Fruits of procrastination

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 $\epsilon$ .