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Functional Analysis 2 -- Bounded Linear Operator

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Functional Analysis 2 -- Bounded Linear Operator

This is course note from Siran Li-Shanghai Jiao Tong University.

Definition: Bounded Linear Operator(BLO):

Let \(X, Y\) be NVS. Let \(T: X \rightarrow Y\) be a linear operator. Define norm of \(T\) by \(||T|| := \sup\{||Tx||_{Y} : ||x||_{X} \le 1\}\). \(T\) is called bounded if \(||T|| < \infty\).

Theorem 2.1

Let \(X, Y\) be NVS. Then \(B(X,T) := \{T: X \rightarrow Y \mid T \ is \ BLO\}\) is a NVS when applied with the operator norm. In addition, if \(Y\) is a Banach space, then so is \((B(X,Y), ||\cdot||)\).

Notation: \[ \mathcal{L}(X,Y) := \{T: X \rightarrow Y \mid T \ is \ linear\}. \] \[ B(X,Y) := \{T: X \rightarrow Y \mid T \ is \ BLO\}. \] If \(X=Y\), \(B(X) = B(X,X)\).

Theorem 2.2

Let X, Y be NVS. TFAE(The Following Are Equivalent):

  1. \(T: X \rightarrow Y\) is a BLO.
  2. \(T: X \rightarrow Y\) is linear and continuous. (2') \(T: X \rightarrow Y\) is linear and continuous at 0.

proof:

\((2)\Rightarrow(2')\) by linearity.

\((2') \Rightarrow (1)\) By continuity at 0, \(\forall \epsilon > 0, \exists \delta >0, s.t. ||x|| \le \delta \Rightarrow ||Tx|| \le \epsilon.\) Thus, for any \(y \in X\), \(||Ty|| = \frac{||y||}{\delta}||T(\frac{\delta y}{||y||})|| \le \frac{\delta}{\epsilon}||y||.\) \(\Rightarrow ||T|| \le \frac{\epsilon}{\delta}\) is bounded.

\((1)\Rightarrow(2')\) \(\forall \epsilon > 0, ||Tx|| \le ||T||||x|| \le \epsilon\) whenever \(||x|| \le \frac{\epsilon}{||T||}\). \(\Box\)

Proposition 2.3:

If \(X, Y, Z\) are NVS, \(T: X \rightarrow Y, S: Y \rightarrow Z\) are BLO. Then \(S\circ C \in B(X,Z).\)

In particular, if \(X = Y = Z, T,S \in B(X)\), we have \(||TS||\le||T||||S||.\)

Definition: Banach Algebra

Let \(A\) be a Banach space \(s.t. \exists \ \cdot: A \times A \rightarrow A, s.t. \forall \lambda \in \mathbb{K}, x,y \in A. (\lambda x)\cdot y = x\cdot (\lambda y) = \lambda(x \cdot y)\) and \(s.t. ||x \cdot y|| \le ||x|| ||y||.\) Then \(A\) is called a Banach algebra.

Example: Let \(X\) be a Banach space, then \(B(X)\) is a Banach algebra.

Examples of BLO([[TODO]):

Definition: Isormorphism

Let \(X,Y\) be NVS.

  1. \(T: X \rightarrow Y\) is a vector space isormorphism iff \(T \in \mathcal{L}(X,Y), T^{-1}\) exists and \(T^{-1} \in \mathcal{L}(X,Y).\)
  2. \(T: X \rightarrow Y\) is a NVS isormorphism iff \(T \in B(X,Y), T^{-1}\) exists and \(T^{-1} \in B(X,Y).\)
  3. \(Aut(X) := \{T \in B(X) \mid T \text{ is a NVS isormorphism}\}\) is called Automorphism.

Theorem 2.4

If \(X\) is a Banach space, then \(Aut(X) \subset B(X).\)

proof:([[TODO]])

Fact: Evert finite dimensional (normed) vector space \(V\) over \(\mathbb{K}\) is (normed) vector space isormorphism to \(\mathbb{K}^d\) with \(d = \dim_k V.\)

This fact classifies finite dimensional (normed) vector space!!!

proof:([[TODO]])

Theorem 2.5

Let \(X\) be a NVS, TFAE:

  1. \(X\) is finite dimensional.
  2. \(B_1 = \{x \in X: ||x|| \le 1\}\) is compact.

Remark: (2) is equivilent to "a set in \(X\) is compact iff \(X\) is closed and bounded".

proof:([[TODO]])

Remark: In metric spaces, topology is compeletely characterized by convergence of subsequence.