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Chapter 1: Problem 1

Check that the space \(L^{1}(Q)\) is not a Hilbert space. Hint: In any Hilbert space \(\|\alpha+\beta\|^{2}+\|\alpha-\beta\|^{2}=2\|\alpha\|^{2}+2\|\beta\|^{2}\). Try this out for \(Q=[0,1], \alpha=1\), and \(\beta=x\). What happens? Try again.

Short Answer

Expert verified
\(L^1(Q)\) is not a Hilbert space because the parallelogram identity fails.

Step by step solution

01

Understand the Given Expression

We need to apply the condition for being a Hilbert space: \(\|\alpha+\beta\|^{2} + \|\alpha-\beta\|^{2} = 2\|\alpha\|^{2} + 2\|\beta\|^{2}\). This needs to be shown to not hold for some functions in \(L^1(Q)\) to prove \(L^1(Q)\) is not a Hilbert space.
02

Define the Functions

We are given the interval \(Q=[0,1]\) and specific functions: \(\alpha=1\) and \(\beta=x\). These are essentially the functions \(f(x) = 1\) and \(g(x) = x\) on \([0,1]\).
03

Compute Norms in \(L^1(Q)\)

In \(L^1(Q)\), the norm of a function \(h(x)\) over interval \([0,1]\) is \(\|h\| = \int_0^1 |h(x)| dx\). We need to calculate \(\|\alpha+\beta\|\), \(\|\alpha-\beta\|\), \(\|\alpha\|\), and \(\|\beta\|\) using this definition.
04

Calculate \(\|\alpha\|\)

\(\|\alpha\| = \int_0^1 |1| dx = \int_0^1 1 \, dx = 1\).
05

Calculate \(\|\beta\|\)

\(\|\beta\| = \int_0^1 |x| dx = \int_0^1 x \, dx = \frac{1}{2}\).
06

Calculate \(\|\alpha + \beta\|\)

\(\|\alpha + \beta\| = \int_0^1 |1 + x| dx = \int_0^1 (1 + x) \, dx = 1 + \frac{1}{2} = \frac{3}{2}\).
07

Calculate \(\|\alpha - \beta\|\)

\(\|\alpha - \beta\| = \int_0^1 |1 - x| dx = \int_0^1 (1 - x) \, dx = 1 - \frac{1}{2} = \frac{1}{2}\).
08

Check Hilbert Space Condition

Substitute the norms into the condition: \((\frac{3}{2})^2 + (\frac{1}{2})^2\) vs. \(2 \, \times 1^2 + 2 \, \times (\frac{1}{2})^2\). Calculate: \(\frac{9}{4} + \frac{1}{4} eq 2 + \frac{1}{2}\). The equality does not hold.

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Key Concepts

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

L1 Space and Its Properties
The \(L^1(Q)\) space consists of all functions whose absolute value's integral over a specified interval is finite.
This means, for any function \(f\), we compute \(|f|\) and integrate it over the interval to check if this integral is finite. In our example, the interval \(Q\) is \[ [0, 1] \].
**Properties of \(L^1(Q)\):**
  • The functions are integrable over the interval; meaning we can sum their absolute values and get a finite result.
  • \(L^1(Q)\) is a complete normed vector space.
  • It uses absolute integrability to define its norm, often represented as \(\|f\|_1\).
However, unlike a Hilbert space, \(L^1(Q)\) does not have an inner product structure that satisfies the parallelogram law required for Hilbert spaces.
Understanding Fourier Series
Fourier series allows us to represent complex periodic functions as a sum of simple sines and cosines.
This transformation is crucial for analyzing functions in a variety of fields, including physics and engineering.
**Key Points About Fourier Series:**
  • They decompose a function into a sum of trigonometric functions that oscillate at different frequencies and amplitudes.
  • In mathematical notation, the series is expressed as: \(f(x) = a_0 + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx))\).
  • The coefficients \(a_n\) and \(b_n\) determine the contribution of each trigonometric component.
Note that while Fourier series excel in periodic intervals, they rely heavily on function properties possible within spaces like \(L^2\), a Hilbert space, but not typically suited for \(L^1\).
Understanding Mathematical Norm
A mathematical norm is a function that assigns a non-negative length or size to vectors in a space.
Norms obey specific properties:
  • Non-negativity: The norm of any vector \(v\), \(|v|\), is always zero or positive.
  • Scalability: Norms obey \(|cv| = |c||v|\), where \(c\) is any scalar.
  • Triangle inequality: \(|u + v| \leq |u| + |v|\) for any vectors \(u\) and \(v\).
  • Identity of indiscernibles: \(|v| = 0\) implies that \(v\) is the zero vector.
In \(L^1(Q)\), we define the norm as the integral of the absolute value over an interval. It provides a measure of the "size" of the function on this interval but lacks the inner product structure needed in a Hilbert space.
The Role of Interval Integration
Interval integration involves computing the area under the curve of a function within a given range.
**Fundamental Concepts of Interval Integration:**
  • Integration over an interval \([a, b]\) gives a precise value that represents the function's "total accumulation" within these bounds.
  • For functions in \(L^1(Q)\), integration is crucial for calculating norms, specifically the \(L^1\) norm.
  • It's pivotal in determining phenomena such as the area under the curve, and it plays a role in many mathematical and practical applications.
Successful integration hinges on the function being continuous, or at least being approximate to some degree, over the interval. The inability to satisfy certain conditions, like in interval integration norms, explains why \(L^1(Q)\) is not a Hilbert space, as seen by unsuccessful adherence to the parallelogram law.

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Most popular questions from this chapter

\(\lim _{n \uparrow \infty} \int f_{n}=\int f\) if \(f_{1} \geqslant f_{2} \geqslant \cdots, \lim _{n \uparrow \infty} f_{n}=f\), and \(\int f_{1}^{+}<\infty\).

.\( Check that for fixed \)\beta\(, the inner product \)(\alpha, \beta)\( is a continuous function of \)\alpha .\( Hint: \)\quad\left(\alpha_{1}, \beta\right)-\left(\alpha_{2}, \beta\right)=\left(\alpha_{1}-\alpha_{2}, \beta\right) ;$ now use Schwarz's inequality.

Show that any linear map \(l\) of \(L^{2}(Q)\) into the complex numbers which is bounded in the sense that $$ |l(f)| \leqslant \text { constant } \times\|f\| $$ where the constant is independent of \(f\), can be expressed as an inner product: $$ l(f)=(f, g) $$ for some \(g \in \mathrm{L}^{2}(Q)\). This is the so-called Riesz representation theorem. Hint. Suppose \(l \not \equiv 0\) and let \(\mathbf{A}=(f: l(f)=0)\). Check that \(\mathbf{B}=\mathbf{A}^{0}\) is of dimension 1 and find a function \(g \in L^{2}(Q)\) so that \(l(f)=0\) iff \((f, g)=0\). Then $$ l(f)=\|g\|^{-2} l(g)(f, g)=\text { constant } \times(f, g) $$ Why? There are many excellent books on Hilbert space; among the best at an advanced level are Akhiezer and Glazman [1961-1963] and Riesz and Sz.-Nagy \([1955] ;\) at an elementary level, Berberian \([1961]\) is recommended.

Check by example that the inclusion of \(L^{2}[0,1]\) in \({ }^{1}[0,1]\) is proper, that is, find a summable function \(f\) with \(\int_{0}^{1}|f|^{2}=\infty\).

Check that if \(f_{n}\) converges to \(f\) in \(\mathrm{L}^{2}(Q)\) and if it also converges to \(f^{0}\) pointwise a.e., then \(f=f^{0}\) a.e.
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