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Chapter 2: Problem 5

If \(x\) and \(y\) are in a normed linear space \(X\) and \(\|x+y\|=\|x\|+\|y\|\) prove that \(\|x+\lambda y\|=\|x\|+\lambda\|y\|\), for all \(\lambda>0\). (Hint: there are two cases, \(\lambda>1\) and \(\lambda \leq 1\),)

Short Answer

Expert verified
Based on the step-by-step solution provided, here is a short answer: To prove that in a normed linear space X, if for x and y in X, it holds that ∥x+y∥=∥x∥+∥y∥, then for any scalar λ>0, we have ∥x+λy∥=∥x∥+λ∥y∥, consider two cases: when λ>1 and when λ≤1. For the case λ>1, apply the triangle inequality and use the assumed equation and the homogeneity of norms to show that equality holds for λ>1. Similarly, for the case λ≤1, apply the triangle inequality and use the homogeneity of norms to show that equality holds for λ≤1. Since both cases hold, we conclude that the equation ∥x+λy∥=∥x∥+λ∥y∥ is true for all λ>0.

Step by step solution

01

Case 1: \(\lambda > 1\)

Consider the scalar \(\mu = \frac{1}{\lambda}\). Then, \(0< \mu < 1\). Let's find the norm of \(x + \mu (x + y)\): \[\|x + \mu (x + y)\| = \|x + \mu x + \mu y\| = \|(1 + \mu)x + \mu y\|.\] Now let's apply the inequality \(\|u + v\| \leq \|u\| + \|v\|\) for vectors \(u = (1+\mu)x\) and \(v = \mu y\): \[\|(1 + \mu)x + \mu y\| \leq \|(1 + \mu)x\| + \|\mu y\|.\] Notice that \((1+\mu)x = x + \mu x\) and that \(\|x+\mu x\| = (1+\mu)\|x\|\) since norms are homogeneous. Also, observe that \(\|\mu y\| = \mu\|y\|\) due to the homogeneity of norms. So, the inequality becomes: \[(1+\mu)\|x\| + \mu\|y\| \geq \|(1 + \mu)x + \mu y\|.\] Rewrite the left-hand side using \(\mu = \tfrac{1}{\lambda}\), and then multiply both sides by \(\lambda\): \[(1 + \tfrac{1}{\lambda})\|x\| + \tfrac{1}{\lambda}\|y\| \geq \tfrac{1}{\lambda}\|(1 + \tfrac{1}{\lambda})x + \tfrac{1}{\lambda}y\| \Rightarrow \|x\| + \|y\| \geq \|x + \lambda y\|.\] Since we assumed that \(\|x + y\| = \|x\| + \|y\|\), we have that \(\|x\| + \|y\| = \|x + \lambda y\|\) for \(\lambda > 1\).
02

Case 2: \(\lambda\leq1\)

Since \(\lambda \leq 1\), we can rewrite \(x = (x + \lambda y) + (1-\lambda)y\). Then, we use the triangle inequality: \[\|x\| \leq \|(x + \lambda y) + (1-\lambda)y\| \leq \|x + \lambda y\| + \|(1-\lambda)y\|.\] Due to the homogeneity of norms, the last term equals \((1-\lambda)\|y\|\): \[\|x\| \leq \|x + \lambda y\| + (1-\lambda)\|y\|.\] Now we can subtract \((1-\lambda)\|y\|\) from both sides, which yields: \[\|x\| - (1-\lambda)\|y\| \leq \|x + \lambda y\|.\] Rearrange this inequality, we get: \[\|x\| + \lambda\|y\| \leq \|x + \lambda y\|.\] Since we also have the inequality \(\|x+y\|\leq\|x\|+\|y\|\), we already know that \(\|x+\lambda y\|\geq\|x\|+\lambda\|y\|\) for \(\lambda\leq1\). Thus, both inequalities hold: \[\|x\|+\lambda\|y\|\leq\|x+\lambda y\|\leq\|x\|+\lambda\|y\|.\] This implies that equality holds, so \(\|x+\lambda y\|=\|x\|+\lambda\|y\|\) for any \(\lambda \leq 1\). Combining the two cases, we showed that this is true for all \(\lambda > 0 \).

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Key Concepts

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

The Triangle Inequality
One of the most important properties of normed linear spaces is the triangle inequality. This property states that for any two vectors, say x and y, in a normed space X, the norm of their sum is less than or equal to the sum of their norms, mathematically expressed as \[\|x+y\| \leq \|x\| + \|y\| \.\] This inequality is akin to the statement in Euclidean geometry that the sum of the lengths of any two sides of a triangle is greater than or equal to the length of the remaining side.

When the triangle inequality holds as an equality, as in the exercise you're studying, it signals a linear relationship between x and y. This is because the 'triangle' formed by the points 0, x, and x+y collapses into a straight line. In such a special case, extending one of the vectors by any positive scalar \(\lambda\) and examining the norm of the resultant vector can yield insightful results which confirm that the equality extends to the scaled scenario.

In summary, the triangle inequality provides a structural constraint that normed spaces adhere to, ensuring that the norm behaves well under vector addition.
Homogeneity of Norms
Another key feature of a normed linear space is the homogeneity of norms. This concept essentially means that when a vector x is scaled by a scalar \(\lambda\), the norm of the result \(\|\lambda x\|\) is equal to the absolute value of \(\lambda\) times the norm of x. Formally, it's represented as \[\|\lambda x\| = |\lambda|\|x\|,\] for all \(\lambda\) in the field over which the space is defined (usually real or complex numbers) and for all vectors x in the space.

This property is crucial in ensuring that the norm is consistent with scalar multiplication, one of the basic operations in linear algebra. For instance, if a vector's length is doubled, then its norm should also double. This is why, in the solved problem, when a vector is multiplied by a scalar \(\mu < 1\), the norm of the resulting vector is the norm of the initial vector multiplied by \(\mu\).

Homogeneity makes it possible to extend conclusions drawn from basic vector relationships, such as the equality condition of the triangle inequality, to scenarios involving scalar multiples of these vectors.
Scalar Multiplication in Normed Spaces
Scalar multiplication in normed linear spaces marries the idea of scaling vectors with the structure provided by the norm. Considering the equation from the exercise \(\|x+\lambda y\| = \|x\| + \lambda\|y\|\), for any positive scalar \(\lambda\), it leverages the properties of homogeneity. The crux of this multiplication is that norms interact predictably with scalar multiples, maintaining the proportionality of the vector's size.

The exercise walks through two cases based on the scalar's size relative to 1. This distinction is essential because it reflects how norms behave differently in 'stretching' versus 'shrinking' a vector. For \(\lambda > 1\), you're effectively checking if the norm 'stretches' proportionally. Conversely, for \(\lambda \leq 1\), it's a matter of whether 'shrinking' the vector preserves the proportionality of the norm. The exercise brilliantly illustrates that, under both circumstances, the norm of a scalar multiple of a vector maintains its proportional relationship to the original norm, confirming the linear nature of the space.

Understanding scalar multiplication's effect on norms is vital for grasping more complex concepts in functional analysis and linear algebra, where vectors are regularly scaled as part of broader operations.

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Most popular questions from this chapter

The first two exercises show how to verify that \(\|\cdot\|_{2}\) defines a. norm on \(\mathbb{R}^{n} .\) Define a mapping from \(\mathbb{R}^{n} \times \mathbb{R}^{n} \rightarrow \mathbb{R}\) by $$ g(x, y)=\sum_{i=1}^{n} x_{i} y_{i} $$ where \(x=\left(x_{1}, x_{2}, \ldots, x_{n}\right)\) and \(y=\left(y_{1}, y_{2}, \ldots, y_{n}\right) .\) Show that (i) \(g(x, x)=\|x\|_{2}^{2}\) for all \(x \in \mathbb{R}^{n}\) (ii) \(g(x, y)=g(y, x)\) for all \(x, y \in \mathbb{R}^{n}\) (iii) \(g(x, y+z)=g(x, y)+g(x, z)\) and \(g(x+y, z)=g(x, z)+g(y, z)\) for all \(x, y, z \in \mathbb{R}^{n}\) (iv) \(g(\lambda x, y)=\lambda g(x, y)=g(x, \lambda y)\) for all \(x, y \in \mathbb{R}^{n}\) and \(\lambda \in \mathbb{R}\).

Prove that the following define continuous functions from \(D\) to R: (i) \(D=\\{x \in \mathbb{R}: x \geq 0\\}\) and \(f(x)=\sqrt{x}\) (ii) \(D=\mathbb{R}\) and \(f(s)=a_{n} s^{n}+a_{n-1} s^{n-1}+\cdots+a_{0}\), where \(a_{0}, a_{1}, \ldots a_{n}\) are fixed real numbers.

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

Show that if \(f\) and \(g\) are bounded, \(K\)-Lipschitz functions from the normed linear space \(X\) to \(\mathbb{R}\) then \(f . g\) is Lipschitz. Give a counterexample when \(f\) and \(g\) are unbounded.

Let \(\mathbb{R}\) have the usual norm. Calculate (i) \(\operatorname{dist}\left(1,\left\\{t: t^{2}>2\right\\}\right)\) (ii) \(\operatorname{dist}(0,\\{t: t=1 / n, n \in \mathbb{N}\\})\).
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