Mathematical Immaturity

1.17 Exercises

• In the linear space of all real polynomials, with inner product $(x, y) = \int_0^1 x(t) y(t) \, dt,$ let $x_n(t) = t^n$ for $n = 0, 1, 2, \dots.$ Prove that the functions \begin{align*} y_0(t) = 1, \quad y_1(t) = \sqrt{3} (2t - 1), \quad y_n(t) = \sqrt{5} (6t^2 - 6t + 1) \end{align*} form an orthonormal set spanning the same subspace as $\{x_0, x_1, x_2, \dots\}.$

• Proof. $\quad$ To show that the subspace spanned by $y_0, y_1, y_2, \dots$ is the same as that spanned by $\{x_0, x_1, x_2\},$ we first note that every element $y$ in the subspace spanned by $y_0, y_1, y_2, \dots$ can be expressed as the sum: \begin{align*} y &= \sum_{k=0}^n a_k y_k \\ &= a_0 + a_1\sqrt{3}(2t - 1) \\ &+ \sum_{k=2}^n a_k \sqrt{5}(6t^2 - 6t + 1) \\ &= \left(a_0 - \sqrt{3}a_1 + \sqrt{5}\sum_{k=2}^n a_k\right)(1) \\ &+ \left(2\sqrt{3} - 6\sqrt{5}\sum_{k=2}^n a_k\right)(t) \\ &+ \left(6\sqrt{5}\sum_{k=2}^n a_k\right)(t^2) \end{align*} As we can see, $y$ is a linear combination of the elements $x_0, x_1,$ and $x_2.$ Hence, every $y$ is in the subspace spanned by $\{x_0, x_1, x_2\}.$ We can make a similar argument to show that each $x$ in the subspace spanned by $\{x_0, x_1, x_2\}$ is a linear combination of $y_0, y_1, y_2, \dots.$ Hence, each $x$ is in the subspace spanned by $\{y_0, y_1, y_2, \dots\}.$

  $\quad$Now, to show that the set of $y_n$ is orthogonal, we wish to show that for every $m,n \geq 0,$ with $m \neq n$ the inner product $(y_m, y_n) = 0.$ We can break this down into three cases:

  $\quad$ Case 1. $\quad m = 0,\ n = 1.\quad $ Then, the inner product $(y_m, y_n)$ is given by the integral: \begin{align*} (y_m, y_n) &= \int_0^1 y_m(t)y_n(t)\,dt \\ &= \sqrt{3}\int_0^1 2t - 1 \,dt \\ &= \sqrt{3}\left(t^2 - t\right)\biggr|_0^1 \\ &= 0 \end{align*}

  $\quad$ Case 2. $\quad m = 0,\ n \geq 2.\quad $ Then, the inner product $(y_m, y_n)$ is given by the integral: \begin{align*} (y_m, y_n) &= \int_0^1 y_m(t)y_n(t)\,dt \\ &= \sqrt{5}\int_0^1 6t^2 - 6t + 1 \,dt \\ &= \sqrt{5}\left(2t^3 - 3t^2 + t\right)\biggr|_0^1 \\ &= 0 \end{align*}

  $\quad$ Case 3. $\quad m = 1,\ n \geq 2.\quad $ Then, the inner product $(y_m, y_n)$ is given by the integral: \begin{align*} (y_m, y_n) &= \int_0^1 y_m(t)y_n(t)\,dt \\ &= \sqrt{15}\int_0^1 \left(2t - 1 \right)\left(6t^2 - 6t + 1\right)\,dt \\ &= \sqrt{15}\int_0^1 \left(12t^3 - 12t^2 + 2t\right) - \left(6t^2 - 6t + 1\right)\,dt \\ &= \sqrt{15}\int_0^1 12t^3 - 18t^2 + 8t - 1\,dt \\ &= \sqrt{15} \left[3t^4 - 6t^3 + 4t^2 - t\right]_0^1 \\ &= 0 \end{align*} As such, we have shown that the functions $y_0, y_1, y_2, \dots$ form an orthogonal set that spans the same subspace as $\{x_0, x_1, x_2\}.$

  $\quad$ To show that the set of functions $y_0, y_1, \dots$ is orthonormal, we wish to show that for each $n = 0, 1, 2, \dots,$ $\|y_n\| = 1.$ Equivalently, we can show that for each $n,$ $(y_n, y_n) = 1.$ Again, we can break this down into three cases:

  $\quad$ Case 1. $\quad n = 0:$ \begin{align*} (y_n, y_n) &= \int_0^1 1\,dt \\ &= 1 \end{align*}

  $\quad$ Case 2. $\quad n = 1:$ \begin{align*} (y_n, y_n) &= 3\int_0^1 4t^2 - 4t + 1\,dt \\ &= 3\left[\frac{4}{3}t^3 - 2t^2 + t\right]_0^1 \\ &= 1 \end{align*}

  $\quad$ Case 3. $\quad n \geq 2:$ \begin{align*} (y_n, y_n) &= 5\int_0^1 (36t^4 - 36t^3 + 6t^2) \\ &- 5\int_0^1(36t^3 - 36t^2 + 6t) \\ &+ 5\int_0^1(6t^2 - 6t + 1)\,dt \\ \\ &= \left(36t^5 - 45t^4 + 10t^3\right)\biggr|_0^1 \\ &- \left(45t^4 - 60t^3 + 15t^2\right) \biggr|_0^1 \\ &+ \left(10t^3 - 15t^2 + 5t\right) \biggr|_0^1 \\ &= 1 \end{align*}

  Thus, we have shown that the functions $y_0, y_1, y_2, \dots$ form an orthonormal set spanning the same space as $\{x_0, x_1, x_2\}. \quad \blacksquare$