Is the Uniqueness of Minimum Norm Elements in $L^2$ Spaces a Consequence of Strict Convexity?

Question

I'm delving into the concept of strict convexity within $L^2$ spaces and its implications for the existence and uniqueness of elements with minimum $L^2$ norms. Given $L^2(\mathbb{R})$'s role as a pivotal space in functional analysis, particularly in optimization and approximation theory, I'm curious about how its structural properties influence such uniqueness.

Mathematical Definition of Strict Convexity: A normed vector space $V$ is said to be strictly convex if, for any two distinct points $x, y \in V$ with $x \neq y$ and any $t \in (0, 1)$, the following inequality holds: $$ \|tx + (1-t)y\| < t\|x\| + (1-t)\|y\|. $$ This definition implies that the line segment connecting any two points in the space lies entirely within the unit ball, except for the endpoints.

Question:

Given the strict convexity of $L^2(\mathbb{R})$, is it correct to assert that every non-empty subset of $L^2(\mathbb{R})$ necessarily contains a unique element with the minimum $L^2$ norm? Specifically, does strict convexity guarantee that for any subset of $L^2(\mathbb{R})$, there cannot be two distinct elements with the same minimum norm?

I'm looking for a mathematical explanation that connects the dots between the strict convexity of $L^2$ spaces and the principle ensuring the uniqueness of minimum norm elements. Any insights, proofs, or references to literature that elaborate on this relationship would be highly appreciated.

Answer 1

As pointed out by BigbearZzz in the comments, no, a general subset of $L^2(\Bbb{R})$ (or indeed, any non-trivial normed space) need not have a unique point of minimum norm, as we can take any non-zero point $x$ in the space, and $\{x, -x\}$ has two distinct points of minimum norm.

The failure of this example is that, while the midpoint of the two points has strictly lesser norm, it is not in the set, and thus does not contradict the minimality of the two points you began with.

If a set is convex (or even just midpoint convex), this is a different story. You can use strict convexity of the norm to (very easily) show that there is at most one point of minimum norm. Indeed, there is a converse to this too: if every convex set attains at most one point of minimum norm, then the normed linear space is strictly convex. This is proven by simply considering the line segment between any two points of the sphere. To summarise,

Theorem 1. A normed linear space $(X, \| \cdot \|)$ is strictly convex if and only if, for all convex sets $C \subseteq X$, there exists at most one point of minimum norm in $C$.

Guaranteeing the existence of a point of minimum norm is another story. We can also construct examples of sets, even convex ones, where no such point exists. Trivially, we could take the empty set. Slightly less trivially, we could consider an open ray pointing at the origin, i.e. $\{\lambda x : \lambda > 1\}$, where $x \neq 0$. This ray does not have a point of minimum norm.

Clearly, in the preceding example, the failure was with the set not being closed; if we had simply included $x$ in the set, then it would be the (unique) point of minimum norm. Will closed sets then guarantee us a point of minimum norm?

No, at least, not in $L^2(\Bbb{R})$. We can choose an infinite orthonormal set $\{e_1, e_2, \ldots\}$. We can then form a set: $$\left\{\left(1 + \frac{1}{1} \right)e_1, \left(1 + \frac{1}{2} \right)e_2, \left(1 + \frac{1}{3} \right)e_3, \ldots\right\},$$ which once again has no minimum norm; it admits norms of the form $1 + \frac{1}{n}$, but no norm of $1$. The set is closed because the distance between any pair of points is greater than $\sqrt{2}$, so any convergent sequences from these points are eventually constant.

Of course, such a set is not convex. If we require that a set is closed, convex, and non-empty, at least in $L^2(\Bbb{R})$, then at least one point of minimum norm must exist! In fact, the following theorem holds true:

Theorem 2. A Banach space $(X, \| \cdot \|)$ is reflexive if and only if every closed, non-empty, convex set admits at least one point of minimum norm.

"Reflexive" is a functional analysis term, with multiple equivalent definitions, usually defined in terms of the natural embedding of $X$ into its second dual $X^{**}$ is surjective, or in terms of the unit ball being weakly compact. Such spaces are necessarily Banach spaces, though as far as I'm aware, the restriction to Banach spaces is necessary for the above theorem to work. All Hilbert spaces, like $L^2(\Bbb{R})$ are reflexive.

So, combining these theorems, the spaces for which closed, non-empty, convex sets contain a unique point of minimum norm, are precisely the reflexive, strictly convex normed spaces, including $L^2(\Bbb{R})$.