Mathematical Immaturity

1.13 Exercises

• In the linear space of all real polynomials, determine whether or not $(f, g)$ is an inner product if $(f, g)$ is defined by the formula given. In case $(f, g)$ is not an inner product, indicate which axioms are violated. In (c), $f'$ and $g'$ denote derivatives.
  $\text{(a)} \quad (f, g) = f(1) g(1).$
  $\text{(b)} \quad (f, g) = \left| \int_0^1 f(t) g(t) \, dt \right|.$
  $\text{(c)} \quad (f, g) = \int_0^1 f'(t) g'(t) \, dt.$
  $\text{(d)} \quad (f, g) = \left( \int_0^1 f(t) \, dt \right) \left( \int_0^1 g(t) \, dt \right).$
• Recall the four 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).

• For each given formula for $(f, g),$ we will verify each of the four axioms of an inner product.
  $\text{(a)}\quad$ $(f, g) = f(1)g(1).$
  $\quad\text{(1)}\quad$ $(x, y) = (y, x)\qquad$ (commutativity, or symmetry). \begin{align*} (f, g) &= f(1)g(1) \\ &= g(1)f(1) \\ &= (g ,f) \quad \blacksquare \end{align*}
  $\quad\text{(2)}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity). \begin{align*} (f, g + h) &= f(1)[g + h](1) \\ &= f(1)[g(1) + h(1)] \\ &= [f(1)g(1)] + [f(1)h(1)] \\ &= (f, g) + (f, h) \quad \blacksquare \end{align*}
  $\quad\text{(3)}\quad$ $c(x, y) = (cx, y)\qquad$ (associativity, or homogeneity). \begin{align*} c(f, g) &= c[f(1)g(1)] \\ &= [cf(1)]g(1) \quad \blacksquare \end{align*}
  $\quad\text{(4)*}\quad$ $(x, x) > 0\quad$ if $x \neq O \qquad$ (positivity).
  $\quad$Counterexample. $\qquad$ Let $f(t) = 1 - t.$ $f \neq O$ but $(f, f) = f^2(1) = 0. \quad \blacksquare$
  
  Conclusion. $\quad$$(f, g) = f(1)g(1)$ violates positivity, and hence is not an inner product.
  
  $\text{(b)}\quad (f, g) = \left| \int_0^1 f(t) g(t) \, dt \right|.$
  $\quad\text{(1)}\quad$ $(x, y) = (y, x)\qquad$ (commutativity, or symmetry). \begin{align*} (f, g) &= \left| \int_0^1 f(t) g(t) \, dt \right| \\ &= \left| \int_0^1 g(t) f(t) \, dt \right| \\ &= (g, f) \quad \blacksquare \end{align*}
  $\quad\text{(2)*}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity). \begin{align*} (f, g + h) &= \left| \int_0^1 f(t) [g+h](t) \, dt \right| \\ &= \left| \int_0^1 f(t) [g(t)+h(t)] \, dt \right| \\ &= \left| \int_0^1 f(t)g(t) + f(t)h(t) \, dt \right| \\ &\leq \left| \int_0^1 f(t) g(t) \, dt \right| + \left| \int_0^1 f(t) h(t) \, dt \right| \\ &= (f, g) + (f, h) \quad \blacksquare \end{align*}
  $\quad\text{(3)*}\quad$ $c(x, y) = (cx, y)\qquad$ (associativity, or homogeneity). \begin{align*} c(f, g) &= c\left| \int_0^1 f(t) g(t) \, dt \right| \\ &\neq |c|\left| \int_0^1 f(t) g(t) \, dt \right| \\ &= \left| \int_0^1 cf(t) g(t) \, dt \right| \\ &= (cf, g) \quad \blacksquare \end{align*}
  $\quad\text{(4)}\quad$ $(x, x) > 0\quad$ if $x \neq O \qquad$ (positivity).
  \begin{align*} (f, f) &= \left| \int_0^1 f(t) f(t) \, dt \right| \\ &= \left| \int_0^1 f^2(t) \, dt \right| \end{align*} Because $f$ is a continuous, real-valued function, $f^2(t) \geq 0$ for all real $t.$ If $f^2(t)$ is constant on $[0, 1]$, that is, if $f^2(t) = a^2_0$ where $a_0$ is some real scalar, then we can see that $(f, f) = 0$ implies $a_0 = 0.$ In other words, $f = O.$
  
  If $f$ is not constant, namely, if there are two points $t_0 \geq 0$ and $t_1 \leq 1$ such that $f^2(t_0) = 0$ and $f^2(t_1) \neq 0,$ respectively, then by the intermediate-value theorem (Volume 1, Section 3.10.), $f^2(t)$ takes on every value between $0$ and $f^2(t_1)$ over the interval $(t_0, t_1).$ But because $f^2(t) \gt 0$ on the interval $(t_0, t_1),$ then by the comparison theorem (Volume 1, Section 1.24), we have: \begin{align*} \int_{t_0}^{t_1} f^2(t)\,dt \gt 0 \end{align*} But because $f^2(t) \geq 0$ on the interval $[0, 1],$ the comparison theorem shows that \begin{align*} \int_{0}^{t_0} f^2(t)\,dt \geq 0; \qquad \int_{t_1}^{1} f^2(t)\,dt \geq 0 \end{align*} Combining the integrals from $t= 0$ to $t = 1,$ we find that \begin{align*} \int_0^1f^2(t)\,dt \gt 0 \end{align*} and by extension, that $(f, f) \gt 0$ if $f \neq O. \quad \blacksquare$
  
  Conclusion. $\quad$$(f, g) = \left| \int_0^1 f(t) g(t) \, dt \right|$ violates distributivity and associativity, and thus is not an inner product.
  
  $\text{(c)} \quad (f, g) = \int_0^1 f'(t) g'(t) \, dt.$
  $\quad\text{(1)}\quad$ $(x, y) = (y, x)\qquad$ (commutativity, or symmetry). \begin{align*} (f, g) &= \int_0^1 f'(t) g'(t) \, dt \\ &= \int_0^1 g'(t) f'(t) \, dt \\ &= (g, f) \quad \blacksquare \end{align*}
  $\quad\text{(2)}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity). \begin{align*} (f, g + h) &= \int_0^1 f'(t) [g+h]'(t) \, dt \\ &= \int_0^1 f'(t) [g'(t) + h'(t)] \, dt \\ &= \int_0^1 f'(t) g'(t) + f'(t)h'(t)\, dt \\ &= \int_0^1 f'(t) g'(t) \, dt + \int_0^1 f'(t)h'(t)\, dt \\ &= (f, g) + (f, h) \quad \blacksquare \end{align*}
  $\quad\text{(3)}\quad$ $c(x, y) = (cx, y)\qquad$ (associativity, or homogeneity). \begin{align*} c(f, g) &= c\int_0^1 f'(t) g'(t) \, dt \\ &= \int_0^1 [cf'(t)] g'(t) \, dt \\ &= (cf, g) \quad \blacksquare \end{align*}
  $\quad\text{(4)*}\quad$ $(x, x) > 0\quad$ if $x \neq O \qquad$ (positivity).
  $\quad$Counterexample. $\quad$ Let $f = a_0,$ where $a_0 \neq 0.$ Then $f \neq O$ but $f' = O$ giving us: \begin{align*} (f, f) &= \int_0^1 [f'(t)]^2 \, dt \\ &= \int_0^1 0 \, dt \\ &= 0 \quad \blacksquare \end{align*}
  
  Conclusion. $\quad$$(f, g) = \int_0^1 f'(t) g'(t) \, dt$ violates positivity, and thus is not an inner product.
  
  $\text{(d)}\quad (f, g) = \left(\int_0^1 f(t)\,dt \right)\left(\int_0^1 g(t)\,dt \right).$
  $\quad\text{(1)}\quad$ $(x, y) = (y, x)\qquad$ (commutativity, or symmetry). \begin{align*} (f, g) &= \left(\int_0^1 f(t)\,dt \right)\left(\int_0^1 g(t)\,dt \right) \\ &= \left(\int_0^1 g(t)\,dt \right)\left(\int_0^1 f(t)\,dt \right) \\ &= (g, f) \quad \blacksquare \end{align*}
  $\quad\text{(2)}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity). \begin{align*} (f, g + h) &= \left(\int_0^1 f(t)\,dt \right)\left(\int_0^1 g(t) + h(t)\,dt \right) \\ &= \left(\int_0^1 f(t)\,dt \right)\left(\int_0^1 g(t)\,dt \right) + \left(\int_0^1 f(t)\,dt \right)\left(\int_0^1 h(t)\,dt \right) \\ &= (f, g) + (f, h) \quad \blacksquare \end{align*}
  $\quad\text{(3)}\quad$ $c(x, y) = (cx, y)\qquad$ (associativity, or homogeneity). \begin{align*} c(f, g) &= c\left(\int_0^1 f(t)\,dt \right)\left(\int_0^1 g(t)\,dt \right) \\ &= \left(\int_0^1 cf(t)\,dt \right)\left(\int_0^1 g(t)\,dt \right) \\ &= (cf, g) \quad \blacksquare \end{align*}
  $\quad\text{(4)*}\quad$ $(x, x) > 0\quad$ if $x \neq O \qquad$ (positivity).
  $\quad$Counterexample. $\quad$ Let $f(t) = 1 - 2t.$ Then $f \neq O$ but \begin{align*} (f, f) &= \left(\int_0^1 f(t) \,dt\right)^2 \\ &= \left(\int_0^1 1 - 2t \, dt\right)^2 \\ &= \left[\left(t - t^2\right)_0^1\right]^2 \\ &= 0 \quad \blacksquare \end{align*}
  
  Conclusion. $\quad$$(f, g) = \left(\int_0^1 f(t)\,dt \right)\left(\int_0^1 g(t)\,dt \right)$ violates positivity, and thus is not an inner product.