Project 3 - MAT-3313 - Tarrence Nichols

Prop: ① Recall that an operator \(A\) on a Hilbert space \(H\) is said to be bounded if there exists a scalar \(\lambda \geq 0\) such that

\[\Large\lVert{Ax}\rVert\leq\lambda\lVert{x}\rVert, \forall x \in H.\] The norm of a bounded operator of \(A\) is defined as the infimum of all scalars \(\lambda\) that satisfy the preceding inequality (i.e., nonnegativity, homogeneity, triangle inequality, etc.) Prove that the concept of norm above is really a norm on the space of operators on \(H\). Prove that \(\lVert{\cdot}\rVert\) is a norm, i.e. it satisfies the norm axioms.

Proof:

Assume that \(A\) is a bounded operator on a Hilbert space \(H\). As a bounded operator, the norm of \(A\) would be defined as follows: \[\Large\lVert{A}\rVert = \inf \{\lambda : \lambda \in \mathbb{R} \ and \ \lVert{Ax}\rVert \leq \lambda\lVert{x}\rVert, \forall x \in H\}\]

It follows that \[\Large A(0) = A(0*x) = 0 * Ax = 0\]

and, as a norm in a Hilbert space,

\[\Large \lVert{A}\rVert \geq 0, \lVert{A}\rVert = 0 \iff Ax = 0, \forall x \in H.\]

Then if, \[\Large\lVert{A}\rVert = 0, \ and \ \lVert{x}\rVert > 0 \implies \exists x > 0 \in H\]

such that \[\Large Ax = 0 \ and \ A = 0\]

\[\Large\therefore \forall \lambda, x \geq 0 \in H, \lVert{Ax}\rVert \leq \lambda\lVert{x}\rVert\] and thus, we have proven that the norm of A is really a norm on the space of operators of H.

\(\Large \blacksquare\)

Prop: ② Prove that if \(A,B\) are operators on a Hilbert space, \(H\), then \[\lVert{AB}\rVert\ \leq \lVert{A}\rVert\lVert{B}\rVert\]

Proof:

First, let \(\Large x \in H.\) Then recall that by definition \[\Large \lVert{AB}\rVert = \sup_{\lVert{x}\rVert=1} \lVert{ABx}\rVert.\]

Then, by the homogeneity property of norms, \[\Large \lVert{ABx}\rVert \leq \lVert{A}\rVert\lVert{Bx}\rVert\] and, \[\Large \lVert{A}\rVert\lVert{Bx}\rVert \leq \lVert{A}\rVert\lVert{B}\rVert\lVert{x}\rVert.\] Since \(\Large \lVert{x}\rVert = 1\),

\[\Large \lVert{ABx}\rVert\ \leq \lVert{A}\rVert\lVert{B}\rVert \implies \sup_{\lVert{x}\rVert = 1}\lVert{ABx}\rVert \leq \lVert{A}\rVert\lVert{B}\rVert.\] And because \(\Large \lVert{x}\rVert = 1\) we can say, \[\Large \lVert{AB}\rVert \leq \lVert{A}\rVert\lVert{B}\rVert.\]

\(\Large \blacksquare\)

Prop: ③ Let A be a linear operator on a Hilbert space,\(H\). Prove that the following conditions are equivalent:

(a) A is bounded.

(b) There does not exist a sequence \(\Large (x_{n})_{n \in \mathbb{N}}\) in the unit sphere of \(H\) such that \[\Large \lVert{Ax_{n}}\rVert \to \infty \ as \ n \to \infty.\]

Proof:

Assume, for the sake of contradiction, that there is a sequence \(\Large (x_{n})_{n \in \mathbb{N}}\) in the unit sphere of \(H\).

Then we have, \[\Large\lVert{x_{n}}\rVert = 1, \forall n \ and \ \lVert{Ax_{n}}\rVert \to \infty \ as \ n \to \infty.\]

If \(\Large\lVert{Ax_{n}}\rVert \to \infty\,\) then \(\Large\exists \ M > 0\) such that \(\Large\lVert{Ax_{n}}\rVert > M, \forall n.\)

A linear operator must satisfy the following property: \(\Large A(cx) = cAx, \forall x \in H \ and \ \forall c \in \mathbb{C}\)

Applying this property, we can assign \[\Large y_{n} = \frac{1}{\lVert{Ax_{n}}\rVert}x_{n}.\]

Then, if we apply A, we get: \[\Large Ay_{n} = A(\frac{1}{\lVert{Ax_{n}}\rVert}x_{n}) = \frac{1}{\lVert{Ax_{n}}\rVert}Ax_{n}\]

Since \(\Large \lVert{Ax_{n}}\rVert \to \infty \ as \ n \to \infty\), we know that \(\Large \frac{1}{\lVert{Ax_{n}}\rVert} \to 0 \ as \ n \to \infty.\)

Then, \[\Large \lVert{Ay_{n}}\rVert = \frac{\lVert{Ax_{n}}\rVert}{\lVert{Ax_{n}}\rVert^{2}} = \frac{1}{\lVert{Ax_{n}}\rVert} \to 0 \ as \ n \to \infty.\]

Given that \(\Large \lVert{Ay_{n}}\rVert \to 0\), the sequence \((Ay_{n})\) is bounded in \(H\). This contradicts the original assertion that \(\Large \lVert{Ax_{n}}\rVert \to \infty.\)

\(\Large \blacksquare\)