Orthogonality and Bounded Linear Maps

Closed Subspaces and Orthogonality

Proposition

In any finite dimensional inner product space, every subspace is closed.

Proof Suppose is a finite dimensional inner product space and is a subspace of . Then is finite dimensional. Since every finite dimensional normed space is complete, it follows that is closed in .

Remark

However, this is not true in general for infinite dimensional spaces. For example, in an infinite dimensional Hilbert space with basis , the subspace , which is the space of all finite linear combinations of the basis vectors, is not closed. To see this, consider the sequence , which converges to , which is not in .

Lemma

Suppose is a closed subspace of a Hilbert space , and . Then

• There exists a unique element which is closest to , in the sense that .
• is perpendicular to . i.e. for all .

Proposition

If is a closed subspace of a Hilbert space , then is a direct sum of and its orthogonal:

Proposition

Suppose is a subspace of an (possibly infinite dimensional) inner product space, then is always closed, and , where denotes the closure of .

Proof For all that converges to , we claim that . Note that for all and some , we have so converges to . Note that since each , it converges to . Hence, for all . This shows that . So is closed. Thus is closed, and . Now for any , since is closed, , so we can write , where and . For any , there holds As , . So for all , which implies , hence . Therefore, , which implies that .

Corollary

Suppose is a subspace of an (infinite dimensional) inner product space, then is dense in iff .

Proof This is straightforward from the previous proposition.

e.g. Consider the Hilbert space , has an orthonormal basis , where

Bounded Linear Operators

Bounded Linear Operator

Suppose and are normed spaces, then a linear operator is bounded if there exists such that In such case, the operator norm of is defined as

Lemma

For any bounded linear map between two Hilbert spaces, we have

Proof By Cauchy-Schwarz inequality, for all with , we have and so . Conversely, hence, we have , which implies that .

Proposition

A linear operator is bounded if and only if it is continuous w.r.t the norm topology.

Proof If is bounded, then for all in , we have for all . Taking the limit as , we have as , so is continuous. Conversely, if is continuous, assume that is unbounded. Then for each , there exists some such that . Let , then , so . Since is continuous at , we have , however, , which is a contradiction. Therefore, must be bounded.

Remark

There are many types of continuity. In fact, we have the following chain of implications: