Question

Let \(Y, Z\) be closed subspaces of a Banach space \(X\) such that \(Y \cap Z=\) \(\\{0\\}\). Denote \(d=\operatorname{dist}\left(S_{Y}, S_{Z}\right) .\) Show that the subspace \(Y+Z\) is closed in \(X\) if and only if \(d>0\)

Short answer

Y+Z is closed iff d>0.

Step-by-step solution

Define given terms

Let Y and Z be closed subspaces of a Banach space X such that their intersection is the set containing only the zero vector, i.e., \(Y \cap Z = \{0\}\). Denote \(d = \text{dist}(S_Y, S_Z)\), where \(S_Y\) and \(S_Z\) are the unit spheres of Y and Z respectively.

Recall definition of distance between sets

The distance between two sets is defined as: \[\text{dist}(S_Y, S_Z) = \text{inf}\{\|y - z\| : y \in S_Y, z \in S_Z \}\]

Show if d>0 then Y+Z is closed

Suppose \(d > 0\). Assume \(w_k\to w\) for some \(w_k\in Y + Z\) and \(w\in X\). We can write each \(w_k = y_k + z_k\), with \(y_k\in Y\) and \(z_k\in Z\). Since \(w_k\to w\), both \(y_k\) and \(z_k\) are bounded sequences. By the Banach space completeness, we extract subsequences converging to some \(y\in Y\) and \(z\in Z\). As \(w_k\to w\), necessarily we have \(w = y+z\) showing \(w\in Y+Z\). Hence Y+Z is closed.

Show if Y+Z is closed then d>0

Assume \(Y + Z\) is closed. Suppose to the contrary that \(d = 0\), then there exist sequences \(y_k\in S_Y\) and \(z_k\in S_Z\) such that \(\|y_k - z_k\| \to 0\). Observe \((y_k - z_k)\in Y + Z \). If \(Y + Z\) is closed and \(d = 0\), \(y_k = z_k\) for large k. Hence \(y_k - z_k = 0\to 0\), or \(y_k = z_k\) when \(\|y_k - z_k\| \leq d\), contradicting the fact LSB spaces' intersection \(\{0\}\). Therefore, \(d > 0\).

Key concepts

Closed subspaces

A subspace in a Banach space is called closed if it contains all its limit points. In simpler terms, if a sequence of vectors in the subspace converges to a limit in the Banach space, that limit should also be within the same subspace.
Closed subspaces are crucial in analysis because they ensure the structure of the space remains intact even when we consider limits, which is a common occurrence in functional analysis.
For example, in the exercise, we have two closed subspaces, Y and Z, in a Banach space X. Since these subspaces are closed, we can apply limits and still remain within these subspaces. This property is useful in proving statements about convergence and completeness.

Unit spheres

In Banach spaces, the unit sphere of a subspace is the set of all vectors with a norm (or length) of 1. Formally, the unit sphere in a subspace Y is defined as:
\( S_Y = \{ y \in Y : \| y \| = 1 \} \)
Unit spheres help us understand the distance and geometry within the space. They provide a standard to measure distances and facilitate proofs involving norms and limits.
In the given exercise, we focus on the unit spheres of two subspaces, S_Y and S_Z, to prove a property about the combined subspace Y+Z. By examining the distance between these unit spheres, we’re able to draw conclusions about the trueness of the subspace formed by the union of Y and Z and its geometric relationship.

Distance between sets

The distance between two sets helps quantify how close the sets are to each other. It's defined as the infimum of the distances between elements from each set. Formally, the distance between sets S_Y and S_Z is:
\[\text{dist}(S_Y, S_Z) = \text{inf}\{\|y - z\| : y \in S_Y, z \in S_Z \}\]
In simpler terms, this is the minimum distance by which you could 'travel' from an element in one set to an element in the other. This concept is pivotal in the provided exercise.
If the distance \( d \) between the unit spheres of Y and Z is greater than zero (i.e., \( d > 0 \)), it implies there is a definitive 'gap' between the elements of the two subspaces. This gap helps ensure the combined subspace Y+Z is closed. Conversely, if \( d = 0 \), the subspaces Y and Z would mix together too closely, suggesting the combined subspace might not be closed.
Understanding the exact distance between sets (subspaces in this case) aids in determining the structural properties of the resulting space.