Functional Analysis 9 -- Main Theorems in Functional Analysis 1

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

Principle Uniform Boundedness

(Banach-Steinhaus Theorem), Steinhaus is the mentor of Banach.

A family of bounded linear operator is uniformly bounded iff it is pointwise bounded.

Let \(X\) be a Banach space and \(Y\) be a NVS. Let \(F \subset B(X,Y).\) If \(\sup {||Tx||_Y: T\in F} < \infty,\) for every \(x \in X\). Then\(\sup \{||T||_{B(X,Y): T \in F} \} < \infty.\)

Rmk: The theorem tells us that \[ ||Tx||_Y \le C_x < \infty \forall x \in X \]

\[ \Rightarrow ||T||_{B(X,Y)} \le C < \infty \] Here \(C_x\) dependent on \(x\), \(C\) independent of everything.

proof: Consider \(G_n = \{x \in X: ||Tx||_Y \le n, \forall T \in F\}.\) By assumption, \(X = \cup G_n.\) As \(G_n = \cap_T T^{-1}(\bar{B_Y(0,n)}),\) we see that \(G_n\) is closed.

But \(X\) is Banach, by BCT, \(\exists n_0 \in \mathbb{N},\) s.t. \(G^o_{n_0} \neq \phi.\) Pick \(V^o_X(x_0, r_0) \subset G_{n_0}.\) For any \(x \in B_X^o(0,r_0),\) we have \(x_0 + x \in B^o_X(x_0, r_0).\) \[ \Rightarrow ||T(x+x_0)||_Y \le n_0 \] \[ \Rightarrow ||Tx||_Y \le ||T(x+x_0)|| + ||Tx_0|| \le 2n_0 \] By linearity, \(||T|| \le \frac{2n_0}{r_0}.\) where the choice of \(n_0, r_0\) is not dependent on \(x.\) Thus we get the uniform bound. \(\Box\)

对于一个赋范空间，要知道其范数信息，通常考虑在圆心的球上的\(T.\)

Corollary: Let \(X,Y\) be Banach spaces and \(\{T_n\} \subset B(X,Y),\) TFAE:

1. \(\exists T \in B(X,Y), T_nx \rightarrow Tx, \forall x \in X.\)

2. \(\{T_n x\} \subset Y\) is convergent \(\forall x \in X.\)

3. \(\exists M < \infty\) and dense \(Z \subset X\) s.t. \(||T_n|| \le M\) and \(\{T_nz\}\) is convergent \(\forall z \in Z.\)

proof:[[TODO]]

Definition: \(\{T_n\} \subset B(X,Y)\) s.t. \(T_n x \rightarrow Tx, \forall x \in X \Leftrightarrow T_n \rightarrow T\) is SOT(Strong Operator Topology).

Rmk: SOT is weaker than operator topology.

Open Mapping Theorem

Let \(X,Y\) be Banach spaces. \(T \in B(X,Y).\) If \(T\) is onto, then \(T\) is an open mapping (from open set to open set).

Definition: Open Mapping Let \(A,B\) be topological spaces, \(f:A\rightarrow B\) is open if \(f(O)\) is open in \(B\) whenever \(O\) is open in \(A.\)

\(f\) is continuous iff \(f^{-1}(O)\) is open in \(A\) whenever \(O\) is open in \(B.\)

Corollary: Inverse Mapping Theorem

If \(T\) is a 1-1 and onto mapping, then \(T\) is a Banach space isomorphism.

Remark: For \(T\) onto, one may consider \(X/\ker T \cong Y.\)

Notation: \(B_X^{\circ} = B_X^{\circ}(0,1), B_X^{\circ}(r) = B_X^{\circ}(0,r).\)

Lemma:

If dist(\(y, T(B^{\circ}_X(0, n_0))\)) < \(\epsilon, \forall y \in B_{Y}^{\circ},\) then \(B_Y^{\circ} \subset T(B_X^{\circ}(0, \frac{n_0}{1-\epsilon}))\).$

proof:

We want to solve given \(y \in B_Y^o,\) the equation \(y = Tx\) for \(x \in B_X^o(\frac{n_0}{1-\epsilon}).\) We shall construct \(x_1, x_2, ... \in X\) s.t. \(x_k \in B^o_X(n_0, \epsilon^k)\) and to prove that \(y = Tx_1 + Tx_2+...\) In this way, the series \(\sum x_i\) will absolutely converge as \(\sum ||x_i|| < \sum n_0\epsilon^k = \frac{n_0}{1-\epsilon}.\) Then \(\sum x_i\) converges to \(x \in X.\)

Thus \(||x|| \le \sum ||x_i|| \le \frac{n_0}{1-\epsilon}, i.e. x \in B^o_X(\frac{n_0}{1-\epsilon}).\) Therefore, \(y = Tx.\)

Next we show how to construct such \(x_1, x_2, ... \in X.\) By assumption we have given $ y B_Y^o,$ dist(\(y, T(B_X^o(n_0))\))< \(\epsilon.\)

Then \(\exists x_0 \in B_X^o(n_0)\) s.t. \(||y - Tx_0||< \epsilon.\) So \(\frac{y-Tx_0}{\epsilon} \in B_Y^o.\) Thus \(\exists x_1 \in B_X^o(n_0),\) s.t. \(||\frac{y-Tx_0}{\epsilon} - Tx_1|| < \epsilon.\) That is, \(||y - Tx_0 - T\tilde{x}_1|| < \epsilon^2\) for \(\tilde{x}_1 = \epsilon x_1.\)

Continue in this way. Assume that \(x_0, \tilde{x_1}, \tilde{x_2}, ... \in X\) are chosen s.t. \(||\tilde{x}_j|| < n_0 \epsilon^j\) and \(||y-Tx_0 - ... - T\tilde{x}_{k}|| < \epsilon^{k}.\)

Then we have \(\frac{y - Tx_0-...-T\tilde{x}_k}{\epsilon^{k+1}} \in B_Y^o.\) So by assumption we have \(\exists x_{k+1} \in B^o_X(n_0),\) s.t. \[ ||\frac{y-Tx_0-...-T\tilde{x}_k}{\epsilon^{k}} - Tx_{k+1}||< \epsilon. \] Hence \[ ||y-Tx_0-...-T\tilde{x}_k - T\epsilon^{k}x_{k+1}||< \epsilon^{k+1}. \] Denote \(\tilde{x}_{k+1} = \epsilon^k x_{k+1}.\) It satisfies the described condition. \(\Box\)

Rmk: This lemma allows us to prove \(\bar{T(B_X^o(n_0))}\) has nonempty interior. \(\Rightarrow T(B_X^o(n_0))\) has nonempty interior, if the NVS \(X\) is complete.

Definition: Let \(f: S_1 \rightarrow S_2\) be a function between two sets \(S_1, S_2.\) Then the graph of \(f\) is defined as \(\Gamma = \{(x, f(x)), x \in S_1\}.\)

Theorem: Closed Graph Theorem

Let \(X,Y\) be Banach spaces and \(T \in L(X,Y).\) Then \(T\in B(X,Y)\) iff the graph of \(T\), \(\Gamma_T\) is closed.

Remark: \(\Gamma_T\) is a subspace of \(X \times Y.\) Here \(X \times Y\) is a Banach space w.r.t \(||(x,y)|| = ||x|| + ||y||.\)

proof: \(\Rightarrow\) exercise \(\Leftarrow\) Suppose that \(\Gamma_T\) is closed. So \(\Gamma_T \le X \times Y\) is a closed subspace. Consider \(pr_1, pr_2\) as two maps projecting \((x,Tx) \in \Gamma_T\) to \(x\), \(Tx\). Thus \(pr_1, pr_2\) are two b.l.o between Banach spaces(\(||pr_1|| = 1, ||pr_2|| = 1\)). Also, \(pr_1\) is also 1-1 and onto. So \(pr_1^{-1}\) is a bounded linear operator. Hence, \(T = pr_2 \circ pr_1^{-1}.\) \(\Box\)

Key Remark: Big Four Banach-Steinhaus Theorem, Open Mapping Theorem, Inverse Mapping Theorem, Closed Graph Theorem are all equivalent!