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Chapter 2: Problem 4

Let \(t_{1}, \ldots, t_{k}\) be distinct points in \([a, b]\), and for \(f \in \mathcal{P}_{n}[a, b]\) define $$ \nu(f)=\sum_{j=1}^{k}\left|f\left(t_{j}\right)\right| $$ Show that \(\nu\) is a norm on \(\mathcal{P}_{n}[a, b]\) if and only if \(k \geq n+1\).

Short Answer

Expert verified
The functional \(u(f)\) is a norm on \(\mathcal{P}_n[a, b]\) if and only if \(k \geq n+1\).

Step by step solution

01

Understand the Norm Criteria

Recall that a norm must satisfy the following properties for all functions f and g in the space and all scalars c: 1) \(u(f) \geq 0\) and \(u(f) = 0 \, \text{if and only if} \, f = 0 \). 2) \(u(cf) = |c| u(f)\). 3) \(u(f + g) \leq u(f) + u(g)\). We need to show that \(u(f) = \sum_{j=1}^{k}|f(t_{j})|\) satisfies these properties if and only if \(k \geq n+1\).
02

Verify Non-negativity and Definiteness

Since \(\left|f(t_j)\right|\geq0\) for all \(j\), \(u(f)\geq 0\). If \(u(f) = 0\), then \(\sum_{j=1}^{k}\left|f(t_j)\right| = 0\), meaning \(\left|f(t_j)\right| = 0\) for all \(j\), and thus \(f(t_j) = 0\) for all \(t_j\). For polynomials, this implies that \(f = 0\) if \(f(t_j)=0\) at \(k=n+1\) distinct points.
03

Check for Absolute Homogeneity

Given a scalar \(c \) and polynomial \(f \, u(cf) = \sum_{j=1}^{k}\left|cf(t_j)\right| = \sum_{j=1}^{k}|c||f(t_j)| = |c| \sum_{j=1}^{k}\left|f(t_j)\right| = |c| \u(f)\). Clearly, \(u(cf) = |c|u(f)\).
04

Verify Subadditivity

To show that \(u(f + g) = \sum_{j=1}^{k}\left|f(t_j) + g(t_j)\right| \leq \sum_{j=1}^{k} (\left|f(t_j)\right| + \left|g(t_j)\right|) = \sum_{j=1}^{k}\left|f(t_j)\right| + \sum_{j=1}^{k}\left|g(t_j)\right| = \u(f) + \u(g)\). This follows directly from the triangle inequality for real numbers.
05

Condition for the Norm Property

For the norm \(u(f) = \sum_{j=1}^{k}\left|f(t_j)\right|\) to imply \(f = 0\) when \(u(f) = 0\), \(f(t_j)\) must be zero at each point \(t_j = 1, 2, \ldots, k\). A polynomial of degree \(n\) has at most \(n\) zeros unless it is the zero polynomial. Therefore, \(k \, \text{must be greater than or equal to} \, n+1\) to ensure that if \(u(f) = 0\), then \f must be the zero polynomial.

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Key Concepts

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

Norm Properties in Polynomial Spaces
In mathematics, a norm is a function that assigns a strictly non-negative length or size to each vector in a vector space, except for the zero vector. For any polynomial function \(f\) and scalar \(c\), it must fulfill three main criteria:
  • Non-negativity and definiteness: \(u(f) \geq 0\) and \(u(f) = 0\) if and only if \(f = 0\).
  • Absolute homogeneity: \(u(cf) = |c| u(f)\).
  • Subadditivity (Triangle Inequality): \(u(f + g) \leq u(f) + u(g)\).
For the functional \(u(f) = \sum_{j=1}^k |f(t_j)|\) to be a norm on the polynomial space \(\mathcal{P}_n[a, b]\), one must prove it satisfies these conditions specifically and that it works only if \(k \geq n+1\).
Polynomial Functions
Polynomial functions are expressions consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents. In a space \(\mathcal{P}_n[a, b]\), polynomials are of degree at most \(n\), meaning the highest power of the variable is \(n\).
An example is \(f(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n\). When dealing with norms in polynomial functions, it鈥檚 vital to understand their behavior at various points, denoted as \(t_1, t_2, ..., t_k\). This helps in analyzing if certain properties like non-negativity, definiteness, homogeneity, and subadditivity hold.
The Zero Polynomial
A zero polynomial is a polynomial whose coefficients are all zero, resulting in every term becoming zero, represented as \(f(x) = 0\). When defining norms, the property of definiteness holds if for \(f\), \(u(f) = 0\) implies \(f = 0\). For example, if our functional \(u(f) = \sum_{j=1}^k |f(t_j)|\) is zero, it must mean that \(|f(t_j)| = 0\) for all points \(t_j\). For polynomials up to degree \(n\), having \(\geq n+1\) distinct points \(t_j\) ensures that \(f\) must indeed be the zero polynomial if its value and norms are zero at these points.
This is because a polynomial of degree \(n\) that has more than \(n\) zeros must be the zero polynomial.
Homogeneity
Homogeneity in the context of norms means that scaling the polynomial by a constant \(c\) scales the norm by \(|c|\). Mathematically, if \(u\) is a norm on a polynomial space \(\mathcal{P}_n[a, b]\), then for any polynomial \(f\) and scalar \(c\), the property \(u(cf) = |c|u(f)\) must hold. For our functional, let鈥檚 verify:
  • \(u(cf) = \sum_{j=1}^k |cf(t_j)|\)
  • = \(\sum_{j=1}^k |c||f(t_j)|\)
  • = \(|c| \sum_{j=1}^k |f(t_j)|\)
  • = \(|c| u(f)\).
This clearly demonstrates absolute homogeneity, meaning our functional meets this criteria of being a norm.
Subadditivity
Subadditivity, commonly known as the triangle inequality, requires that for all polynomial functions \(f\) and \(g\) in the space \(\mathcal{P}_n[a, b]\):
\(u(f+g) \leq u(f) + u(g)\).
For the functional \(u(f) = \sum_{j=1}^k |f(t_j)|\), this property can be verified as follows:
  • \(u(f+g) = \sum_{j=1}^k |f(t_j) + g(t_j)|\)
  • \leq \(\sum_{j=1}^k (|f(t_j)| + |g(t_j)|)\)
  • = \(\sum_{j=1}^k |f(t_j)| + \sum_{j=1}^k |g(t_j)|\)
  • = \(u(f) + u(g)\).
This inequality stems from the triangle inequality for real numbers and confirms that our functional satisfies the subadditivity property, completing the norm verification for the polynomial space.

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Most popular questions from this chapter

Let \(\mathcal{N}\) be the set of all norms on a linear space \(X\) such that \(\sup _{\nu \in \mathcal{N}} \nu(x)<\infty\) for every \(x \in X\). Show that \(x \mapsto \sup _{\nu \in \mathcal{N}} \nu(x)\) is a norm on \(X\).

Let \(X\) be a linear space, \(Y\) be a normed linear space, and \(A: X \rightarrow Y\) be a linear operator Suppose \(y_{0} \in R(A)\) and \(r>0\) are such that the equation \(A x=y\) has a solution for every \(y \in B_{Y}\left(y_{0}, r\right)\). Show that \(A x=y\) has a solution for every \(y \in Y\).

Let \(X\) be a normed linear space. Consider the norms $$ \begin{array}{l} (x, y) \mapsto\|x\|+\|y\|, \quad(x, y) \in X \times X, \\ (\alpha, x) \mapsto|\alpha|+\|x\|, \quad(\alpha, x) \in \mathbb{K} \times X \end{array} $$ on \(X \times X\) and \(\mathbb{K} \times X\), respectively. Show that the maps $$ (x, y) \mapsto x+y, \quad x \mapsto-x, \quad(\alpha, x) \mapsto \alpha x $$ are continuous functions. Show also that for each \(u \in X\) and nonzero \(\alpha \in \mathbb{K}\), the maps $$ x \mapsto x+u, \quad x \mapsto \alpha x $$ are homeomorphisms from \(X\) onto itself, i.e., these maps are bijective, continuous, and their inverses are also continuous.

Let \(X\) be an inner product space over \(\mathbb{C}\) and \(x, y \in X\). Show that \(x \perp y\) if and only if \(\|\alpha x+\beta y\|^{2}=\|\alpha x\|^{2}+\|\beta y\|^{2}\) for every \(\alpha, \beta \in \mathbb{C}\) [Hint: Look at the relation when \(\alpha=1, \beta=\langle x, y\rangle .]\)

Let \(X=C[-1,1]\) be equipped with \(L^{2}\) -inner product, $$ \begin{array}{l} S_{1}=\\{f \in X: f(-t)=f(t) \quad \forall t \in[-1,1]\\} \\ S_{2}=\\{f \in X: f(-t)=-f(t) \quad \forall t \in[-1,1]\\} \end{array} $$ Show that \(S_{1} \perp S_{2}\), i.e., \((f, g\rangle=0\) for every \((f, g) \in S_{1} \times S_{2}\). Also, show that $$ S_{1}^{\perp}=S_{2}, \quad S_{2}^{\perp}=S_{1}, \quad X=S_{1}+S_{2} $$
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