The mysterious linear bounded mapping

by ayushkhaitan3437

What exactly is a linear bounded mapping? The definition says T is called a linear bounded mapping if \|Tx\|/\|x\|\leq c. When you hear the word “bounded”, the first thing that strikes you is that the mappings can’t exceed a particular value. That all image points are within a finite outer covering. That, unfortunately, is not implied by a linear bounded mapping.

The image points can lie anywhere in infinite space. It is just that the vectors with norm 1 are mapped to vectors whose norms have a finite upper bound (here the upper bound is c. One may also refer to it as \|T\|). Say a is a vector which is mapped to Ta. Then sa will be mapped to sTa, where s is a scalar. This is how the mapping is both bounded (for vectors with norm 1) and linear (T(\alpha x+\beta y)=\alpha Tx+\beta Ty).

Could this concept be generalized? Of course! We could have quadratic bounded mappings: vectors with norm 1 are mapped in a similar way as linear bounded mappings, and T(\alpha x)=\alpha^2 Tx. What about T(\alpha x+\beta y)? Let r be a scalar such that \|r(\alpha x+\beta y)\|=1. Then T(\alpha x+\beta y)=\frac{1}{r^2} T(r(\alpha x+\beta y)). Similarly we could have cubic bounded mappings, etc.

Then why are linear bounded mappings so important? Why haven’t we come across quadratic bounded mappings, or the like? This is because linear bounded mappings are definitely continuous, whist not much can be said about other bounded mapppings. Proof: \|T(x-x_0)\|\leq c\|x-x_0\|\implies \|Tx-Tx_0\|\leq c\|x-x_0\| implies that linear bounded mappings are continuous. Does this definitely prove that quadratic bounded mappings are not continuous? No. All that is shown here is that we can’t use this method to prove that quadratic or other bounded mappings are continuous.