/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Let \(Y\) be a closed subspace o... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(Y\) be a closed subspace of a normed space \(X\). Show that if \(Y\) and \(X / Y\) are both Banach spaces, then \(X\) is a Banach space. Note: A property \(\mathcal{P}\) is said to be a three-space property if the following holds: Let \(Y\) be a closed subspace of a space \(X\). If \(Y\) and \(X / Y\) have \(\mathcal{P}\), then \(X\) has \(\mathcal{P}\) (see, e.g., [CaGo]). Thus, the property of being complete is a three-space property in the class of normed linear spaces. Hint: If \(\left\\{x_{n}\right\\}\) is Cauchy in \(X\), there is \(x \in X\) such that \(\hat{x}_{n} \rightarrow \hat{x}\). There are \(\left\\{y_{n}\right\\}\) in \(Y\) such that \(\left\\{x_{n}-x-y_{n}\right\\} \rightarrow 0\). Thus \(\left\\{y_{n}\right\\}\) is Cauchy, so \(y_{n} \rightarrow y\) and \(x_{n} \rightarrow x+y\).

Short Answer

Expert verified
Since every Cauchy sequence in \(X\) converges to a limit in \(X\), \(X\) is a Banach space.

Step by step solution

01

Understand the Given Information

You're given that both subspace \(Y\) and quotient space \(X / Y\) are Banach spaces. A Banach space is a complete normed vector space. We need to show that \(X\) is also a Banach space.
02

Define Completeness

Recall that a space is complete if every Cauchy sequence in that space converges to a limit within the space. So, we need to show that every Cauchy sequence in \(X\) converges to a limit in \(X\).
03

Consider a Cauchy Sequence in \(X\)

Let \(\{x_{n}\}\) be a Cauchy sequence in \(X\). Since \(\{x_{n}\}\) is Cauchy, for every \(\epsilon > 0\), there exists an \(N\) such that for all \(m, n \geq N\), we have \(\|x_{n} - x_{m}\| < \epsilon\).
04

Image of a Cauchy Sequence in the Quotient Space

Consider the image of the sequence \(\{x_{n}\}\) in \(X / Y\). Denote this image by \(\{\hat{x}_{n}\}\), where \(\hat{x}_{n}\) is the equivalence class of \(x_{n}\). Since the natural projection \(\pi : X \rightarrow X / Y\) is continuous, \(\{\hat{x}_{n}\}\) is a Cauchy sequence in \(X / Y\).
05

Convergence in the Quotient Space

Given that \(X / Y\) is Banach, every Cauchy sequence in \(X / Y\) converges. Thus, there exists an element \(\hat{x} \in X / Y\) such that \(\hat{x}_{n} \rightarrow \hat{x}\).
06

Choose Representatives

With \(\hat{x}_{n} \rightarrow \hat{x}\), we can choose representatives from each equivalence class: there exist elements \(y_n \in Y\) such that \(x_n - y_n\) converges to some \(x \in X\).
07

Verify \(y_n\) forms a Cauchy Sequence in \(Y\)

Since \(y_n = x_n - x\) and \(\{x_{n} - x_{m}\}\) is Cauchy, it follows that \(\{y_n\}\) must also be a Cauchy sequence in the closed subspace \(Y\). Since \(Y\) is Banach, \(\{y_n\}\) converges to some \(y \in Y\).
08

Convergence in \(X\)

Given \(x_n - y_n \rightarrow x\) and \(y_n \rightarrow y\), adding these two convergent sequences, we get \(x_n \rightarrow x + y\). Hence, \(x_n\) converges to an element in \(X\), which is \(x + y\).
09

Conclude Completeness of \(X\)

Since every Cauchy sequence \(\{x_{n}\}\) in \(X\) converges to \(x + y \in X\), this implies \(X\) is complete. Therefore, \(X\) is a Banach space.

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Key Concepts

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

Closed subspace
A closed subspace is an important concept in functional analysis and topology. In a normed space, a subspace is closed if it contains all its limit points. This means if a sequence within the subspace converges, its limit is also within the subspace.
Imagine it like a completed jigsaw puzzle: every piece that fits from the space stays within the puzzle’s frame. In our exercise, the subspace in question is denoted as \(Y\). Since \(Y\) is a closed subspace of \(X\), any sequence of points in \(Y\) that converges will do so within \(Y\) itself.
Normed space
A normed space is a vector space on which a norm is defined. The norm measures the 'size' or 'length' of vectors in the space. For a vector \(x\) in a normed space \(X\), the norm is often denoted as \(\|x\|\).
A good analogy would be considering the vectors as paths in a city, where the norm is like a GPS system measuring the length of each path.
In our problem, both \(X\) and \(Y\) are normed spaces, and we can measure distances (or norms) of vectors in these spaces.
Three-space property
The three-space property is a concept that helps in understanding the structure of spaces when they are broken down into subspaces and quotient spaces. Essentially, if a subspace \(Y\) and the quotient space \(X / Y\) have a property \(\mathcal{P}\), then the whole space \(X\) should also have property \(\mathcal{P}\).
Think of it like building a house: if both the basement (\(Y\)) and the upper floors (\(X / Y\)) are strong and stable, then the whole house (\(X\)) is also strong and stable.
As mentioned, completeness is such a three-space property in the context of normed linear spaces.
Cauchy sequence
A Cauchy sequence is a sequence where the elements get arbitrarily close to each other as the sequence progresses. For a sequence \(\{x_n\}\) in a normed space to be Cauchy, for every \(\epsilon > 0\), there must be an index \(N\) so that for all \(m, n \geq N\), the distance \(\|x_n - x_m\| < \epsilon\).
Picture it like a group of friends gradually coming together closer and closer until they almost stand on the same spot.
In our context, we consider a Cauchy sequence in the space \(X\), and show how elements in this sequence eventually converge within \(X\).
Completeness
Completeness is a critical property for Banach spaces. A normed space is complete if every Cauchy sequence converges to a limit within the space. If the Cauchy sequence \(\{x_n\}\) converges to an element \(x\) in the space, this space is considered complete.
Imagine the process of a sculptor finishing a statue: every chisel move (Cauchy sequence element) leads to a complete statue.
In the exercise, we demonstrate that if both \(Y\) and \(X / Y\) are Banach spaces (complete normed spaces), then \(X\) must also be a complete space and hence a Banach space.

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Most popular questions from this chapter

Let \(X\) be the normed space obtained by taking \(c_{0}\) with the norm \(\|x\|_{0}=\sum 2^{-i}\left|x_{i}\right| .\) Show that \(X\) is not a Banach space. Note that this shows that \(\|\cdot\|_{0}\) is not an equivalent norm on \(c_{0}\). Hint: The sequence \(\\{(1,1, \ldots, 1,0, \ldots)\\}_{n=1}^{\infty}\) is Cauchy and not convergent since the only candidate for the limit would be \((1,1, \ldots) \notin c_{0}\).

Let \(f \in L_{p_{0}}[0,1]\) for some \(p_{0}>1 .\) Show that \(\lim _{p \rightarrow 1^{+}}\|f\|_{L_{p}}=\|f\|_{L_{1}} .\) If \(f \in L_{\infty}[0,1]\), then \(\lim _{p \rightarrow \infty}\|f\|_{L_{p}}=\|f\|_{L_{\infty}}\) Let \(x \in \ell_{q}\) for some \(q \geq 1\). Show that \(\lim _{p \rightarrow \infty}\|x\|_{\ell_{p}}=\|x\|_{\ell_{\infty}}\).

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). If there is \(\delta>0\) such that \(\|T(x)\| \geq \delta\|x\|\) for all \(x \in X\), then \(T(X)\) is closed in \(Y .\) Moreover, \(T\) is an isomorphism from \(X\) into \(Y\). Hint: Let \(\left\\{T\left(x_{n}\right)\right\\} \subset T(X)\) be such that \(T\left(x_{n}\right) \rightarrow y\). Then \(\left\\{T\left(x_{n}\right)\right\\}\) is Cauchy and hence also \(\left\\{x_{n}\right\\}\) is Cauchy. By completeness, \(x_{n} \rightarrow x\), so \(T\left(x_{n}\right) \rightarrow T(x)\), by the uniqueness of the limit we get \(y=T(x)\). The inequality we assume implies that \(T\) is one-to-one, so the inverse \(T^{-1}: T(X) \rightarrow X\) is defined and \(\left\|T^{-1}\right\| \leq \frac{1}{\delta}\)

Show that a normed space \(X\) is a Banach space if and only if \(\sum y_{n}\) converges whenever \(\left\|y_{n}\right\| \leq 2^{-n}\) for every \(n\). Hint: Use Lemma 1.15. Note that if \(\sum x_{k}\) is absolutely convergent, there are \(N_{n}\) so that \(\sum_{k>N_{n}}\left\|x_{k}\right\| \leq 2^{-k} .\) Set \(y_{n}=\sum_{k=N_{n}+1}^{N_{k+1}} x_{k}\), then \(\left\|y_{n}\right\| \leq 2^{-n}\), and if \(M>N_{m}\), then \(\left\|\sum_{n=1}^{m} y_{n}-\sum_{k=1}^{M} x_{k}\right\| \leq 2^{-m}\).

Let \(X, Y\) be normed spaces and \(T \in \mathcal{B}(X, Y)\). Show that $$ \|T\|=\sup \left\\{\|T(x)\|_{Y} ;\|x\|_{X}<1\right\\}=\sup \left\\{\|T(x)\|_{Y} ;\|x\|_{X}=1\right\\} $$ Hint: Clearly, both suprema are not greater than \(\|T\|\). Given \(\varepsilon>0\), find \(x \in B_{X}\) such that \(\|T(x)\|_{Y} \geq \sqrt{1-\varepsilon}\|T\| .\) Then \(\|\sqrt{1-\varepsilon} x\|<1\) and \(\frac{x}{\|x\|_{x}} \in S_{X}\), and both vectors give \(\|T(y)\|_{Y} \geq(1-\varepsilon)\|T\|\).

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