by ayushkhaitan3437

Today, I will discuss this research paper by Javed Ali, Professor of Topology and Analysis, BITS Pilani.

What exactly is a proximinal set? It is the set of elements X in which for any x\in X, you can find the nearest point(s) to it in X. More formally, for each x\in X, \exists y\in X such that d(x,y)=\inf \{d(x,K)\}.

This article says a Banach space is a complete vector space with a norm. One might find the difference between a complete metric space and a complete vector space to be minimal. However, the difference that is crucial here is that not every metric space is a vector space. For example, let (X,d) be a metric space, satisfying the relevant axioms. However, for x,y\in X, x+y not being defined is possible. However, if X is a vector space, then $x+y\in X$. Hence, every normed vector space is a metric space if one were to define \|x-y\|=d(x,y), but the converse is not necessarily true.

What is a convex set? This article says a convex set is one in which a line joining any two points in the set lies entirely inside the set. But how can a set containing points contain a line? Essentially, the the convex property implies that every point the line passes through is contained within the convex space. Convexity introduces a geometrical flavor to Banach spaces. It is difficult to imagine what such a line segment would be in the Banach space of matrices (with the norm \|\mathbf{A}-\mathbf{B}\|=\mathbf{A}-\mathbf{B}.

What is a uniformly convex set? This paper says that a uniform convex space is defined thus: \forall<\epsilon\leq 2\in\Bbb{R}, \exists \delta(\epsilon)>0 such that for \|x\|=\|y\|=1 and \|x-y\|\geq \epsilon, \frac{\|x+y\|}{2}\leq 1-\delta(\epsilon). Multiplying by -1 on both sides, we get 1- \frac{\|x+y\|}{2}\geq \delta(\epsilon). What does this actually mean? The first condition implies that x and y cannot lie in the same direction. Hence, \|x+y\|<2. As a result, we get \left\|\frac{x+y}{2}\right\|<1, or 1-\left\|\frac{x+y}{2}\right\|>0. As \delta(\epsilon)>0, and as 1-\left\|\frac{x+y}{2}\right\| is bounded, \delta(\epsilon) can be the lower bound of 1-\left\|\frac{x+y}{2}\right\|.

But what is uniform about this condition? It is the fact that \delta(\epsilon) does not change with the unit vector being considered, and depends only on \epsilon.

Now we go on to prove that every closed convex set of every uniformly convex Banach space is proximinal.

Advertisements