Mathematical Immaturity

1.13 Exercises

• In the linear space $P_n$ of all real polynomials of degree $\leq n,$ define $$(f, g) = \sum_{k=0}^n f\left(\frac{k}{n}\right) g\left(\frac{k}{n}\right).$$
  $\text{(a)}\quad$ Prove that $(f, g)$ is an inner product for $P_n.$
  $\text{(b)}\quad$ Compute $(f, g)$ when $f(t) = t$ and $g(t) = at + b.$
  $\text{(c)}\quad$ If $f(t) = t,$ find all linear polynomials $g$ orthogonal to $f.$
• Recall the axioms of an inner product from Section 11:

  Definition. $\quad$ A real linear space $V$ is said to have an inner product if for each pair of elements $x$ and $y$ in $V$ there corresponds a unique real number $(x, y)$ satisfying the following axioms for all choices of $x,$ $y,$ and $z$ in $V$ and all real scalars $c.$
  $\quad\text{(1)}\quad$ $(x, y) = (y, x)\qquad$ (commutativity, or symmetry).
  $\quad\text{(2)}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity).
  $\quad\text{(3)}\quad$ $c(x, y) = (cx, y)\qquad$ (associativity, or homogeneity).
  $\quad\text{(4)}\quad$ $(x, x) > 0\quad$ if $x \neq O \qquad$ (positivity).

• 
  $\text{(a)} \quad$ To prove that \begin{align*} (f, g) &= \sum_{k=0}^n f\left(\frac{k}{n}\right) g\left(\frac{k}{n}\right) \end{align*} is an inner product, we must show that $(f, g)$ satisfies all four axioms as defined above.
  
  $\text{(1)}\quad$ $(x, y) = (y, x)\qquad$ (commutativity, or symmetry).
  Since $f$ and $g$ are real-valued polynomials, we can use the commutative property of multiplication to get: \begin{align*} (f, g) &= \sum_{k=0}^n f\left(\frac{k}{n}\right) g\left(\frac{k}{n}\right) \\ &= \sum_{k=0}^n g\left(\frac{k}{n}\right) f\left(\frac{k}{n}\right) \\ &= (g, f) \quad \blacksquare \end{align*}
  
  $\text{(2)}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity).
  Let $f,$ $g,$ and $h$ be elements of $P_n.$ We can use the distributive property of multiplication to get: \begin{align*} (f, g + h) &= \sum_{k=0}^n f\left(\frac{k}{n}\right) (g + h)\left(\frac{k}{n}\right) \\ &= \sum_{k=0}^n f\left(\frac{k}{n}\right) \left[g\left(\frac{k}{n}\right) + h\left(\frac{k}{n}\right)\right] \\ &= \sum_{k=0}^n f\left(\frac{k}{n}\right) g\left(\frac{k}{n}\right) + f\left(\frac{k}{n}\right)h\left(\frac{k}{n}\right) \\ &= \sum_{k=0}^n f\left(\frac{k}{n}\right) g\left(\frac{k}{n}\right) +\sum_{k=0}^nf\left(\frac{k}{n}\right)h\left(\frac{k}{n}\right) \\ &= (f, g) + (f, h) \quad \blacksquare \end{align*}
  
  $\text{(3)}\quad$ $c(x, y) = (cx, y)\qquad$ (associativity, or homogeneity).
  If $f$ and $g$ are two elements of $P_n,$ we have: \begin{align*} c(f, g) &= c\sum_{k=0}^n f\left(\frac{k}{n}\right) g\left(\frac{k}{n}\right) \\ &= \sum_{k=0}^n cf\left(\frac{k}{n}\right) g\left(\frac{k}{n}\right) \\ &= (cf, g) \quad \blacksquare \end{align*}
  
  $\text{(4)}\quad$ $(x, x) > 0\quad$ if $x \neq O \qquad$ (positivity).
  Let $f$ be an element of $P_n.$ Then, we have: \begin{align*} (f, f) &= \sum_{k=0}^n f\left(\frac{k}{n}\right) f\left(\frac{k}{n}\right) \\ &= \sum_{k=0}^n f^2\left(\frac{k}{n}\right) \end{align*} Because $f$ is a real-valued function, each term of the sum must be $\geq 0.$ Now, suppose $(f, f) = 0.$ We will now show that this implies $f = O,$ contradicting our given information. Setting $(f, f) = 0,$ we get \begin{align*} \sum_{k=0}^n f^2\left(\frac{k}{n}\right) &= 0 \end{align*} Subtracting any of the $n+1$ terms from both sides, we find that every term in the sum is zero. For example: \begin{align*} \sum_{k=1}^{n} f^2\left(\frac{k}{n}\right) &= -f^2(0) \end{align*} As we can see, the left-hand side must be $\geq 0$ and the right-hand side must be $\leq 0.$ This is satisfied when both sides are equal to zero. But if $f^2(k/n) = 0$ for $k = 0, \dots, n,$ then $f(k/n) = 0$ for each $k.$ But because the components $(1, x, \dots, x^n)$ of $f$ are independent, this implies that the coefficients $a_0, \dots, a_n$ are all zero. In other words, $(f, f) = 0$ implies $f = O.$ As such, if $f \neq O,$ we have $(f, f) \gt 0.\quad \blacksquare$
  
  $\text{(b)} \quad$ If $f(t) = t$ and $g(t) = at + b,$ the inner product $(f, g)$ can be written as: \begin{align*} (f, g) &= \sum_{k=0}^n f\left(\frac{k}{n}\right)g\left(\frac{k}{n}\right) \\ &= \sum_{k=0}^n \frac{k}{n}\left(\frac{ak}{n} + b \right) \\ &= \frac{a}{n^2}\sum_{k=0}^n k^2 + \frac{b}{n}\sum_{k=0}^n k \end{align*} From the introductory chapter of Volume 1 (Sections I-1.3 and I-4.4), we recall that the sums $\sum_{k=0}^n k$ and $\sum_{k=0}^n k^2$ have the closed-form solutions: \begin{align*} \sum_{k=0}^n k &= \frac{n(n+1)}{2} \quad \text{and} \quad \sum_{k=0}^n k^2 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \end{align*} Thus, for $f(t) = t$ and $g(t) = at + b,$ the inner product $(f, g)$ evaluates to \begin{align*} (f, g) &= \frac{a}{n^2}\sum_{k=0}^n k^2 + \frac{b}{n}\sum_{k=0}^n k \\ &= \frac{a}{n^2}\left(\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\right) + \frac{b}{n} \frac{n(n + 1)}{2} \\ &= \frac{(n + 1)(2n + 1)}{6n}\,a + \frac{n + 1}{2}\,b. \quad \blacksquare \end{align*}
  
  $\text{(c)} \quad$ To find all linear polynomials $g$ orthogonal to $f(t) = t,$ we refer back to the result of part (b). We found that when $f(t) = t$ and $g(t) = at + b$ for real scalars $a$ and $b,$ the inner product $(f, g)$ is \begin{align*} (f, g) &= \frac{(n + 1)(2n + 1)}{6n}\,a + \frac{n + 1}{2}\,b. \end{align*} Setting this inner product equal to zero, we find that this relation is satisfied when $b = -(2n + 1)a/3n.$ Hence, for any real $a,$ the linear polynomial $g(t)$ orthogonal to $f(t) = t$ is given by $$g(t) = a\left(t - \frac{2n + 1}{3n}\right). \quad \blacksquare$$