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Chapter 3: Problem 9

Let \(E \subset L^{2}(-\pi, \pi)\) be the set of all functions $$ f_{m, n}(t)=e^{i m t}+m e^{i n t} $$ where \(m, n\) are integers and \(0 \leq m

Short Answer

Expert verified
In summary, we found that there are no elements in \(E_3\), as it is empty. All elements in the weak closure \(E_*\) of \(E\) have the property that there exists a sequence \(\{f_{m_k,n_k}\}\) such that \(\lim_{k\to\infty} \langle f_{m_k,n_k}(t),g(t) \rangle=\langle g(t),g(t) \rangle.\) We also showed that 0 is in \(E_w\) but not in \(E_1\) and lies in the weak sequential closure of \(E_1\).

Step by step solution

01

Write the function \(f_{m,n}(t)\)

\(f_{m,n}(t)=e^{i m t}+m e^{i n t}\), where \(0 \leq m<n\). Let's find out the inner product of \(f_{m,n}(t)\) with itself:
02

Compute the inner product

\(\langle f_{m,n}(t), f_{m,n}(t) \rangle=\int_{-\pi}^{\pi} |f_{m,n}(t)|^2 dt=\int_{-\pi}^{\pi} (e^{i m t}+m e^{i n t})(e^{-i m t}+m e^{-i n t}) dt.\) Now, we can calculate the integral:
03

Evaluate the integral

Simplifying the integral, we have: \(\int_{-\pi}^{\pi}(1+m^2+2m\cos((n-m)t))dt=2\pi(1+m^2)\), since the integral of \(\cos((n-m)t)\) over \([-\pi,\pi]\) is 0 when \(n\neq m\). Now let's handle each part of the exercise: (a) Find all \(g \in E_{3}\). For finding elements in \(E_3\), we have to check if there is a sequence in \(E\) that converges weakly to \(g\).
04

Check if a sequence in \(E\) converges weakly to \(g\)

We have to check if for every \(g\) in \(E_3\), there exists a sequence \(\{f_{m_k,n_k}\}\) s.t \(\lim_{k\to\infty} \langle f_{m_k,n_k}(t),g(t) \rangle=\langle g(t),g(t) \rangle\). In this case, since \(E_3\) is empty, there are no elements to find. Therefore, the answer is \(\emptyset\). (b) Find all \(g\) in the weak closure \(E_*\) of \(E\). Based on the properties found in the elements of \(E\), all \(g\) in the weak closure \(E_*\) of \(E\) must have the following property:
05

Properties of elements in \(E_*\)

For every \(g\) in \(E_*\), there exists a sequence \(\{f_{m_k,n_k}\}\) such that \(\lim_{k\to\infty} \langle f_{m_k,n_k}(t),g(t) \rangle=\langle g(t),g(t) \rangle.\) (c) Show that 0 is in \(E_{w}\) but not in \(E_1\) and lies in the weak sequential closure of \(E_1\).
06

Show 0 is in \(E_{w}\)

To show \(0 \in E_{w}\), we need to find a sequence \(\{f_{m_k,n_k}\}\) such that \(\lim_{k\to\infty} \langle f_{m_k,n_k}(t),0(t) \rangle=0.\) This can be easily done by choosing the sequence \(f_{m,n}(t)\) with \(m=n\), then \(\lim_{k\to\infty} \langle f_{m,m}(t),0(t)\rangle=0.\)
07

Show 0 is not in \(E_1\)

To show \(0 \notin E_1\), simply notice that \(E_1\) is given as the weak sequential closure of \(E\) and all elements in \(E\) are non-zero, so \(0\) cannot be in \(E_1\).
08

Show 0 is in the weak sequential closure of \(E_1\)

Since \(0 \in E_{w}\), it means there exists a sequence in \(E\) that converges weakly to \(0\). Therefore, we can conclude that \(0\) is in the weak sequential closure of \(E_1\).

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Key Concepts

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

Functional Analysis
Functional analysis is a branch of mathematics, particularly within the scope of analysis, concerned with the study of vector spaces and the linear operators acting upon them.

It provides a framework for discussing concepts like convergence, continuity, and linearity of functions, but in a much broader sense than you might have encountered in calculus. One of the primary spaces studied within functional analysis is the L2 space, often denoted as \( L^{2} \), which consists of square-integrable functions. This essentially means that if you square the function and integrate it over its domain, you will end up with a finite value.

Within this field, the notion of convergence is refined beyond the simple pointwise convergence familiar from calculus to include types of convergences such as uniform, strong, and weak convergence. These distinctions are important since various types of convergence may lead to vastly different outcomes when analyzing infinite-dimensional spaces like function spaces.
Weak Convergence
Weak convergence is a type of convergence that is less stringent than pointwise or uniform convergence. A sequence of functions \( \{f_n\} \) is said to converge weakly to a function \( f \) if, for every continuous linear functional \( L \), the sequence of real numbers \( \{L(f_n)\} \) converges to \( L(f) \).

In the L2 space, a good way to think about this is in terms of inner product. For a sequence of functions to converge weakly to a function \( g \), the inner products \( \langle f_n, h \rangle \) must converge to \( \langle g, h \rangle \) for every other function \( h \) in the space. It's less about the functions getting 'close' to each other everywhere and more about their 'average' behavior with respect to the rest of the space approaching that of the limit function.

This subtlety explains the distinction between weak sequential closure and weak closure. Weak sequential closure considers only sequences, while weak closure considers all possible 'net' convergences, a more general concept than sequences.
L2 Space
The \( L^2 \) space, often encountered in functional analysis, represents the space of square-integrable functions over a given domain. For instance, in the context of this problem, the domain is \( [-\textbackslash pi, \textbackslash pi] \).

A function \( f \) belongs to \( L^2 \) if the integral of \( f^2 \) is finite, meaning you can square the function and integrate it over its entire domain without producing an infinite result. In mathematical terms, \( \int |f|^2 dt < \textbackslash infty \). This is important because it guarantees that every function in the L2 space has a well-defined norm, which is a measure of its 'size' or 'length' in a sense analog to Euclidean space.

Furthermore, \( L^2 \) is a Hilbert space, which means it has a complete and orthonormal basis and a well-defined inner product, allowing us to discuss concepts like orthogonality, projection, and distance within the space. The exercise mentioned involves functions within this space, and the associated inner product plays a crucial role in discussing their weak convergence.

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