Mathematical Immaturity

1.13 Exercises

• Prove that the space of all complex-valued functions continuous on an interval $[a, b]$ becomes a unitary space if we define an inner product by the formula $$(f, g) = \int_a^b w(t) f(t) \overline{g(t)} \, dt,$$ where $w$ is a fixed positive function, continuous on $[a, b].$
• Recall from Section 11 that the term unitary space is used to describe a complex linear space with an inner product.

  To show that the space of all complex-valued functions continuous on an interval $[a, b]$ becomes a complex Euclidean (unitary) space with the following inner product: $$(f, g) = \int_a^b w(t)f(t)\overline{g(t)}\,dt,$$ we must show that the function $(f, g)$ satisfies the axioms of an inner product in a complex linear space:

  $\text{(1')}\quad$ $(x, y) = \overline{(y, x)}\qquad$ (commutativity, or Hermitian symmetry).
  Proof.$\quad$ Note that because $w(t)$ is a positive function for all $t,$ the imaginary part of $w(t)$ is always zero, thus $\overline{w(t)} = w(t).$ \begin{align*} (f, g) &= \int_a^b w(t)f(t)\overline{g(t)}\,dt \\ \end{align*} We can express the complex functions $f(t)$ and $g(t)$ in terms of their real and imaginary components as follows: $$f(t) = a_1(t) + b_1(t)i, \qquad g(t) = a_2(t) + b_2(t)i.$$ Where $a_1(t),\ a_2(t),\ b_1(t),\ b_2(t)$ are real-valued functions. As such, the complex conjugate of $g(t)$ is given by $\overline{g(t)} = a_2(t) - b_2(t)i$. Given this decomposition, we can rewrite $(f, g)$ as follows: \begin{align*} (f, g) &= \int_a^b w(t)\left[a_1a_2(t) - a_1b_2(t)i + b_1a_2(t)i + b_1b_2(t)\right]\,dt \end{align*} Because the functions $w,\ a_1,\ a_2,\ b_1,$ and $b_2$ are real-valued and continuous on the interval $[a,b],$ we can treat this integral as a complex number. Namely, we have: \begin{align*} (f, g) &= \overline{\int_a^b w(t)\left[a_1a_2(t) + a_1b_2(t)i - b_1a_2(t)i + b_1b_2(t)\right]\,dt} \\ &= \overline{\int_a^b w(t)\left[a_1(t) - b_1(t)i\right]\left[a_2(t) + b_2(t)i\right]\,dt} \\ &= \overline{\int_a^b w(t)\left[a_2(t) + b_2(t)i\right]\left[a_1(t) - b_1(t)i\right]\,dt} \\ &= \overline{\int_a^b w(t)g(t)\overline{f(t)}} \\ &= \overline{(g, f)} \quad \blacksquare \end{align*}

  $\text{(2)}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity).
  Proof.$\quad$ Defining function addition in the typical way, we have: \begin{align*} (f, g + h) &= \int_a^b w(t)f(t)\overline{(g + h)(t)}\,dt \\ &= \int_a^b w(t)f(t)\overline{g(t) + h(t)}\,dt \end{align*} If we define: \begin{align*} f(t) &= a_1(t) + b_1(t)i \\ g(t) &= a_2(t) + b_2(t)i \\ h(t) &= a_3(t) + b_3(t)i \end{align*} where each coefficient function is real-valued and continuous on the interval $[a, b],$ the above integral becomes \begin{align*} (f, g + h) &= \int_a^b w(t)\left[a_1(t) + b_1(t)i\right]\left[\left(a_2(t) - b_2(t)i\right) + \left(a_3(t) - b_3(t)i\right)\right]\,dt \\ &= \int_a^b w(t)\left[a_1(t) + b_1(t)i\right]\left[a_2(t) - b_2(t)i\right]\,dt \\ &+ \int_a^b w(t)\left[a_1(t) + b_1(t)i\right]\left[a_3(t) - b_3(t)i\right]\,dt \\ &= \int_a^b w(t)f(t)\overline{g(t)}\,dt + \int_a^b w(t)f(t)\overline{h(t)}\,dt \quad \blacksquare \end{align*}

  $\text{(3)}\quad$ $c(x, y) = (cx, y)\qquad$ (associativity, or homogeneity).
  Proof.$\quad$ \begin{align*} c(f, g) &= c\int_a^b w(t)f(t)\overline{g(t)}\,dt \\ &= \int_a^b w(t)\left[cf(t)\right]\overline{g(t)}\,dt \\ &= (cf, g) \quad \blacksquare \end{align*}

  $\text{(4)}\quad$ $(x, x) > 0\quad$ if $x \neq O \qquad$ (positivity).
  Proof.$\quad$ \begin{align*} (f, f) &= \int_a^b w(t)f(t)\overline{f(t)}\,dt \\ &= \int_a^b w(t)|f(t)|^2\,dt \end{align*} By definition, $|f(t)|^2 \geq 0$ and $w(t) \gt 0$ for all $t$ in $[a, b].$ As such, if $\int_a^b w(t)|f(t)|^2\,dt = 0,$ this would imply that $|f(t)|^2 = 0$ for all $t$ in $[a, b],$ meaning $f = O$ on $[a, b].$ Thus, if $f \neq O$ on $[a, b],$ then $(f, f) \gt 0. \quad \blacksquare$