Mathematical Immaturity

1.13 Exercises

• In the real linear space $C(1, e)$, define an inner product by the equation $$(f, g) = \int_1^e (\log x) f(x) g(x) \, dx.$$
  $\text{(a)}\quad$ If $f(x) = \sqrt{x},$ compute $\|f\|.$
  $\text{(b)}\quad$ Find a linear polynomial $g(x) = a + bx$ that is orthogonal to the constant function $f(x) = 1.$

• $\text{(a)}\quad$ If $f(x) = \sqrt{x},$ then $(f, f)$ is given by the integral \begin{align*} (f, f) &= \int_1^e x\log x \, dx \end{align*} Integrating by parts, with $u = \log x$ and $dv = x\,dx$ we get \begin{align*} (f, f) &= \int_1^e x\log x \, dx \\ &= uv\,\biggr|_1^e - \int_1^e v\,du \\ &= \frac{x^2\log x}{2}\Biggr|_1^e - \frac{1}{2}\int_1^e x\,dx \\ &= \frac{e^2}{2} - \frac{e^2 - 1}{4} \\ &= \frac{e^2 + 1}{4} \end{align*} Then, $\|f\| = \sqrt{(f, f)} = \frac{1}{2}\sqrt{e^2 + 1}. \quad \blacksquare$

  $\text{(b)}\quad$ We want to find a linear polynomial $g(x) = a + bx$ orthogonal to $f(x) = 1.$ In other words, we want to find real scalars $a$ and $b$ such that \begin{align*} (f, g) &= \int_1^e (\log x)(a + bx) \, dx \\ &= a\int_1^e (\log x)\, dx + b\int_1^e x(\log x)\, dx \\ &= 0 \end{align*} To find the first integral, we can integrate by parts with $u = a\log x$ and $dv = dx$ to give us \begin{align*} a\int_1^e (\log x)\, dx &= uv\,\biggr|_1^e - \int_1^e v\,du \\ &= ax\log x\biggr|_1^e - ax\biggr|_1^e \\ &= a \end{align*} We know from part (a) that the value of the second integral is $\frac{b}{4}(e^2 + 1).$ This means that $(f, g) = 0$ if $a = \frac{-b}{4}(e^2 + 1),$ where $b$ is arbitrary. Thus, for any real $b,$ the polynomial $g(x)$ orthogonal to the constant function $f(x) = 1$ is \begin{align*} g(x) &= b\left(x - \frac{e^2 + 1}{4}\right). \quad \blacksquare \end{align*}