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Let \(Y\) be a subspace of a Banach space \(X\) and \(\|\cdot\|\) be an equivalent norm on \(Y\). Show that \(\|\cdot\|\) can be extended to an equivalent norm on \(X\). Hint: Let \(B_{2}\) be the unit ball of the original norm of \(X .\) Assume without loss of generality that the unit ball \(B_{1}\) of \(\|\cdot\|\) on \(Y\) contains \(B_{2} \cap Y\). The Minkowski functional of the set \(\operatorname{conv}\left(B_{1} \cup B_{2}\right)\) yields the desired norm.

Short Answer

Expert verified
The Minkowski functional of the convex hull of the union of the unit balls of the original norm and the new norm on Y extends the norm to the entire Banach space X, ensuring equivalence.

Step by step solution

01

Understand the Problem

We need to extend an equivalent norm \(\forall y \in Y: \|y\|\) to an equivalent norm on the entire Banach space \(\forall x \in X: \|x\|_{new}\). The hint suggests using the Minkowski functional of the convex hull of the union of the unit balls of the original and the new norm.
02

Use the Unit Balls

Denote \(B_{1}\) and \(B_{2}\) as the unit balls of the norm \( \|\cdot\| \) on \(Y\) and the original norm on \(X\), respectively. Assume \(B_{1}\) includes \(B_{2} \cap Y\).
03

Construct the Convex Hull

Consider the convex hull \(\text{conv}(B_{1} \cup B_{2})\). This set will allow us to take advantage of structures of both norms and combine them.
04

Define the Minkowski Functional

The Minkowski functional of a set \(C\) is defined as \(\rho_{C}(x) = \inf \{ \lambda > 0 : x \in \lambda \cdot C \}\). For \(C = \text{conv}(B_{1} \cup B_{2})\), this defines a new norm.
05

Prove the Norm is Equivalent

Show that for some constants \(a, b > 0\), we have \(a \|x\| \leq \|x\|_{new} \leq b \|x\|\) for all \(x \in X\). This ensures that \(\rho_{\text{conv}(B_{1} \cup B_{2})}\) is an equivalent norm on \(X\).

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Key Concepts

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

Banach space
A Banach space is a specific type of vector space. It is not only a vector space but also a complete normed space.

This means every Cauchy sequence (sequences where the elements become arbitrarily close to each other as the sequence progresses) in this space converges to a limit within the space.

Imagine you are measuring distances in a room, where every path follows a straight line or 'norm.' If every measurement you take adheres to a set of rules and you can detect the smallest differences accurately, you are essentially working in a Banach space.

In our context, the Banach space is represented by \(X\) and contains a subspace \(Y\). The problem requires extending a norm from \(Y\) to \(X\).

Important properties of Banach spaces include:
  • Completeness: Every Cauchy sequence has a limit in the space.
  • Normed structure: Distances and lengths follow specific rules defined by the norm.
Equivalent norm
An equivalent norm essentially modifies how we measure distances but retains the core properties of those measurements.

Formally, two norms \(orm{\bullet}_1\) and \(orm{\bullet}_2\) on a vector space are equivalent if there exist two positive constants \(a\) and \(b\) such that for every vector \(x\):
\[a orm{x}_1 \leq orm{x}_2 \leq b orm{x}_1\]

This simply means that although the two norms might measure differently, they scale with each other within fixed bounds. In the problem, we aim to extend the equivalent norm defined on \(Y\) to the entire space \(X\).

Key points about equivalent norms include:
  • Scalability: They keep measurements within consistent bounds.
  • Interchangeability: Calculations remain unaltered under different but equivalent norms.
Minkowski functional
The Minkowski functional is a tool that helps us define norms using sets.

Consider a convex set \(C\). The Minkowski functional \(\rho_C(x)\) of this set is defined as:
\[\rho_C(x) = \inf \{ \lambda > 0 : x \in \lambda \cdot C\}\]

This formula essentially finds the smallest \(\lambda\) such that \(x\) is within the scaled version of the set \(C\).

In our solution, \(C\) is chosen as the convex hull of the union of two unit balls of norms from \(X\) and \(Y\). This approach allows deriving a new norm for \(X\) that respects the original and new norm properties.

Essential properties:
  • Infimum property: It uses the smallest effective scaling factor.
  • Unit ball: Typically associated with the norms' unit ball for defining distances.
Convex hull
The convex hull is the smallest convex set containing a given set of points or shapes.

A set is convex if for any two points within the set, the entire line segment connecting them also lies within the set.

Mathematically, if you have sets \(B_1\) and \(B_2\), the convex hull, denoted as \(\text{conv}(B_1 \cup B_2)\), captures all the linear combinations \(\theta_1x + \theta_2y\) where \(\theta_1, \theta_2 \geq 0\) and \(\theta_1 + \theta_2 = 1\).

In the context of our exercise, constructing the convex hull of the union of two unit balls \((B_1 \cup B_2)\) helps in blending the characteristics of both norms for defining a new equivalent norm on \(X\).

Important features include:
  • Minimal Convex Set: Smallest set to encompass all desired points.
  • Linear Combinations: It includes all points forming via linear combinations within the boundaries.

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Most popular questions from this chapter

Let \(X\) be a normed space with two norms \(\|\cdot\|_{1}\) and \(\|\cdot\|_{2}\) such that \(X\) in both of them is a complete space. Assume that \(\|\cdot\|_{1}\) is not equivalent to \(\|\cdot\|_{2}\). Let \(I_{1}\) be the identity map from \(\left(X,\|\cdot\|_{1}\right)\) onto \(\left(X,\|\cdot\|_{2}\right)\) and \(I_{2}\) be the identity map from \(\left(X,\|\cdot\|_{2}\right)\) onto \(\left(X,\|\cdot\|_{1}\right) .\) Show that neither \(I_{1}\) nor \(I_{2}\) are continuous. Hint: The Banach open mapping theorem.

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). Show that if \(T\) maps bounded closed sets in \(X\) onto closed sets in \(Y\), then \(T(X)\) is closed in \(Y\). Hint: Assume \(T\left(x_{n}\right) \rightarrow y \notin T(X)\). Put \(M=\operatorname{Ker}(T)\), set \(d_{n}=\operatorname{dist}\left(x_{n}, M\right)\) and find \(w_{n} \in M\) such that \(d_{n} \leq\left\|x_{n}-w_{n}\right\| \leq 2 d_{n} .\) If \(\left\\{x_{n}-w_{n}\right\\}\) is bounded, then \(T\left(x_{n}-w_{n}\right) \rightarrow y \in T(X)\), since the closure of \(\left\\{x_{n}-w_{n}\right\\}\) is mapped onto a closed set containing \(y\). If \(\left\|x_{n}-w_{n}\right\| \rightarrow \infty\), then since \(T\left(x_{n}-w_{n}\right) \rightarrow y\), we have \(T\left(\frac{x_{n}-w_{n}}{\left\|x_{n}-w_{n}\right\|}\right) \rightarrow 0 .\) By the hypothesis, \(M\) must contain a point \(w\) from the closure of \(\left\\{\frac{x_{n}-w_{n}}{\left\|x_{n}-w_{n}\right\|}\right\\}\) since 0 lies in the closure of the image of this sequence. Fix \(n\) so that \(\left\|\frac{x_{n}-w_{n}}{\left\|x_{n}-w_{n}\right\|}-w\right\|<1 / 3\). Then \(\left\|x_{n}-w_{n}-\right\| x_{n}-w_{n}\|w\| \leq \frac{1}{3}\left\|x_{n}-w_{n}\right\|<(2 / 3) d_{n}\) and \(w_{n}+\left\|x_{n}-w_{n}\right\| w \in M\) a contradiction.

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). Show that if \(Y\) is separable and \(T\) is onto \(Y\), then there is a separable closed subspace \(Z\) of \(X\) such that \(T(Z)=Y\).

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). If \(T\) is an isomorphism into \(Y\), is \(T^{*}\) necessarily an isomorphism into \(X^{*}\) ? Hint: No, embed \(\mathbf{R}\) into \(\mathbf{R}^{2}\).

Let \(X=\mathbf{R}^{2}\) with the norm \(\|x\|=\left(\left|x_{1}\right|^{4}+\left|x_{2}\right|^{4}\right)^{\frac{1}{4}} .\) Calculate directly the dual norm on \(X^{*}\) using the Lagrange multipliers. Hint: The dual norm of \((a, b) \in X^{*}\) is \(\sup \left\\{a x_{1}+b x_{2} ; x_{1}^{4}+x_{2}^{4}=1\right\\} .\) Define \(F\left(x_{1}, x_{2}, \lambda\right)=a x_{1}+b x_{2}-\lambda\left(x_{1}^{4}+x_{2}^{4}-1\right)\) and multiply by \(x_{1}\) and \(x_{2}\), respectively, the equations you get from \(\frac{\partial F}{\partial x_{1}}=0\) and \(\frac{\partial F}{\partial x_{2}}=0\)

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