.

Let's define a linear functional h on the subspace M by:

h(x) = <f₀, x>

For x in M, we have:

|h(x)| = |<f₀, x>| ≤ α ||x||

This shows that h is a bounded linear functional on M with ||h|| ≤ α. Now comes the magic of Hahn-Banach! The Hahn-Banach theorem guarantees that we can extend h to a bounded linear functional h̃ on the entire space E such that ||h̃|| = ||h|| and h̃(x) = h(x) for all x in M.

Since ||h̃|| = ||h|| ≤ α, we know that for any x in E with ||x|| ≤ 1, we have:

|h̃(x)| ≤ ||h̃|| ||x|| ≤ α

Now, let's define g = f₀ - h̃. We claim that g belongs to M^⊥. To see this, let x be any element in M. Then:

<g, x> = <f₀ - h̃, x> = <f₀, x> - <h̃, x> = <f₀, x> - h(x) = <f₀, x> - <f₀, x> = 0

Thus, g annihilates M, so g is in M^⊥. Now, let's estimate ||f₀ - g||:

||f₀ - g|| = ||f₀ - (f₀ - h̃)|| = ||h̃|| ≤ α

This is almost what we wanted! To get the ε, we need a slight modification. Let's consider (1 + ε/α)h if α is not 0 (if α is 0, the result is trivial so we can skip some steps). The norm is now bounded by α + ε and extending again with Hahn-Banach gives us the final inequality. Thus, for any ε > 0, we've found a g in M^⊥ such that:

||f₀ - g|| ≤ α + ε

Taking the infimum over all g in M^⊥, we get:

inf{||f₀ - g|| : g ∈ M<sup>⊥</sup>} ≤ α = sup{|<f₀, x>| : ||x|| ≤ 1, x ∈ M}

This is the second inequality we needed!

Conclusion

By proving both inequalities, we've successfully shown that:

inf{||f₀ - g|| : g ∈ M<sup>⊥</sup>} = sup{|<f₀, x>| : ||x|| ≤ 1, x ∈ M}

This result is a beautiful illustration of the interplay between norms, annihilators, and the power of the Hahn-Banach theorem in functional analysis. It highlights how we can relate the distance between a functional and an annihilator to the supremum of its values on a subspace. This kind of result is super useful in optimization problems, approximation theory, and many other areas of math.

So, there you have it! We've conquered Brezis Exercise 3.9. I hope this deep dive has been insightful and has helped you better understand these fundamental concepts. Keep exploring the fascinating world of functional analysis, guys!