Mathematical Immaturity

1.17 Exercises

• In the real linear space $C(0, \pi),$ with inner product $(x, y) = \int_0^\pi x(t) y(t) \, dt,$ let $x_n(t) = \cos nt$ for $n = 0, 1, 2, \dots.$ Prove that the functions $y_0, y_1, y_2, \dots,$ given by \begin{align*} y_0(t) = \frac{1}{\sqrt{\pi}}, \quad y_n(t) = \frac{\sqrt{2}}{\sqrt{\pi}} \cos nt \quad \text{for} \quad n \geq 1, \end{align*} form an orthonormal set spanning the same subspace as $x_0, x_1, x_2, \dots.$

• Proof. $\quad$ First, we note that $y_n$ is a scalar multiple of $x_n$ for $n = 0, 1, 2, \dots.$ Thus, the subspace spanned by the set of $y_n$ is the same as that spanned by $x_0, x_1, x_2, \dots.$

  $\quad$Now, we wish to show that for every $m,n \geq 0,$ with $m \neq n$ the inner product $(y_m, y_n) = 0.$ Suppose $m = 0,$ then for some $n \geq 1,$ we have: \begin{align*} (y_0, y_n) &= \frac{\sqrt{2}}{\pi}\int_0^{\pi} \cos nt \,dt \\ &= \frac{\sqrt{2}}{n\pi}\sin nt \biggr|_{t=0}^{\pi} \\ &= 0 \end{align*} If $m, n \geq 1$ with $m \neq n,$ we have: \begin{align*} (y_m, y_n) &= \frac{2}{\pi}\int_0^{\pi} \cos mt \cos nt \,dt \end{align*} From Volume 1, Theorem 2.3 (f), we recall the cosine addition formula: $$\cos(x + y) = \cos x \cos y - \sin x \sin y.$$ Applying this to our equation for $(y_m, y_n),$ we get: \begin{align*} \int_0^{\pi} \cos mt \cos nt \,dt &= \int_0^{\pi} \cos [(m+n)t] + \sin mt \sin nt\, dt \end{align*} But as we saw in the case where $m = 0,$ the integral $\int_0^{\pi}\cos nt\, dt = 0$ for any integer $n \geq 1,$ hence \begin{align*} \int_0^{\pi} \cos mt \cos nt \,dt &= \int_0^{\pi}\sin mt \sin nt\, dt \end{align*} Evaluating the integral on the left-hand side using integration by parts with $u = \cos mt$ and $dv = \cos nt\,dt,$ we get: \begin{align*} \int_0^{\pi} \cos mt \cos nt \,dt &= \frac{1}{n}\cos mt \sin nt\,\biggr|_0^{\pi} + \frac{m}{n}\int_0^{\pi} \sin mt \sin nt \,dt \\ &= \frac{m}{n}\int_0^{\pi} \sin mt \sin nt \,dt \end{align*} But because $\int_0^{\pi} \cos mt \cos nt \,dt = \int_0^{\pi} \sin mt \sin nt \,dt,$ this means that for all integers $m, n \geq 1,$ with $m \neq n,$ we have: \begin{align*} \frac{m}{n}\int_0^{\pi} \sin mt \sin nt \,dt = \int_0^{\pi} \sin mt \sin nt \,dt \end{align*} implying that $\int_0^{\pi}\cos mt \cos nt\,dt = 0$ and that for all integers $m, n \geq 0,$ the inner product $(y_m, y_n) = 0.$

  $\quad$To complete the proof, we wish to show that $\|y_n\| = 1$ for $n = 0, 1, 2, \dots.$ Equivalently, we can show that $(y_n, y_n) = 1$ for each $n.$ When $n = 0,$ we have $\int_0^{\pi}\frac{1}{\pi}\,dt = 1.$ When $n \geq 1,$ we have: \begin{align*} (y_n, y_n) &= \frac{2}{\pi}\int_0^{\pi}\cos^2nt\,dt \end{align*} Integrating by parts, with: $u = \cos nt,\ dv = \cos nt\,dt$ we get: \begin{align*} (y_n, y_n) &= \frac{2}{\pi}\int_0^{\pi}\cos^2nt\,dt \\ &= \frac{2}{n\pi}\sin nt \cos nt\biggr|_0^{\pi} + \frac{2}{\pi}\int_0^{\pi}\sin^2 nt\,dt \\ &= \frac{2}{\pi}\int_0^{\pi}\sin^2 nt\,dt \end{align*} But if $\int_0^{\pi}\cos^2 nt\,dt = \int_0^{\pi}\sin^2 nt\,dt,$ we can use the Pythagorean identity to give us: \begin{align*} 2(y_n, y_n) &= \frac{2}{\pi}\int_0^{\pi}\cos^2 nt\,dt + \frac{2}{\pi}\int_0^{\pi}\sin^2 nt\,dt \\ &= \frac{2}{\pi}\int_0^{\pi}\left(\cos^2 nt + \sin^2 nt \right)\,dt \\ &= \int_0^{\pi}\frac{2}{\pi}\,dt \\ &= 2 \end{align*} Hence, for each $n = 0, 1, 2, \dots,$ $(y_n, y_n) = 1$ and equivalently, $\|y_n\| = 1.$ Thus, we have shown that the set of functions $\{y_0, y_1, y_2, \dots\}$ is an orthonormal set spanning the same subspace as $x_0, x_1, x_2, \dots. \quad \blacksquare$