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Chapter 1: Problem 3

Let \(C[-1,2]\) denote the space of continuous complex-valued functions \(f:[-1,2] \rightarrow \mathbb{C}\). Which of the following define an inner product on \(C[-1,2]\), and which do not? Explain. (a) \(\langle f, g\rangle=\int_{-1}^{2}|f(t)+g(t)| d t\) (b) \(\langle f, g\rangle=\int_{-1}^{2} f(t) \overline{g(t)} d t+f\left(-\frac{1}{2}\right) \overline{g\left(-\frac{1}{2}\right)}\) (c) \(\langle f, g\rangle=3 \int_{-1}^{2} f(t) \overline{g(t)} d t\) (d) \(\langle f, g\rangle=f(0) \overline{g(0)}+f(1) \overline{g(1)}\)

Short Answer

Expert verified
Only options (b), (c), and (d) define an inner product on \(C[-1,2]\).

Step by step solution

01

Define Inner Product Axioms

An inner product on a vector space must satisfy the following axioms: 1) Conjugate Symmetry: \( \langle f, g \rangle = \overline{\langle g, f \rangle} \) for all \( f, g \); 2) Linearity: \( \langle af+bh, g \rangle = a\langle f, g \rangle + b\langle h, g \rangle \) for all scalars \( a, b \) and functions \( f, h \); 3) Positive-definiteness: \( \langle f, f \rangle \geq 0 \), and \( \langle f, f \rangle = 0 \) if and only if \( f = 0 \).
02

Analyze Option (a)

The expression is \( \langle f, g \rangle = \int_{-1}^{2} |f(t)+g(t)| dt \). This does not satisfy the linearity condition because the absolute value operator \( | \cdot | \) does not preserve linear combinations of \( f \) and \( g \). Therefore, it is not an inner product.
03

Check Conjugate Symmetry for (a)

Check if \( \langle f, g \rangle = \overline{\langle g, f \rangle} \). For option (a), conjugate symmetry is not guaranteed because the expression is inherently scalar (real-valued) due to the absolute value, thus not complex conjugating \( g(t) \). Instead, it provides a real number which does not exhibit the required behavior of switching roles of \( f \) and \( g \). Thus, it does not exhibit conjugate symmetry.
04

Analyze Option (b)

The expression is \( \langle f, g \rangle = \int_{-1}^{2} f(t) \overline{g(t)} dt + f\left(-\frac{1}{2}\right) \overline{g\left(-\frac{1}{2}\right)} \). This satisfies conjugate symmetry, linearity due to the properties of integrals and the sum of products of functions, and also can be shown to be positive definite for \( \langle f, f \rangle \). Thus, it defines an inner product.
05

Analyze Option (c)

The expression is \( \langle f, g \rangle = 3 \int_{-1}^{2} f(t) \overline{g(t)} dt \). Any scalar multiple of an inner product (where scalar is positive) remains an inner product because it retains conjugate symmetry, linearity, and positive-definiteness due to positive scalar multiplication properties. Therefore, this is a valid inner product.
06

Analyze Option (d)

The expression is \( \langle f, g \rangle = f(0) \overline{g(0)} + f(1) \overline{g(1)} \). This satisfies conjugate symmetry, linearity, and positive-definiteness because if \( \langle f, f \rangle = 0 \) then \( f(0) = 0 \) and \( f(1) = 0 \), suggesting \( f(t) = 0 \) for the required points, although not explicitly over the whole interval unless additional rules are imposed. It does satisfy conditions for subset analysis so it defines an inner product.

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Key Concepts

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

Vector Spaces
In mathematics, a vector space is essentially a collection of objects, known as vectors, which can be added together and multiplied by scalars. These scalars can be real or complex numbers, depending on the type of vector space you are dealing with. In our exercise, we are focusing on the space of continuous complex-valued functions on the interval \([-1, 2]\), denoted as \(C[-1,2]\). This set qualifies as a vector space because it satisfies two essential properties:
  • Closure under addition: If you take any two functions \(f(t)\) and \(g(t)\) from this space, then \(f(t) + g(t)\) is also in this space.
  • Closure under scalar multiplication: Multiplying a function by any scalar \(a\) results in another function \(af(t)\), which also belongs to the same space.
The space is important because it allows us to apply transformations, operations, and, in this exercise, determine which expressions can qualify as inner products.
Conjugate Symmetry
Conjugate symmetry is a vital property for an expression to be considered an inner product. It involves switching the roles of two functions, say \(f\) and \(g\), and expecting the result to be the complex conjugate of the original. Mathematically, this is represented as:
\[\langle f, g \rangle = \overline{\langle g, f \rangle}\]
This property ensures that the inner product behaves well under transformations such as reflection, maintaining the symmetry of the system involved.
  • The expression in option (b), \(\int_{-1}^{2} f(t) \overline{g(t)} dt + f\left(-\frac{1}{2}\right) \overline{g\left(-\frac{1}{2}\right)}\), clearly maintains conjugate symmetry since each term separately respects the property.
  • Option (a), \(\int_{-1}^{2} |f(t) + g(t)| dt\), however, fails here as the absolute value in the integrand does not retain conjugate behavior when switching \(f\) and \(g\).
Linearity
Linearity deals with how an expression reacts to addition and scalar multiplication. An inner product must respect linear transformations, ensuring consistency in scaling and summing functions:
\[\langle af + bh, g \rangle = a\langle f, g \rangle + b\langle h, g \rangle\]Here, \(a\) and \(b\) are scalars, and \(f\), \(h\), and \(g\) are functions from the vector space. This property ensures that the operation scales functions as expected and is distributive.
  • Expression (b) honors linearity because integrals naturally respect addition and scalar multiplication.
  • On the other hand, (a) falters under linearity due to the absolute value applied, disrupting linear combinations of \(f\) and \(g\).
Linearity is essential for the predictability and functionality of inner products, allowing coherent and consistent operations within the vector space.
Positive-definiteness
A key requirement for functions to define an inner product space is positive-definiteness. This condition demands that the inner product of any function with itself should always result in a non-negative value. Furthermore, it only reaches zero if the function is essentially the zero function:
\[\langle f, f \rangle \geq 0\]
Moreover, if \(\langle f, f \rangle = 0\), it must mean that \(f = 0\).
  • Expressions like (b) and (c) are positive-definite. Option (b) subtly checks all values of the function over the interval and at specific points, while (c) simply scales the integral, preserving its positivity as a result of the positive scalar.
  • Option (d) focuses on select points (0 and 1), and efficiently checks positivity for subsets within \(C[-1,2]\).
Positive-definiteness provides a measure of magnitude or norm for vectors or, in this case, functions, paving the way for us to measure lengths and angles, analogously to physical vectors.

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Most popular questions from this chapter

Let \(N\) be a natural number. For each \(m=0,1, \ldots, N-1\), we define a vector in \(\mathbb{C}^{N}\) by $$ u_{m}=\left(1, e^{\frac{-2 \pi i m}{N}}, e^{\frac{-4 \pi i m}{N}}, \ldots, e^{\frac{-2(N-1) \pi i m}{N}}\right) . $$ Prove that: (a) The system of vectors \(\left\\{u_{m}\right\\}_{m=0}^{N-1}\) is an orthogonal system in \(\mathbb{C}^{N}\) with the standard inner product. (b) For each \(m=0,1, \ldots, N-1,\left\|u_{m}\right\|^{2}=N\). (c) For every \(u \in \mathbb{C}^{N}\), we have \(u=\frac{1}{N} \sum_{m=0}^{N-1}\left\langle u, u_{m}\right\rangle u_{m}\).

Let \(C[-1,1]\) be the space of continuous functions \(f:[-1,1] \rightarrow \mathbb{C}\) with inner product $$ \langle f, g\rangle=\int_{-1}^{1} f(x) \overline{g(x)} d x $$ (a) Let \(P_{0}(x)=1, P_{1}(x)=x\), and \(P_{2}(x)=1-3 x^{2}\). Prove that this set of polynomials is orthogonal in \(C[-1,1]\). (b) Find constants \(a, b\), and \(c\) such that the polynomial $$ P_{3}(x)=a+b x+c x^{2}+x^{3} $$ is orthogonal (perpendicular) to each of the previous polynomials.

Let \(\mathcal{P}_{2}\) be the space of real polynomials of degree less than or equal to 2 . To each \(f_{1} g \in \mathcal{P}_{2}\), define $$ \langle f, g\rangle=\int_{0}^{\infty} f(x) g(x) e^{-x} d x $$ (a) Prove that this is an inner product on \(\mathcal{P}_{2}\). (b) Show that the set \(\left\\{1,1-x, 1-2 x+\frac{1}{2} x^{2}\right\\}\) is an orthonormal system with respect to this inner product.

Let \(V=C^{2}[-\pi, \pi]\) be the space of real-valued twice continuously differentiable functions defined on the interval \([-\pi, \pi]\). Set $$ \langle f, g\rangle=f(-\pi) g(-\pi)+\int_{-\pi}^{\pi} f^{\prime \prime}(x) g^{\prime \prime}(x) d x . $$ Is this an inner product on \(V\) ?

Let \(C^{1}[0,1]\) be the space of continuous functions \(f:[0,1] \rightarrow \mathbb{C}\) with a continuous first derivative on \([0,1]\). (a) Prove that $$ \langle f, g\rangle=f(0) \cdot \overline{g(0)}+\int_{0}^{1} f^{\prime}(x) \overline{g^{\prime}(x)} d x $$ is an inner product on \(C^{1}[0,1]\). (b) Find an orthonormal system \(\left\\{h_{1}, h_{2}, h_{3}\right\\}\) in \(C^{1}[0,1]\), with respect to this inner product, for which $$ \operatorname{span}\left\\{h_{1}, h_{2}, h_{3}\right\\}=\operatorname{span}\left\\{1, x, x^{2}\right\\} . $$
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