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Let \(X=\mathbb{R}^{2}\) with the norm defined by $$ \|(s, t)\|=|s|+|t| $$ Let \(A=\\{(s, t): s=t\\} .\) What is dist \((y, A)\), when \(y=(0,1) ?\) Describe the set of best approximations to \(y\) from \(A\).

Short Answer

Expert verified
Answer: The distance between the point y=(0,1) and the set A is 2, and the set of best approximations to y from A is {(1,1)}.

Step by step solution

01

Find the projection of y onto A

To find the projection of y=(0,1) onto A={(s,t): s=t}, we first need to identify the line representing A. Since s=t, we can represent A as the line y=x in the xy-plane. The projection of y onto A is the closest point on A to y. We can find this point by considering a general point on A, given by (s,s), and then minimizing the distance from this point to y.
02

Calculate the distance between y and the projection

Let's consider an arbitrary point (s,s) on A. We now calculate the distance between this point and y using the given norm: \(\|(s, s) - (0, 1)\| = \|(s, s-1)\| = |s| + |s-1|\) To find the distance between y and the projection, we need to find the minimum value of this expression. The minimum value occurs when s is as close as possible to 1. In this case, let s = 1. This gives: \(\|(1, 0) - (0, 1)\| = \|(1, -1)\| = |-1| + |1| = 2\) Thus, the distance between y and the projection onto A (or dist(y, A)) is 2.
03

Describe the set of best approximations to y from A

Since (1,1) is the point on A that is closest to y, it is the best approximation. In this case, there is only one point that minimizes the distance between y and A. Therefore, the set of best approximations to y from A is: {(1,1)}

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Key Concepts

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

Best Approximations
When we talk about best approximations in a normed vector space, we are referring to the points in a given set that are closest to a particular point outside the set. These approximations are essential in various mathematical and engineering applications, such as optimization and signal processing.

In our exercise, the set A represents a line in \(\mathbb{R}^{2}\), and we want to find which point on this line is nearest to the point y=(0,1). This point is referred to as the projection of y onto the set A and represents our best approximation. The concept relies on minimizing the distance, measured by the norm, between the point y and the points on the line A. In simpler terms, we are looking for a point in the set A that is the least 'far away' from point y when measured in a straight line, which, as we've seen in the solution, turns out to be the point (1,1). This single point, (1,1), makes up the whole set of best approximations of y from A as there are no other points in A that are closer to y.
Distance in Normed Spaces
Distance in normed spaces refers to the measure of how 'far apart' two elements are, with respect to the given norm. Each normed space comes with a unique method of measuring distances between its elements. The norm on \(\mathbb{R}^{2}\) given in our exercise is defined by \(\|(s, t)\| = |s| + |t|\), which is not the standard Euclidean norm but another valid way to measure distances.

Using this norm, to calculate the distance between any two points, you take the sum of the absolute values of the differences of their respective coordinates. From the exercise, the calculated distance between the point y and its projection on A is 2. This distance effectively quantifies how 'far' y=(0,1) is from the set A. Understanding the notion of distance in various norms is crucial, especially when norms provide different 'shapes' of distance, which might be relevant in areas like approximation theory, computer science, and physics.
Projections onto a Set
The concept of projections onto a set in a normed vector space is similar to the intuitive idea of a shadow cast by an object onto a surface. Mathematically, when we project a point onto a set, we're searching for the point within the set that is closest to the original point, according to the space's norm. This close point is what we call the projection.

In our exercise, we looked for the projection of the point y onto the set A by minimizing the norm-defined distance. This process is very common in applications such as numerical analysis and computer graphics, where projections are used to reduce dimensions, minimize errors, and simplify complicated models. The methodology can differ slightly depending on the norm or the structure of the set we project onto, but fundamentally, it revolves around the concept of finding the nearest neighbor in a mathematical sense.

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Most popular questions from this chapter

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be continuous on the interval \([a, b]\). Define \(g:[a, b] \rightarrow \mathbb{R}\) by \(g(s)=\max \\{f(t): t \in[a, s]\\}\) if \(s \neq a\) and \(g(a)=f(a)\). Show that \(g\) is continuous on \([a, b]\). We shall conclude this section by examining the continuity of functions. on the linear space \(\mathbb{R}^{n}\). Recall that \(\mathbf{R}^{n}\) is the set of \(\mathrm{n}\)-tuples of the form $$ \left(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\right) $$ where each entry \(x_{i}, 1 \leq i \leq n\), is a real number. The usual definitions of addition and multiplication by scalars are adopted, so that addition of two n-tuples is effected by adding the individual components and multiplication by a scalar simply means that each member of the ntuple is multiplied by that scalar. Hence, $$ \left(x_{1}, x_{2}, \ldots, x_{n}\right)+\left(y_{1}, y_{2}, \ldots, y_{n}\right)=\left(x_{1}+y_{1}, x_{2}+y_{2}, \ldots, x_{n}+y_{n}\right) $$ and $$ \alpha\left(x_{1}, x_{2}, \ldots, x_{n}\right)=\left(\alpha x_{1}, \alpha x_{2}, \ldots, \alpha x_{n}\right) $$ There are three principal definitions which we adopt for the norm in \(\mathbb{R}^{n}\). If we assume that \(x=\left(x_{1}, x_{2}, \ldots, x_{n}\right)\) then these norms are defined as follows: $$ \|x\|_{2}=\left\\{\sum_{i=1}^{n} x_{i}^{2}\right\\}^{\frac{1}{2}}, \quad\|x\|_{1}=\sum_{i=1}^{n}\left|x_{i}\right|, \quad\|x\|_{\infty}=\max \left\\{\left|x_{i}\right|: 1 \leq i \leq n\right\\} $$ It is perhaps a little surprising that the verification that the first of these definitions really has the properties required of a norm is quite difficult. The other two are straightforward, but one of the exercises guides you through the proof of the first case. There are sensible historical reasons for the subscripting of these norms, which come from a much larger family of norms on \(\mathbb{R}^{n}\). Throughout this section the symbol \(\|\cdot\|\) is used to denote any one of the three norms above and unless mention is made of a particular norm any statement about convergence or continuity should be understood to refer to each of the three norms. The reason for this ambiguity is contained in the next lemma

Suppose \(f, g: \mathbb{R}^{n} \rightarrow \mathbb{R}\) are continuous at \(x \in \mathbb{R}^{n}\), and \(g(x) \neq 0\). Show that \(f / g\) is continuous at \(x\).

Let \(X=C[-1,1]\) with norm \(\|x\|=\sup \\{|x(t)|: t \in[-1,1]\\}\), for \(x \in X\). Let \(\mathrm{A}\) be the set of constant functions in \(C[-1,1]\). Let \(x \in X\) be defined by \(x(t)=t^{2}\). Compute dist \((x, A)\). Does \(x\) have a best approximation in \(A\) ?

Suppose in a normed linear space \(X, x_{n} \rightarrow x\) and \(y_{n} \rightarrow y\). Show that \(\left\|x_{n}\right\| \leq\left\|y_{n}\right\|\) for all \(n \in \mathbb{N}\) implies \(\|x\| \leq\|y\|\). What can be concluded if \(\left\|x_{n}\right\|<\left\|y_{n}\right\|\) for all \(n \in \mathbb{N}\) ? What happens if \(\left\|x_{n}\right\|=\left\|y_{n}\right\|\) for all \(n \in \mathbb{N}\) ?

Show that \(f: \mathbb{R} \rightarrow \mathbb{R}\) defined by \(f(s)=s^{2}\) is not uniformly continuous on \(\mathbb{R}\).

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