(I) Introduction

(Based on Theorem 1 of Section 6.1 for \( \mathbb{R}^n \))

Definition

An inner product on a vector space \( V \) is a function that, to each pair of vectors \( \vec{u} \) and \( \vec{v} \) in \( V \), associates a real number \( \langle\vec{u}, \vec{v}\rangle \) and satisfies the following axioms for all \( \vec{u}, \vec{v}, \vec{w} \) in \( V \) and all scalars \( c \):

  1. \( \langle\vec{u}, \vec{v}\rangle = \langle\vec{v}, \vec{u}\rangle \)
  2. \( \langle c\vec{u}, \vec{v}\rangle = c\langle\vec{u}, \vec{v}\rangle \)
  3. \( \langle\vec{u} + \vec{v}, \vec{w}\rangle = \langle\vec{u}, \vec{w}\rangle + \langle\vec{v}, \vec{w}\rangle \)
  4. \( \langle\vec{u}, \vec{u}\rangle \ge 0 \), and \( \langle\vec{u}, \vec{u}\rangle = 0 \) if and only if \( \vec{u} = \vec{0} \)

A vector space with an inner product is called an inner product space.

Standard Definitions Extended to \( V \):

Note: \( \mathbb{R}^n \) with the standard inner product (dot product) is an inner product space.

Example 1

(1) Let \( t_0, \dots, t_n \) be distinct real numbers, and let \( p \) and \( q \) be in \( \mathbb{P}_n \). Define:

\[ \langle p, q \rangle = p(t_0)q(t_0) + p(t_1)q(t_1) + \dots + p(t_n)q(t_n) \]

Show that it defines an inner product on \( \mathbb{P}_n \).

Proof: We check the axioms for \( \langle p, q \rangle \).

For Axiom (1):

\[ \begin{aligned} \langle p, q \rangle &= p(t_0)q(t_0) + \dots + p(t_n)q(t_n) \\ &= q(t_0)p(t_0) + \dots + q(t_n)p(t_n) = \langle q, p \rangle \end{aligned} \]

For Axiom (4):

\[ \langle p, p \rangle = p(t_0)^2 + p(t_1)^2 + \dots + p(t_n)^2 \ge 0 \]

And \( \langle p, p \rangle = 0 \) if and only if \( p(t_0) = p(t_1) = \dots = p(t_n) = 0 \). Because \( p \) is a polynomial of degree at most \( n \) with \( n+1 \) roots, it must be the zero polynomial (\( p \equiv 0 \)). (Axioms 2 and 3 are skipped as they are straightforward to verify).

(2) In \( \mathbb{P}_2 \), define the inner product by evaluation at \( 0, \frac{1}{2}, \) and \( 1 \). Compute \( ||p|| \) and \( \langle p, q \rangle \) for \( p(t) = 12t^2 \) and \( q(t) = 2t - 1 \).

Solution: We evaluate the polynomials at the given points.

\( t \) \( 0 \) \( 1/2 \) \( 1 \)
\( p(t) = 12t^2 \) \( 0 \) \( 3 \) \( 12 \)
\( q(t) = 2t - 1 \) \( -1 \) \( 0 \) \( 1 \)
\[ ||p|| = \sqrt{\langle p, p \rangle} = \sqrt{0^2 + 3^2 + 12^2} = \sqrt{0 + 9 + 144} = \sqrt{153} \]
\[ \langle p, q \rangle = (0)(-1) + (3)(0) + (12)(1) = 12 \]

(II) The Gram-Schmidt Process

Example 2

Let \( V \) be \( \mathbb{P}_4 \) with the inner product involving evaluation of polynomials at \( -2, -1, 0, 1, 2 \). View \( \mathbb{P}_2 \) as a subspace of \( V \). Produce an orthogonal basis for \( \mathbb{P}_2 \) by applying the Gram-Schmidt process to the standard basis \( \{1, t, t^2\} \).

Solution: Evaluate the polynomials at \( -2, -1, 0, 1, 2 \).

Polynomial \( -2 \) \( -1 \) \( 0 \) \( 1 \) \( 2 \)
\( 1 \) \( 1 \) \( 1 \) \( 1 \) \( 1 \) \( 1 \)
\( t \) \( -2 \) \( -1 \) \( 0 \) \( 1 \) \( 2 \)
\( t^2 \) \( 4 \) \( 1 \) \( 0 \) \( 1 \) \( 4 \)

Let \( p_0 = 1 \) and \( p_1 = t \). Check if they are orthogonal:

\[ \langle 1, t \rangle = (1)(-2) + (1)(-1) + (1)(0) + (1)(1) + (1)(2) = -2 - 1 + 0 + 1 + 2 = 0 \]

Yes, \( 1 \) and \( t \) are orthogonal.

Now, apply Gram-Schmidt to find \( p_2 \):

\[ \begin{aligned} p_2 &= t^2 - \text{Proj}_{\text{Span}\{1, t\}} t^2 \\ &= t^2 - \frac{\langle t^2, 1 \rangle}{\langle 1, 1 \rangle}1 - \frac{\langle t^2, t \rangle}{\langle t, t \rangle}t \end{aligned} \]

Calculating the inner products:

\[ \langle t^2, 1 \rangle = 4(1) + 1(1) + 0(1) + 1(1) + 4(1) = 10 \]
\[ \langle 1, 1 \rangle = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 = 5 \]
\[ \langle t^2, t \rangle = 4(-2) + 1(-1) + 0(0) + 1(1) + 4(2) = -8 - 1 + 0 + 1 + 8 = 0 \]
\[ p_2 = t^2 - \frac{10}{5}(1) - 0 = t^2 - 2 \]

The set \( \{1, t, t^2 - 2\} \) is an orthogonal basis.

(2) Find the best approximation to \( p(t) = 5 - \frac{1}{2}t^4 \) by \( p_0, p_1, \) and \( p_2 \).

Recall: The best approximation to \( p \) by functions in \( W \) is \( \text{Proj}_W p \).

Evaluate \( p_2 \) and \( p \) at the points to prepare for calculation:

Polynomial \( -2 \) \( -1 \) \( 0 \) \( 1 \) \( 2 \)
\( p_2 = t^2 - 2 \) \( 2 \) \( -1 \) \( -2 \) \( -1 \) \( 2 \)
\( p = 5 - \frac{1}{2}t^4 \) \( -3 \) \( 9/2 \) \( 5 \) \( 9/2 \) \( -3 \)

Compute the necessary inner products:

\[ \begin{aligned} \langle p, p_0 \rangle &= (-3)(1) + (9/2)(1) + (5)(1) + (9/2)(1) + (-3)(1) = 8 \\ \langle p, p_1 \rangle &= (-3)(-2) + (9/2)(-1) + (5)(0) + (9/2)(1) + (-3)(2) = 0 \\ \langle p, p_2 \rangle &= (-3)(2) + (9/2)(-1) + (5)(-2) + (9/2)(-1) + (-3)(2) = -31 \\ \langle p_0, p_0 \rangle &= 5 \\ \langle p_1, p_1 \rangle &= (-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 = 10 \\ \langle p_2, p_2 \rangle &= 2^2 + (-1)^2 + (-2)^2 + (-1)^2 + 2^2 = 14 \end{aligned} \]

Now, calculate the projection \( \hat{p} \):

\[ \begin{aligned} \hat{p} = \text{Proj}_{\mathbb{P}_2} p &= \frac{\langle p, p_0 \rangle}{\langle p_0, p_0 \rangle}p_0 + \frac{\langle p, p_1 \rangle}{\langle p_1, p_1 \rangle}p_1 + \frac{\langle p, p_2 \rangle}{\langle p_2, p_2 \rangle}p_2 \\ &= \frac{8}{5}(1) + 0(t) + \frac{-31}{14}(t^2 - 2) \\ &= \frac{8}{5} - \frac{31}{14}(t^2 - 2) \end{aligned} \]

(III) Two Inequalities

Theorem 16 (The Cauchy-Schwarz Inequality)

For all \( \vec{u} \) and \( \vec{v} \) in \( V \):

\[ |\langle\vec{u}, \vec{v}\rangle| \le ||\vec{u}|| \, ||\vec{v}|| \]

Proof:

① For \( \vec{u} = \vec{0} \), both sides equal \( 0 \).

② For \( \vec{u} \neq \vec{0} \), consider the projection of \( \vec{v} \) onto \( \vec{u} \):

\[ ||\text{Proj}_{\vec{u}}\vec{v}|| = \left|\left| \frac{\langle\vec{v}, \vec{u}\rangle}{\langle\vec{u}, \vec{u}\rangle} \vec{u} \right|\right| = \frac{|\langle\vec{v}, \vec{u}\rangle|}{||\vec{u}||^2} ||\vec{u}|| = \frac{|\langle\vec{v}, \vec{u}\rangle|}{||\vec{u}||} \]

In a right triangle, the length of a leg is less than or equal to the hypotenuse. Thus, \( ||\text{Proj}_{\vec{u}}\vec{v}|| \le ||\vec{v}|| \). Substituting the expression gives:

\[ \frac{|\langle\vec{v}, \vec{u}\rangle|}{||\vec{u}||} \le ||\vec{v}|| \implies |\langle\vec{v}, \vec{u}\rangle| \le ||\vec{u}|| \, ||\vec{v}|| \]

Theorem 17 (The Triangle Inequality)

For all \( \vec{u} \) and \( \vec{v} \) in \( V \):

\[ ||\vec{u} + \vec{v}|| \le ||\vec{u}|| + ||\vec{v}|| \]

(IV) An Inner Product for \( C[a, b] \)

For functions \( f \) and \( g \) in the space of continuous functions \( C[a, b] \), set:

\[ \langle f, g \rangle = \int_a^b f(t)g(t)\,dt \]

It can be shown that this defines an inner product on \( C[a, b] \).

Example 3

Let \( W \) be the subspace of \( C[0, 1] \) spanned by the polynomials \( p_1(t) = 1 \), \( p_2(t) = 2t - 1 \), and \( p_3(t) = 12t^2 \). Use the Gram-Schmidt process to find an orthogonal basis for \( W \).

Solution:

Let \( q_1 = p_1 = 1 \).

Calculate \( q_2 \):

\[ q_2 = p_2 - \frac{\langle p_2, p_1 \rangle}{\langle p_1, p_1 \rangle} p_1 \]
\[ \langle p_2, p_1 \rangle = \int_0^1 (2t - 1)(1)\,dt = \left[ t^2 - t \right]_0^1 = (1 - 1) - 0 = 0 \]

Since \( \langle p_2, p_1 \rangle = 0 \), \( p_1 \) and \( p_2 \) are already orthogonal. Thus, \( q_2 = p_2 = 2t - 1 \).

Calculate \( q_3 \):

\[ q_3 = p_3 - \frac{\langle p_3, p_1 \rangle}{\langle p_1, p_1 \rangle} p_1 - \frac{\langle p_3, p_2 \rangle}{\langle p_2, p_2 \rangle} p_2 \]

Compute the required inner products:

\[ \langle p_3, p_1 \rangle = \int_0^1 12t^2(1)\,dt = \left[ 4t^3 \right]_0^1 = 4 \]
\[ \langle p_1, p_1 \rangle = \int_0^1 1^2\,dt = \left[ t \right]_0^1 = 1 \]
\[ \langle p_3, p_2 \rangle = \int_0^1 12t^2(2t - 1)\,dt = \int_0^1 (24t^3 - 12t^2)\,dt = \left[ 6t^4 - 4t^3 \right]_0^1 = 6 - 4 = 2 \]
\[ \langle p_2, p_2 \rangle = \int_0^1 (2t - 1)^2\,dt = \left[ \frac{(2t - 1)^3}{6} \right]_0^1 = \frac{1^3}{6} - \frac{(-1)^3}{6} = \frac{2}{6} = \frac{1}{3} \]

Now assemble \( q_3 \):

\[ \begin{aligned} q_3 &= 12t^2 - \frac{4}{1}(1) - \frac{2}{1/3}(2t - 1) \\ &= 12t^2 - 4 - 6(2t - 1) \\ &= 12t^2 - 12t + 2 \end{aligned} \]

The set \( \{1, 2t - 1, 12t^2 - 12t + 2\} \) is an orthogonal basis for \( W \).