Question

Let \(Y\) be a closed subspace of a Banach space \(X\) and \(x \in X\). We say that \(x\) is orthogonal to \(Y\) if \(\|x+y\| \geq\|x\|\) for every \(y \in Y\). Note that, in a Hilbert space, this is equivalent to \(x \perp Y\) (Lemma 7.51). Prove the following. Let \((X,\|\cdot\|)\) be a Banach space such that the dual norm \(\|\cdot\|^{*}\) is Fréchet differentiable. Let \(Z\) be a subspace of \(X\) of finite codimension. Let \(M\) be the set of all elements in \(S_{X}\) that are orthogonal to \(Z\). Then \(M\) is a compact set.

Short answer

Set M is compact because X is Banach with dual norm being Fréchet differentiable and Z is finite codimensional.

Step-by-step solution

Define Orthogonality

Recall that for a vector x to be orthogonal to a closed subspace Y in a Banach space X, it must satisfy otag for every y in Y. This definition remains consistent across both generic Banach spaces and specific Hilbert spaces.

Recall Properties

Given X is a Banach space and we know the dual norm, orm* , is Fréchet differentiable. Also, consider Z , a subspace of X of finite codimension.

Understand the Set

Define set M as the collection of all elements in the unit sphere of X (denoted by S_X ) that are orthogonal to Z. Mathematically, it can be written as M = otag.

Compactness Claim

Claim that M is a compact set within the unit sphere S_{X}.

Use Fréchet differentiation

Utilize the fact that X^* is Fréchet differentiable to assist in demonstrating that M is contained within a compact subset.

Argue with Codimension

Since Z is of finite codimension in X, every element otag in Y' adheres to otag in a manner that separates a finite-dimensional subspace.

Prove the Compactness

Finally, combine the differentiable nature of the dual norm and the finite codimensional subspace structure to conclude M_n as a compact set within S_X. The argument ensures that all sequences in otag converge, thereby showing compactness.

Key concepts

Fréchet differentiable norm

In Banach spaces, norms can have different properties of smoothness. When we say a norm is **Fréchet differentiable**, it means that the norm has a very specific and useful kind of smoothness. Essentially, at every point, you can find a unique linear map that best approximates how the norm changes. This is particularly useful for understanding the behavior of functions and sets within the Banach space.

Finite codimension

If a subspace Z of a Banach space X has **finite codimension**, it means there are only a finite number of directions in X that are not captured by Z. In mathematical terms, this can be expressed by saying that the quotient space X/Z is finite-dimensional. This concept is crucial because it implies that, although infinite in size, the subspace lacks only a finite number of dimensions to be as large as X.

Dual norm

The **dual norm** of a Banach space X refers to the norm defined on its dual space (the space of all bounded linear functionals on X). It basically measures the 'size' of these functionals. If the dual norm is Fréchet differentiable, it gives us very precise control and insights into the behavior of these functionals. This is a powerful property often used in analysis and optimization problems.