Chapter 11: Problem 30
Consider the identity injection \(I\) from \(C[0,1]\) to \(L_{2}[0,1]\). Show that it is continuous but not weakly compact.
Short Answer
Expert verified
The identity injection \( I \) is continuous because uniform convergence in \( C[0,1] \) implies convergence in \( L_2[0,1] \), but it is not weakly compact because the sequence \( x^n \) does not have a weakly convergent subsequence in \( L_2[0,1] \).
Step by step solution
01
- Understand the identity injection
The identity injection, denoted by \( I \), maps each function in \( C[0,1] \) (the space of continuous functions on \( [0,1] \)) to itself in the Hilbert space \( L_2[0,1] \) (the space of square-integrable functions on \( [0,1] \)). This means that for a function \( f \in C[0,1] \), we have \( I(f) = f \).
02
- Prove the continuity of the injection
To show that \( I \) is continuous, we need to show that for every convergent sequence \( f_n \rightarrow f \) in \( C[0,1] \), the sequence \( I(f_n) \rightarrow I(f) \) in \( L_2[0,1] \). Since \( I(f_n) = f_n \), this means proving \( \| f_n - f \|_{L_2} \rightarrow 0 \). Given that \( f_n \rightarrow f \) uniformly (the norm in \( C[0,1] \) implies uniform convergence), it helps to use the fact that uniform convergence implies pointwise convergence and thus gives a bound on the \( L_2 \) norm.
03
- Calculate the transition between norms
We know that if \( f_n \rightarrow f \) uniformly, for all \( \epsilon > 0 \), there exists an \( N \) such that for all \( n > N \), \( | f_n(x) - f(x) | < \epsilon \) for all \( x \in [0,1] \). Consequently, \[ \| f_n - f \|_{L_2}^2 = \int_0^1 | f_n(x) - f(x) |^2 dx < \epsilon^2 \. \] Thus, \( \| f_n - f \|_{L_2} \rightarrow 0 \), proving the continuity of \( I \).
04
- Understand weak compactness
A set in a Banach space is called weakly compact if every sequence in the set has a subsequence that converges weakly. We need to show that the identity injection \( I \) is not weakly compact.
05
- Show that \( I \) is not weakly compact
Consider the sequence of functions \( f_n(x) = x^n \) in \( C[0,1] \). This sequence does not have a weakly convergent subsequence in \( L_2[0,1] \). Intuitively, while \( f_n \rightarrow 0 \) pointwise for \( x \in [0,1) \), at \( x = 1 \) the functions do not tend towards any single value uniformly. Therefore, in \( L_2[0,1] \) terms, they do not have a subsequence converging weakly because their norm does not tend to 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuity
Continuity is a fundamental concept in mathematics, especially in functional analysis. A function is continuous if small changes in the input result in small changes in the output. When we say that the identity injection from \(C[0,1]\) to \(L_2[0,1]\) is continuous, it means for any sequence \(f_n\) converging to \(f\) in \(C[0,1]\), the sequence \(I(f_n) = f_n\) converges to \(I(f) = f\) in \(L_2[0,1]\).
In mathematical terms: \
In mathematical terms: \
- For every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( \|f_n - f\|_{C[0,1]} < \delta \), then \( \|f_n - f\|_{L_2[0,1]} < \epsilon \).
Weak Compactness
Weak compactness is the property that every sequence in a set has a weakly convergent subsequence. In the context of Banach spaces, a set is weakly compact if every sequence has a subsequence that converges weakly. For the identity injection from \(C[0,1]\) to \(L_2[0,1]\), demonstrating non-weak compactness involves showing the absence of such subsequences.
If we take a sequence like \(f_n(x) = x^n\) in \(C[0,1]\), it points to the challenge in having a weakly convergent subsequence in \(L_2[0,1]\). The sequence \(f_n\) converges pointwise to zero for all \(x\) in [0,1), but it does not uniformly converge because at \(x = 1\), it remains at one. This irregular behavior prevents the existence of a weakly convergent subsequence in \(L_2[0,1]\). Simply put, the norms of \(f_n\) do not tend to zero uniformly, proving \(I\) is not weakly compact.
If we take a sequence like \(f_n(x) = x^n\) in \(C[0,1]\), it points to the challenge in having a weakly convergent subsequence in \(L_2[0,1]\). The sequence \(f_n\) converges pointwise to zero for all \(x\) in [0,1), but it does not uniformly converge because at \(x = 1\), it remains at one. This irregular behavior prevents the existence of a weakly convergent subsequence in \(L_2[0,1]\). Simply put, the norms of \(f_n\) do not tend to zero uniformly, proving \(I\) is not weakly compact.
Norms
Norms provide a measure of the size or length of a vector in a vector space. In the context of function spaces like \(C[0,1]\) and \(L_2[0,1]\), they help us understand convergence and continuity.
The norm in \(C[0,1]\) is the supremum norm (uniform norm) defined as: \
For \(L_2[0,1]\), the norm is the \(L_2\) norm, which is given by: \
Uniform convergence in \(C[0,1]\) norm implies convergence in the \(L_2\) norm. This is essential in proving the continuity of the identity injection because it ensures that small changes in functions with regard to the \(C[0,1]\) norm lead to small changes in the \(L_2\) norm.
The norm in \(C[0,1]\) is the supremum norm (uniform norm) defined as: \
- \( \|f\|_{C[0,1]} = \sup_{x \in [0,1]} |f(x)| \)
For \(L_2[0,1]\), the norm is the \(L_2\) norm, which is given by: \
- \( \|f\|_{L_2} = \left( \int_0^1 |f(x)|^2 dx \right)^{1/2} \)
Uniform convergence in \(C[0,1]\) norm implies convergence in the \(L_2\) norm. This is essential in proving the continuity of the identity injection because it ensures that small changes in functions with regard to the \(C[0,1]\) norm lead to small changes in the \(L_2\) norm.
L2 Space
The \(L_2[0,1]\) space is a Hilbert space consisting of square-integrable functions on the interval [0,1]. A function \(f\) belongs to \(L_2[0,1]\) if \
When considering the identity injection from \(C[0,1]\) to \(L_2[0,1]\), it means mapping a continuous function on [0,1] into the Hilbert space where we can measure it using the \(L_2\) norm. Since continuous functions are integrable, they fit well into \(L_2[0,1]\). This space allows for a different perspective on convergence and continuity compared to purely pointwise or uniform viewpoints.
- \( \int_0^1 |f(x)|^2 dx < \infty \)
When considering the identity injection from \(C[0,1]\) to \(L_2[0,1]\), it means mapping a continuous function on [0,1] into the Hilbert space where we can measure it using the \(L_2\) norm. Since continuous functions are integrable, they fit well into \(L_2[0,1]\). This space allows for a different perspective on convergence and continuity compared to purely pointwise or uniform viewpoints.
Uniform Convergence
Uniform convergence is when a sequence of functions \(f_n\) converges to a function \(f\) such that the speed of convergence does not depend on the point in the domain. In mathematical terms, \
Uniform convergence is stronger than pointwise convergence and ensures continuity of the limit function.
In the context of the identity injection from \(C[0,1]\) to \(L_2[0,1]\), uniform convergence is significant because it guarantees that not only do the functions converge pointwise, but also the norms \( \|f_n - f\|_{L_2}\) converge to zero, ensuring continuity. This plays a vital role in proving the continuity part of the assertion. Uniform convergence helps transition easily between different norms due to its strong convergence properties.
- \( f_n \rightarrow f \) uniformly if for every \( \epsilon > 0 \), there exists an \( N \) such that for all \( n > N \) and \( x \in [0,1] \), \( |f_n(x) - f(x)| < \epsilon \).
Uniform convergence is stronger than pointwise convergence and ensures continuity of the limit function.
In the context of the identity injection from \(C[0,1]\) to \(L_2[0,1]\), uniform convergence is significant because it guarantees that not only do the functions converge pointwise, but also the norms \( \|f_n - f\|_{L_2}\) converge to zero, ensuring continuity. This plays a vital role in proving the continuity part of the assertion. Uniform convergence helps transition easily between different norms due to its strong convergence properties.