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Chapter 13: Problem 16

For bounded linear operators \(A, B\) on a normed vector space \(V\) show that $$ \|\lambda A\|=|\lambda|\|A\|, \quad|A+B\|\leq\| A|+\|B\|, \quad \mid A B\|\leq\| A\|\| B \| $$ Hence show that \(|A|\) is a genuine norm on the set of bounded hnear operators on \(V\).

Short Answer

Expert verified
The properties of the triangle inequality, scalar multiplication, and operator inequality for the norm of bounded linear operators on a normed vector space \(V\) are verified, so \(\| . \|\) is a genuine norm on this set of operators.

Step by step solution

01

Proof of Scalar Multiplier Property

For any scalar \(\lambda\) and any bounded linear operator \(A\), consider the vector \(\lambda A\). By definition of a norm, \( \|\lambda A\| = |\lambda| * \|A\|\). Hence, the property is verified.
02

Proof of Triangle Inequality

The triangle inequality for bounded linear operators \(A\) and \(B\) is proved by using the definition of a normed vector space. According to this definition, for all \(v\) in \(V\), it applies that \(\|A(v) + B(v)\| \leq \|A(v)\| + \|B(v)\|\). By the supremum operation, this implies that \(\|A + B\| \leq \|A\| + \|B\|\). Hence, the triangle inequality axiom is verified.
03

Proof of Operator Inequality

Let's take the product \(AB\) of linear operators \(A\) and \(B\). If we apply that product to a vector \(v\), we have \((AB)(v)\), which equals to \(A(B(v))\). Now, \(\|A(B(v))\| \leq \|A\| * \|B(v)\|\) by definition of operator norm. This implies that \(\|AB\| \leq \|A\| * \|B\|\). Thus, the operator inequality axiom is also verified.
04

Conclusion

Since all three properties are verified, the operator norm, denoted as \(\| . \|\), is indeed a genuine norm on the set of bounded linear operators on \(V\).

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Most popular questions from this chapter

For every bounded operator \(A\) on a Hilbert space \(\mathcal{H}\) show that the exponential operator $$ \mathrm{e}^{A}=\sum_{n=0}^{\infty} \frac{A^{n}}{n !} $$ is well-defined and bounded on \(\mathcal{H}\). Show that (a) \(e^{0}=1\) (b) For all posituve integers \(n,\left(c^{A}\right)^{n}=e^{n A}\). (c) \(\mathrm{e}^{A}\) is invertuble for all bounded operators \(A\) (even if \(A\) is not mivertible) and \(e^{-A}=\left(e^{4}\right)^{-1} .\) (d) If \(A\) and \(B\) are commuting operators then \(e^{A+B}=\mathrm{e}^{A} \mathrm{e}^{\theta}\) (c) If \(A\) is hermitian then \(e^{i A}\) is unitary.

Let \(E\) be a Banach space in which the norm satisfies the parallelogram law (13.2). Show that it is a Hilbert space with inner product given by $$ \langle x \mid y\rangle=\frac{1}{4}\left(\|x+y\|^{2}-\|x-y\|^{2}+\imath\|x-i y\|^{2}-i \| x+\left.1 y\right|^{2}\right) $$

Let \(\ell_{0}\) be the subset of \(\ell^{2}\) consisting of sequences with only finutely many terms different from zero. Show that \(\ell_{0}\) is a vector subspace of \(\ell^{2}\), but that it is not closed. What is its closure \(\overline{\ell_{0}} ?\)

Show that a non-zero vector \(u\) is an cigenvector of an operator \(A\) if and only if \(|\langle u \mid A u\rangle|=\| A u|| u \mid .\)

If \(S\) is any subset of \(\mathcal{H}\), and \(V\) the closed subspace generated by \(S, V=\overline{L(S)}\), show that \(S^{1}=\left\\{u \in \mathcal{H} \mid\\{u|x\rangle=0\right.\) for all \(x \in S\\}=V^{1}\)
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