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Chapter 2: Problem 20

If \(\psi(\cdot, \cdot)\) is a semi-inner product on a linear space \(X\) and if $$ Y=\\{x \in X: \psi(x, x)=0\\} $$ then show that \(Y\) is a subspace of \(X\) and $$ \|x+Y\|_{*}:=\sqrt{\psi(x, x)}, \quad x+Y \in X / Y $$ defines a norm on \(X / Y\).

Short Answer

Expert verified
Y is a subspace of X. The norm \(\|x + Y\|_* = \sqrt{\psi(x, x)}\) is well-defined and satisfies positivity, homogeneity, and the triangle inequality.

Step by step solution

01

Prove Y is a Subspace

To show that Y is a subspace of X, verify that Y is closed under addition and scalar multiplication. Take any two elements \(y_1, y_2 \in Y\). Since \(y_1, y_2 \in Y\), we have \(\psi(y_1, y_1) = 0\) and \(\psi(y_2, y_2) = 0\). Use the properties of semi-inner products to show \(\psi(y_1 + y_2, y_1 + y_2) = 0\), implying \(y_1 + y_2 \in Y\). Next, for any scalar \(a\) and element \(y \in Y\), show \(\psi(ay, ay) = 0\) so \(ay \in Y\). Finally, check that 0 is in Y because \(\psi(0, 0) = 0\).
02

Define the Quotient Norm

Define the quotient space norm \(\|x + Y\|_* = \sqrt{\psi(x, x)}\). Ensure this norm is well-defined. If \(x_1 + Y = x_2 + Y\), then \(x_1 - x_2 \in Y\). This implies \(\psi(x_1 - x_2, x_1 - x_2) = 0\). Then show \(\psi(x_1, x_1) = \psi(x_2, x_2)\), guaranteeing \(\|x_1 + Y\|_* = \|x_2 + Y\|_*\).
03

Prove Norm Properties - Positivity

Show that the quotient norm is positive. If \(x + Y eq x + Y\), \(\psi(x, x) > 0\). Therefore, \(\|x + Y\|_* = \sqrt{\psi(x, x)} > 0\). If \(x + Y = Y\), the norm is zero: \(\|Y\|_* = 0\), since \(\psi(x, x) = 0\).
04

Prove Norm Properties - Homogeneity

For any scalar \(a\) and \(x \in X\), show \(\|ax + Y\|_* = |a| \|x + Y\|_*\). Calculate \(\|ax + Y\|_* = \sqrt{\psi(ax, ax)} = \sqrt{|a|^2 \psi(x, x)} = |a|\sqrt{\psi(x, x)} = |a| \|x + Y\|_*\).
05

Prove Norm Properties - Triangle Inequality

For any \(x, y \in X\), show \(\|x + y + Y\|_* \leq \|x + Y\|_* + \|y + Y\|_*\). This follows from the subadditivity property of the semi-inner product: \(\psi(x + y, x + y) \leq (\sqrt{\psi(x, x)} + \sqrt{\psi(y, y)})^2\). Taking square roots on both sides yields the desired inequality.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Semi-Inner Product
A semi-inner product is a function \(\psi(x, y)\) that assigns a complex number to each pair of vectors in a linear space \(\ X\). While similar to an inner product, a semi-inner product need not satisfy all the properties of an inner product. \(\psi(x, y)\) must be:
  • Linearly dependent on its first argument: \(\psi(ax + by, z) = a\psi(x, z) + b\psi(y, z)\)
  • Conjugate symmetric: \(\psi(x, y) = \overline{\psi(y, x)}\)
  • Positive semi-definite: \(\psi(x, x) \geq 0\)
We use semi-inner products to help define properties and structures in linear spaces that are less restrictive than those requiring a full inner product. This flexibility is useful in functional analysis and other mathematical contexts.
Quotient Space
Given a linear space \(\ X\) and a subspace \(\ Y\), the quotient space \(\ X / Y\) consists of all cosets of the form \(\ x + Y\) where \(\ x\) is in \(\ X\).
In this space, we consider two elements equivalent if their difference lies in \(\ Y\). So, \(\ x_1 + Y = x_2 + Y \iff x_1 - x_2 \in Y\).
Defining a norm on the quotient space requires showing that this construction is well-defined and respects the equivalence relation. The norm is given as \(\|x + Y\|_* = \sqrt{\psi(x, x)}\) and is foundational for analysis in quotient spaces.
Norm Properties
A norm must satisfy three properties: positivity, homogeneity, and the triangle inequality.
  • Positivity: For any \(\ x + Y\), \(\|x + Y\|_* = \sqrt{\psi(x, x)}\) is always non-negative, and is zero if and only if \(\ x \in Y\).
  • Homogeneity: For any scalar \(\ a\) and vector \(\ x\), \(\|ax + Y\|_* = |a|\|x + Y\|_*\). Squaring both sides and using the properties of \(\ \psi \), this follows directly.
  • Triangle Inequality: For any vectors \(\ x\) and \(\ y\), \(\|x + y + Y\|_*\leq \|x +Y \|_* + \|y + Y\|_*\). This follows from semi-inner product subadditivity.
Proving these properties ensures that the defined function behaves as a norm and can be used in analysis.
Subspaces
A subspace \(\Y\) of a linear space \(\X\) is a subset that is also a linear space. To prove \(\ Y \) is a subspace, we need to check:
  • Closure under addition: For any \(\ y_1, y_2 \in Y\), \(\ y_1 + y_2 \in Y\). This follows if \(\ \psi(y_1 + y_2, y_1 + y_2) = 0 \).
  • Closure under scalar multiplication: For any scalar \(\ a\) and \(\ y \in Y\), \(\ a y \in Y\) if \(\ \psi(ay, ay) = 0\).
  • Zero vector: The zero vector, \(\ 0 \), must be in \(\ Y\), which it is if \(\ \psi(0, 0) = 0\).
These conditions ensure that \(\ Y\) behaves like a vector space within \(\ X\).
Triangle Inequality
The triangle inequality is essential in defining a norm. For any vectors \(\ x\) and \(\ y\) in a normed space, the inequality states:
  • \

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Most popular questions from this chapter

Let \(X\) be an inner product space and \(\left(x_{n}\right)\) be a sequence in \(X\). For \(x \in X\), show that \(\left\|x_{n}-x\right\| \rightarrow 0\) as \(n \rightarrow \infty\) whenever \(\left\|x_{n}\right\| \rightarrow\|x\|\) and \(\left\langle x_{n}, x\right\rangle \rightarrow\|x\|^{2}\) as \(n \rightarrow \infty\).

Give an example to illustrate that if \(X_{1}\) and \(X_{2}\) are closed subspaces of a normed linear space \(X\), then $$ X_{1}+X_{2}:=\left\\{x_{1}+x_{2}: x_{1} \in X_{1}, x_{2} \in X_{2}\right\\} $$ need not be a closed subspace.

Let \(t_{1}, \ldots, t_{n}\) be distinct points in \([a, b] .\) For \(x \in C[a, b]\), let $$ \nu_{p}=\left\\{\begin{array}{ll} \left(\sum_{j=1}^{n}\left|x\left(t_{j}\right)\right|^{p}\right)^{1 / p} & \text { if } 1 \leq p<\infty \\ \max \left\\{\left|x\left(t_{j}\right)\right|: j=1, \ldots, n\right\\} & \text { if } p=\infty \end{array}\right. $$

Suppose \(X\) is a linear space and \(E \subseteq X\) is such that it is (a) convex, i.e., for every \(x, y \in E, \lambda x+(1-\lambda) y \in E\) for \(0<\lambda<1\), (b) balanced, i.e., \(\alpha x \in E\) whenever \(x \in E\) and \(|\alpha| \leq 1 ;\) (c) absorbing, i.e., \(\forall x \in X, \exists \lambda>0\) such that \(\lambda x \in E\), and (d) there is no subspace \(Y \neq\\{0\\}\) such that \(Y \subseteq E\). Show that \(x \mapsto\|x\|_{E}:=\inf \left\\{\lambda>0: \lambda^{-1} x \in E\right\\}\) is a norm on \(X\).

Let \(X\) be a linear space, \(Y\) be a normed linear space, and \(A: X \rightarrow Y\) be a linear operator Suppose \(y_{0} \in R(A)\) and \(r>0\) are such that the equation \(A x=y\) has a solution for every \(y \in B_{Y}\left(y_{0}, r\right)\). Show that \(A x=y\) has a solution for every \(y \in Y\).
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