Following the idea due to Brezis in his note that we had already mentioned once, it seems that we can show that the following equation
in with has no positive -solution. Here we require .
Indeed, to see this, we first denote
with . Clearly the function is monotone increasing in its domain. Using this, we rewrite our equation as follows
I want to propose an alternative way to calculate the Jacobian of the stereographic projection . In Cartesian coordinates on the sphere and on the plane, the projection and its inverse are given by the formulas
It is well-known that the Jacobian of the stereographic projection is
The way to calculate its Jacobian is to compare the ratio of volumes. First pick two arbitrary points and denote and .
The Euclidean distance between and is
Of interest in this note is the simplicity of the first eigenvalue of the following problem
where and satisfy
with . A simple variational argument shows that exists and can be characterized by
It is well-known that every continuous functions admits antiderivative. In this note, we show how to prove existence of antiderivative of some discontinuous functions.
A typical example if the following function
By taking to different sequences and we immediately see that is discontinuous at . However, we will show that admits as its antiderivative.
To this end, we first consider the following function
First we show that is differentiable. Clearly whenever , we obtain
The aim of this note is to derive some connections between topologies of normed spaces in terms of the equivalency of norms and the convergence of sequences.
Topological space and its topology: First, we start with a topological space, call . Its topology, say is the collection of subsets of which satisfies certain conditions. In the literature, each member of the collection is called an open set.
Regarding to topologies we have the following basic facts:
- Given two topologies and on , we say that is stronger (or finer or richer) than if .
- Given a sequence in , we say that converges to in topology of if for any neighborhood of , there exists some large number such that for all . (Here by the neighborhood of we mean that there exists an open set of , i.e. is a member of the topology , such that .)
The key ingredient to compare topologies is to make use of the identity map. In the following part, we state a result which shall be used frequently in this note.
Topologies under the identity map: Given two topologies and on a topological space , we are interested in comparing and in terms of the identity map .
Lemma 1. The identity map is continuous if and only if is stronger than .
The proof is relatively easy. Indeed, if the map is continuous, then the preimage of any is also a member of which immediately implies that includes .
Having Lemma 1 in hand, we now try to compare topologies using norms.