Problem 28
Question
Show that the following are all norms in the vector space \(\mathbb{R}^{2}\) : $$ \begin{aligned} &\|\mathbf{u}\|_{1}=\sqrt{\left(u_{1}\right)^{2}+\left(u_{2}\right)^{2}} \\ &\|\mathbf{u}\|_{2}=\max \left[\left|u_{1}\right|,\left|u_{2}\right|\right\\} \\\ &\|\mathbf{u}\|_{3}=\left|u_{1}\right|+\left|u_{2}\right| \end{aligned} $$ What are the shapes of the open balls \(B_{a}(\mathrm{u})\) ? Show that the topologes generated by these norms are the same.
Step-by-Step Solution
Verified Answer
The given expressions are valid norms on \(\mathbb{R}^{2}\) and the shapes of the open balls in these norms are respectively disks, squares, and diamonds. The topologies generated by these norms are the same.
1Step 1: Validate Norms
A function is called a norm if it satisfies the four properties – positive definiteness, absolute scalability, triangle inequality, and positive homogeneity. Let's verify these properties for each of the given norms. \(\|\mathbf{u}\|_{1}=\sqrt{\left(u_{1}\right)^{2}+\left(u_{2}\right)^{2}}, \(\|\mathbf{u}\|_{2}=\max \left[\left|u_{1}\right|,\left|u_{2}\right|\right)\), and \(\|\mathbf{u}\|_{3}=\left|u_{1}\right|+\left|u_{2}\right|\) all satisfy these properties and hence are valid norms.
2Step 2: Find Shapes of Open Balls
An open ball centered at \(u\), of radius \(a\) in a normed space with norm \(\|\cdot\|\) is the set \(B_a(u) = \{v \in V : \|v - u\| < a\}\). For the given norms, the shapes of the open balls are as follows:1. For \(\|\mathbf{u}\|_{1}\, the open balls are disks; (B_{a}(\mathrm{u}) is the set of points that lie within a circle of radius \(a\))2. For \(\|\mathbf{u}\|_{2}\, the open balls are squares; (B_{a}(\mathrm{u}) is the set of points that lie within a square with sides of length \(2a\))3. For \(\|\mathbf{u}\|_{3}\, the open balls are diamonds; (B_{a}(\mathrm{u}) is the set of points that lie within a rhombus with diagonals of length \(2a\))
3Step 3: Verify Same Topology
Two norms on a space give the same topology if and only if there is a positive linear relationship between them. Let's prove this relation for the given norms: (1) \(\|\mathbf{u}\|_{2} \leq \|\mathbf{u}\|_{1} \leq \sqrt{2}\|\mathbf{u}\|_{2}\)(2) \(\|\mathbf{u}\|_{3} \leq \|\mathbf{u}\|_{1} \leq \sqrt{2}\|\mathbf{u}\|_{3}\)(3) \(\|\mathbf{u}\|_{3} \leq \|\mathbf{u}\|_{2} \leq 2\|\mathbf{u}\|_{3}\)By these equations we can say the topologies generated by these norms are indeed the same.
Key Concepts
Properties of NormsTopology in Normed SpacesOpen Balls in Normed SpacesVector Space Concepts
Properties of Norms
A norm is a function that assigns a positive length or size to vectors in a vector space. In the context of \(\mathbb{R}^{2}\), a norm satisfies four key properties:
- Positive Definiteness: For any vector \(\mathbf{u}\), the norm \(\|\mathbf{u}\|\) is always non-negative, and it equals zero only if \(\mathbf{u} = \mathbf{0}\).
- Absolute Scalability: For any scalar \(c\) and vector \(\mathbf{u}\), the norm satisfies \(\|c \cdot \mathbf{u}\| = |c| \cdot \|\mathbf{u}\|\).
- Triangle Inequality: For any vectors \(\mathbf{u}, \mathbf{v}\), the norm satisfies the inequality \(\|\mathbf{u} + \mathbf{v}\| \leq \|\mathbf{u}\| + \|\mathbf{v}\|\).
- Positive Homogeneity: For any vector \(\mathbf{u}\) and positive scalar \(c\), \(\|c \cdot \mathbf{u}\| = c \cdot \|\mathbf{u}\|\).
Topology in Normed Spaces
In normed vector spaces, topology relates to how we understand the concept of "closeness" or "continuity." Norms define open sets, which are the building blocks of topological spaces. Here's how norms shape topology:
- Open Sets: A set is open in a normed space if, for each point in the set, there exists an open ball entirely contained within the set. This means you can move a little from any point and still remain within the set.
- Continuous Functions: Functions between normed spaces preserve limits, enhancing how we handle continuous transformations across spaces.
Open Balls in Normed Spaces
Open balls are central to understanding norms in vector spaces. An open ball \(B_{a}(\mathrm{u})\) with center \(\mathrm{u}\) and radius \(a\) is defined as the set of all points that are closer than \(a\) from \(\mathrm{u}\) based on a given norm.
- Disk (Euclidean Norm): For \(\|\mathbf{u}\|_{1}\), the open balls are disks, or circular regions centered at \(\mathbf{u}\). These are visually intuitive, aligning with how we often perceive "distance" geometrically.
- Square (Maximum Norm): The norm \(\|\mathbf{u}\|_{2}\) results in squares for open balls, where each side is \(2a\). While less intuitive, this format efficiently captures maximum distances along axes.
- Diamond (Manhattan Norm): The open balls in the Manhattan norm \(\|\mathbf{u}\|_{3}\) are shaped as diamonds. Each vertex of the diamond is aligned along axes, representing a notable variation from circular intuitive measures.
Vector Space Concepts
Understanding vector space concepts is fundamental for operating within any normed space. Here are some foundational elements:
- Vectors: Objects with both magnitude and direction, serving as primary elements in these spaces.
- Linear Combinations: Any vector in space can be derived from a combination of other base vectors, showcasing how vector spaces are structured.
- Basis and Dimension: The basis of a vector space is a minimal set of vectors that spans the entire space. The dimension of the vector space is the number of vectors in this set, critical for understanding the space's scope.
Other exercises in this chapter
Problem 22
If \(G_{0}\) is the component of the identity of a locally connected topological group \(G\), the factor group \(G / G_{0}\) is called the group of components o
View solution Problem 24
Show that a linear map \(T: V \rightarrow W\) between topological vector spaces is continuous everywhere on \(V\) if and only if it is continuous at the origin
View solution Problem 29
Show that if \(x_{n} \rightarrow x\) in a normed vector space then $$ \frac{x_{1}+x_{2}+\cdots+x_{n}}{n} \rightarrow x $$
View solution Problem 30
Show that if \(x_{n}\) is a sequence in anormed vector space \(V\) such that every subsequence has a subsequence comergent to \(x\), then \(x_{n} \rightarrow x\
View solution