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Chapter 0: Problem 24

Suppose \(\mathfrak{Q}\) is a vector space. Let \(s(f, g)\) be a sesquilinear form on \(\mathfrak{Q}\) and \(q(f)=s(f, f)\) the associated quadratic form. Show that the Cauchy-Schwarz inequality \((0.56)\) \(|s(f, g)| \leq q(f)^{1 / 2} q(g)^{1 / 2}\) holds if \(q(f) \geq 0\).

Short Answer

Expert verified
The Cauchy-Schwarz inequality holds if the quadratic form \(q(f)\) is non-negative.

Step by step solution

01

Understand the Definitions

A sesquilinear form \(s(f, g)\) is a complex function that is linear in one argument and conjugate linear in the other. The associated quadratic form \(q(f) = s(f, f)\) arises from applying the form to the same element twice.
02

Analyze the Cauchy-Schwarz Inequality

The inequality \(|s(f, g)| \leq q(f)^{1/2} q(g)^{1/2}\) is what we want to show is true. This relates the absolute value of the sesquilinear form to the geometric mean of the values it takes on single elements.
03

Establish a Case with Zero Form

Consider first the case when \(q(f) = 0\). By definition, \(q(f) = |s(f, f)| = 0\) implies \(s(f, g) = 0\) for any \(g\), thus \(|s(f, g)| = 0\) which is less than or equal to \(q(f)^{1/2} q(g)^{1/2} = 0\) when \(q(g) \geq 0\).
04

Assume Non-zero Quadratic Forms

Assume both \(q(f) > 0\) and \(q(g) > 0\). This means the sesquilinear form is non-degenerate for \(f\) and \(g\). We will apply properties of the sesquilinear form to prove the inequality.
05

Use Linear and Conjugate Linear Properties

For \(t \, \in \, \mathbb{C}\), consider the form \[ q(f + tg) = s(f + tg, f + tg) = s(f, f) + t s(f, g) + \overline{t} s(g, f) + |t|^2 s(g, g) \].
06

Apply the Non-negativity Condition

Since \(q(f + tg) \geq 0\) for any \(t\), we focus on the real and imaginary parts separately, ensuring that this expression is non-negative.
07

Derive Quadratic Form Equation

Write the expression \(q(f + tg)\) as a real quadratic in terms of real and imaginary parts of \(t\), ensuring non-negativity leads to real roots if discriminant \(|s(f,g)|^2 - q(f) q(g)\) is negative or zero, hence \(|s(f, g)| \leq q(f)^{1/2} q(g)^{1/2}\).

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Key Concepts

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

Sesquilinear Form
A sesquilinear form, denoted as \( s(f, g) \), is a type of mathematical expression defined on vector spaces. It is particularly significant in complex spaces. Let's break this down easily:
A sesquilinear form typically takes in two vector arguments. It operates in a manner that is linear in one argument while being conjugate linear in the other.
  • Linear in one argument: This means if you scale one input by a scalar and it affects the form in a predictable, linear way.
  • Conjugate Linear in the other: Here, scaling by a complex number not only scales but also applies the complex conjugate to the argument.
This functionality makes sesquilinear forms very useful. They can generalize inner products in complex vector spaces, often used for applications in physics and engineering where complex numbers arise naturally.
Quadratic Form
A quadratic form is derived directly from a sesquilinear form by considering what happens when you put the same vector as both arguments. For our problem, the quadratic form is \( q(f) = s(f, f) \).
This quadratic form has some fascinating characteristics:
  • Non-negative Values: If the vector space and the sesquilinear form are set up properly, square terms ensure that \( q(f) \) is non-negative for any vector \( f \).
  • Homogeneous Nature: This form behaves in a "squared" manner, meaning scaling the input by a scalar scales the output by the square of that scalar.
The quadratic form is prominent in many areas, such as optimization and geometry, helping to define curves, surfaces, and influence the distance measurements in various spaces.
Vector Space
Understanding vector spaces is critical to grasping sesquilinear and quadratic forms. A vector space is a collection of objects called vectors, which can be added together and multiplied (scaled) by numbers, called scalars. The set of numbers usually comes from a field, such as the real numbers or complex numbers.
Here are some key points that define a vector space:
  • Additive Identity: There is a zero vector \( \mathbf{0} \) in the vector space such that adding it to any vector \( \mathbf{v} \) keeps \( \mathbf{v} \) unchanged.
  • Scalar Multiplication and Vector Addition: Any vector in the space can be adjusted by scalars to obtain another vector in the same space.
  • Linear Combinations: Vectors can be combined through addition and scalar multiplication to express a wide range of vectors, supporting rich mathematical exploration.
A vector space provides a framework in which sesquilinear and quadratic forms are defined, allowing for complex interactions and transformations of vectors within the space.

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Most popular questions from this chapter

Find a sequence \(f_{n}\) which converges to 0 in \(L^{P}([0,1], d x)\), \(1 \leq p<\infty\), but for which \(f_{n}(x) \rightarrow 0\) for a.e. \(x \in[0,1]\) does not hold. (Hint: Every \(n \in \mathbb{N}\) can be uniquely written as \(n=2^{m}+k\) with \(0 \leq m\) and \(0 \leq k<2^{m}\). Now consider the characteristic functions of the intervals \(\left.I_{m, k}=\left[k 2^{-m},(k+1) 2^{-m}\right] .\right)\)

Let \(I\) be a compact interval. Show that the set of differentiable functions \(C^{1}(I)\) becomes a Banach space if we set \(\|f\|_{\infty, 1}=\) \(\max _{x \in I}|f(x)|+\max _{x \in I}\left|f^{\prime}(x)\right|\).

Show that \(\ell^{\infty}(\mathbb{N})\) is not separable.

Suppose \(\mathfrak{Q}\) is a vector space. Let \(s(f, g)\) be a sesquilinear form on \(\mathfrak{Q}\) and \(q(f)=s(f, f)\) the associated quadratic form. Prove the parallelogram law $$ q(f+g)+q(f-g)=2 q(f)+2 q(g) $$ and the polarization identity \((0.55) \quad s(f, g)=\frac{1}{4}(q(f+g)-q(f-g)+\mathrm{i} q(f-\mathrm{i} g)-\mathrm{i} q(f+\mathrm{i} g))\). Show that \(s(f, g)\) is symmetric if and only if \(q(f)\) is real-valued.

In a Banach space, the unit ball is conver by the triangle inequality. A Banach space \(X\) is called uniformly convex if for every \(\varepsilon>0\) there is some \(\delta\) such that \(\|x\| \leq 1,\|y\| \leq 1\), and \(\left\|\frac{x+y}{2}\right\| \geq 1-\delta\) imply \(\|x-y\| \leq \varepsilon\). Geometrically this implies that if the average of two vectors inside the closed unit ball is close to the boundary, then they must be close to each other. Show that a Hilbert space is uniformly convex and that one can choose \(\delta(\varepsilon)=1-\sqrt{1-\frac{\mathrm{E}^{2}}{4}}\), Draw the unit ball for \(\mathbb{R}^{2}\) for the norms \(\|x\|_{1}=\) \(\left|x_{1}\right|+\left|x_{2}\right|,\|x\|_{2}=\sqrt{\left|x_{1}\right|^{2}+\left|x_{2}\right|^{2}}\), and \(\|x\|_{\infty}=\max \left(\left|x_{1}\right|,\left|x_{2}\right|\right)\). Which of these norms makes \(\mathbb{R}^{2}\) uniformly convex?
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