Question

Let \(X\) be a Banach space. Assume that every separable closed subspace of \(X\) admits an equivalent uniformly convex norm. Show that \(X\) has an equivalent uniformly convex norm.

Short answer

An equivalent uniformly convex norm on \(X\) is constructed by extending a uniformly convex norm from a separable closed subspace.

Step-by-step solution

Understand the Definitions

Recall that a Banach space is a complete normed vector space. A norm is uniformly convex if for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all vectors \(\|x\| = 1\) and \(\|y\| = 1\) with \(\|x - y\| \geq \epsilon\), we have \(\left\|\frac{x + y}{2}\right\| \leq 1 - \delta\). A space is separable if it has a countable dense subset.

Consider a Dense Subspace

Given that every separable closed subspace of \(X\) admits an equivalent uniformly convex norm, take a countable dense subset \(Q = \{q_n\}_{n \in \mathbb{N}}\) of \(X\). The norm closure of \(Q\) in \(X\) is a separable closed subspace, say \(Y\).

Uniformly Convex Norm on Subspace

By assumption, there exists an equivalent uniformly convex norm \(\| \cdot \|_Y\) on the separable closed subspace \(\overline{Q} = Y\).

Define an Equivalent Norm on the Whole Space

Using the M-structure of Banach spaces, extend \(\| \cdot \|_Y\) from \(\overline{Q}\) to an equivalent norm \(\| \cdot \|'\) on \(X\) that preserves uniform convexity. This extension is possible because uniformly convex norms are stable under linear isomorphisms and extensions.

Verification

Verify that the new norm \(\| \cdot \|'\) on \(X\) is indeed uniformly convex. This follows from the uniform convexity of \(\| \cdot \|_Y\) and the properties of equivalent norms.

Conclusion

Thus, \(X\) has an equivalent uniformly convex norm.

Key concepts

Uniformly Convex Norms

A Banach space is more than just a complete normed vector space; it can have special kinds of norms. One important type is the uniformly convex norm.
Uniform convexity means that the space's norm reacts strongly to vectors drifting apart. Specifically, for every \(\backslash\backslash\backslash\text{epsilon} > 0\), there exists a \(\backslash\backslash\backslash\text{delta} > 0\) so that for any two unit vectors \(x\) and \(y\) with \(\backslash\backslash\backslash|x - y\backslash\backslash\backslash| \backslash\backslash\backslash\text{geq} \backslash\backslash\backslash\text{epsilon}\), the midpoint \(\backslash\backslash\backslash\text{left}\backslash\backslash\backslash|\backslash\backslash\backslash\text{frac}\backslash\backslash\backslash{x+y}\backslash\backslash\backslash{2}\backslash\backslash\backslash |\text{right}\backslash\backslash\backslash| \backslash\backslash\backslash\text{less} \backslash\backslash\backslash| cated)}\).
If every separable subspace of a Banach space admits such a norm, then a uniformly convex norm can extend to the entire Banach space. This property is extremely helpful in functional analysis.

Separable Subspaces

In a Banach space, separable subspaces play a crucial role. They are subspaces that have a countable dense subset.
This means you can approximate any element in the space as closely as you want by elements from a countable set of points. This property simplifies many theoretical arguments and makes certain types of analysis feasible.
In the exercise's context, every separable subspace being able to adopt a uniformly convex norm allows us to use these norms to eventually cover the entire space.

Dense Subsets

A subset is dense in a Banach space if every point in the space is either in the subset or arbitrarily close to points in the subset.
A concrete example of a dense subset is the set of rational numbers within the real numbers. This is a countable dense subset, meaning a broad understanding can be built from just a small set of rational numbers.
The key to solving the exercise lies in finding these dense subsets. When a dense subset of separable spaces can take on a uniformly convex norm, this extends to the entire space, ensuring that the whole Banach space inherits this useful property.