Mathematical Immaturity

2.4 Exercises

29. $\quad$ Let $V$ denote the linear space of all real functions continuous on $[-\pi,\pi].$ Let $S$ be the subset of $V$ consisting of all $f$ satisfying \begin{align*} \int_{-\pi}^{\pi} f(t)\,dt &= 0,\quad \int_{-\pi}^{\pi} f(t)\cos t\,dt = 0,\quad \int_{-\pi}^{\pi} f(t)\sin t\,dt = 0. \end{align*}
(a) $\quad$ Prove that $S$ is a subspace of $V.$
(b) $\quad$ Prove that $S$ contains the functions $f(x)=\cos nx$ and $f(x)=\sin nx$ for each $n=2,3,\dots.$
(c) $\quad$ Prove that $S$ is infinite-dimensional.

Let $T:V\to V$ be the linear transformation defined as follows: If $f\in V,$ $g=T(f)$ means that \begin{align*} g(x) &= \int_{-\pi}^{\pi}\{1+\cos(x - t)\}f(t)\,dt. \end{align*} (d) $\quad$ Prove that $T(V),$ the range of $T,$ is finite-dimensional and find a basis for $T(V).$
(e) $\quad$ Determine the null space of $T.$
(f) $\quad$ Find all real $c\neq0$ and all nonzero $f\in V$ such that $T(f)=c\,f.$ (Note that such an $f$ lies in the range of $T.$)

Solution.

(a) $\quad$ Proof.$\quad$ We recall from Theorem 1.4 that a nonempty subset $S$ of a linear space $V$ is a subspace if and only if it satisfies the closure axioms. Then, since $S$ is a subset of $V,$ we know that $f + g$ and $cf$ (where $c$ is a scalar) are also elements of $V.$ To show that $S$ satisfies the closure axioms, we must show that $f + g$ and $cf$ are elements of $S.$ In other words, $f + g$ and $cf$ must satisfy \begin{align*} \int_{-\pi}^{\pi} [f + g](t)\,dt &= 0,\quad \int_{-\pi}^{\pi} [f + g](t)\cos t\,dt = 0,\quad \int_{-\pi}^{\pi} [f + g](t)\sin t\,dt = 0. \\ \\ \int_{-\pi}^{\pi} cf(t)\,dt &= 0,\quad \int_{-\pi}^{\pi} cf(t)\cos t\,dt = 0,\quad \int_{-\pi}^{\pi} cf(t)\sin t\,dt = 0. \end{align*} Rearranging terms, noting that $f$ and $g$ satisfy the above equations, we have \begin{align*} \int_{-\pi}^{\pi} [f + g](t)\,dt &= \int_{-\pi}^{\pi} f(t)\,dt + \int_{-\pi}^{\pi} g(t)\,dt \\ &= 0 \\ \\ \int_{-\pi}^{\pi} [f + g](t)\cos t\,dt &= \int_{-\pi}^{\pi} f(t)\cos t\,dt + \int_{-\pi}^{\pi} g(t)\cos t\,dt \\ &= 0 \\ \\ \int_{-\pi}^{\pi} [f + g](t)\sin t\,dt &= \int_{-\pi}^{\pi} f(t)\sin t\,dt + \int_{-\pi}^{\pi} g(t)\sin t\,dt \\ &= 0 \\ \\ \int_{-\pi}^{\pi} cf(t)\,dt &= c\int_{-\pi}^{\pi} f(t)\,dt \\ &= 0 \\ \\ \int_{-\pi}^{\pi} cf(t)\cos t\,dt &= c\int_{-\pi}^{\pi} f(t)\cos t\,dt \\ &= 0 \\ \\ \int_{-\pi}^{\pi} cf(t)\sin t\,dt &= c\int_{-\pi}^{\pi} f(t)\sin t\,dt \\ &= 0 \end{align*} As we can see, $S$ satisfies the closure axioms, making it a subspace of $V. \quad \blacksquare$

(b) $\quad$ Proof. $\quad$ Setting $f(x) = \cos(nx)$ for $n = 2, 3, \dots,$ we have \begin{align*} \int_{-\pi}^{\pi}\cos nx\,dx &= \frac{1}{n}\left[\sin n\pi - \sin(-n\pi)\right] \\ &= 0 \\ \\ \int_{-\pi}^{\pi}\cos nx\cos x\,dx &= \int_{-\pi}^{\pi}\cos[(n + 1)x]\,dx + \int_{-\pi}^{\pi}\sin nx \sin x\,dx \\ &= 0 - \left[\sin nx \cos x\right]_{-\pi}^{\pi} + n\int_{-\pi}^{\pi}\cos nx \cos x \,dx \\ &= n\int_{-\pi}^{\pi}\cos nx \cos x \,dx \end{align*} If we subtract $\int_{-\pi}^{\pi}\cos nx \cos x\,dx$ from both sides, we find that \begin{align*} (n - 1)\int_{-\pi}^{\pi}\cos nx \cos x\,dx = 0 \end{align*} But since $n \geq 2,$ we can divide both sides by $n - 1$ to give us $$\int_{-\pi}^{\pi}\cos nx \cos x \,dx = 0 \quad \text{for} \quad n = 2, 3, \dots$$ Finally, we have \begin{align*} \int_{-\pi}^{\pi}\cos nx\sin x\,dx &= n\int_{-\pi}^{\pi}\sin(nx)\sin x\,dx - \left[\cos(nx)\cos(x)\right]_{-\pi}^{\pi} \\ &= n\int_{-\pi}^{\pi}\cos nx \cos x - n\int_{-\pi}^{\pi}\cos[(n + 1)x]\,dx \end{align*} But as we saw before, $\int_{-\pi}^{\pi}\cos(nx)\cos x\,dx = 0$ and $\int_{-\pi}^{\pi}\cos[(n + 1)x]\,dx = 0$ for $n = 2, 3, \dots,$ giving us \begin{align*} \int_{-\pi}^{\pi}\cos nx\sin x\,dx &= 0 \end{align*} This proves that $f(x) = \cos(nx)$ is an element of $S$ for $n= 2, 3, \dots.$

$\quad$ Now, if we set $f(x) = \sin(nx)$ for $n = 2, 3, \dots,$ we can use the above results to give us: \begin{align*} \int_{-\pi}^{\pi}\sin nx &= \frac{1}{n}\left[\cos nx\right]_{\pi}^{-\pi} \\ &= \frac{1}{n}\left[\cos(-n\pi) - \cos(n\pi)\right] \\ &= 0 \\ \\ \int_{-\pi}^{\pi}\sin nx \cos x\,dx &= \left[\sin nx \sin x\right]_{-\pi}^{\pi} - n\int_{-\pi}^{\pi}\cos nx \sin x\,dx \\ &= 0 \\ \\ \int_{-\pi}^{\pi}\sin nx \sin x\,dx &= \int_{-\pi}^{\pi}\cos nx \cos x - \int_{-\pi}^{\pi}\cos[(n + 1)x]\,dx \\ &= 0 \end{align*} As we can see, $f(x) = \sin nx$ is also an element of $S$ for each $n = 2, 3, \dots.$ This completes the proof. $\quad \blacksquare$

(c) $\quad$ Proof. $\quad$ The span of $S$ includes all linear combinations of the independent elements $\cos nx$ and $\sin nx$ for $n = 2, 3, \dots.$ As such, the basis of $S$ contains infinitely many elements, making $S$ infinite-dimensional.

(d) $\quad$ Proof. $\quad$ The range of $T$ is the set of functions $g(x),$ given by \begin{align*} g(x) &= \int_{-\pi}^{\pi}\{1+\cos(x - t)\}f(t)\,dt \\ &= \int_{-\pi}^{\pi}f(t)\,dt + \int_{-\pi}^{\pi}f(t)\cos(x - t)\,dt \\ &= \int_{-\pi}^{\pi}f(t)\,dt + \cos x\int_{-\pi}^{\pi}f(t)\cos t\,dt + \sin x \int_{-\pi}^{\pi}f(t)\sin t\,dt \end{align*} But as proven in Theorem 3.2 of Volume 1, since $f(t)$ is continuous on $[-\pi, \pi],$ so are $f(t)\cos t$ and $f(t) \sin t.$ And from Theorem 3.14 of Volume 1, this means that $f(t),$ $f(t)\cos t,$ and $f(t) \sin t$ are integrable on $[-\pi, \pi].$ As such, we can treat the integrals \begin{align*} a &=\int_{-\pi}^{\pi}f(t)\,dt \\ b &= \int_{-\pi}^{\pi}f(t)\cos t\,dt \\ c &= \int_{-\pi}^{\pi}f(t)\sin t\,dt \end{align*} as scalar constants. But then, we can rewrite the range of $T$ as the set of real functions $g,$ continuous on $[-\pi, \pi],$ given by $$g(x) = a + b\cos x + c\sin x$$ In other words, $T(V)$ is spanned by the independent set $\{1, \cos x, \sin x\},$ making $T(V)$ finite-dimensional with $\dim T(V) = 3. \quad \blacksquare$

(e) $\quad$ As we saw in (d), $T(V)$ is the set of real functions $g(x),$ continuous on $[-\pi, \pi]$ given by the equation $$g(x) = a + b\cos x + c\sin x$$ where $a, b, c$ are scalars given: \begin{align*} a &=\int_{-\pi}^{\pi}f(t)\,dt \\ b &= \int_{-\pi}^{\pi}f(t)\cos t\,dt \\ c &= \int_{-\pi}^{\pi}f(t)\sin t\,dt \end{align*} Accordingly, the null set of $T$ is the set of functions $f$ in $V$ such that $a, b,$ and $c$ evaluate to zero, making $g(x) = 0$ for all $x$ in $[-\pi, \pi]$. But as we encountered in parts (a) through (c), the set of all such functions in $V$ is the subspace $S. \quad \blacksquare$

(f) $\quad$ Let $g(x) = T(f)$ be the linear transformation of the element $f.$ From (d), we know that each element of $T(V)$ can be expressed as: \begin{align*} g(x) = c_1 + c_2\cos x + c_3\sin x \end{align*} where the scalars $c_1, c_2, c_3$ are given by: \begin{align*} c_1 &= \int_{-\pi}^{\pi} f(t)\,dt \\ c_2 &= \int_{-\pi}^{\pi}f(t)\cos t\,dt \\ c_3 &= \int_{-\pi}^{\pi}f(t)\sin t\,dt \end{align*} Now, if $g = cf$ for some scalar $c \neq 0,$ we have: \begin{align*} f(x) &= \frac{c_1}{c} + \frac{c_2}{c}\cos x + \frac{c_3}{c}\sin x \end{align*} Substituting this into our equation for $c_1,$ we get: \begin{align*} c_1 &= \int_{-\pi}^{\pi}\frac{c_1}{c} + \frac{c_2}{c}\cos x + \frac{c_3}{c}\sin x\,dx \\ &= 2\pi\frac{c_1}{c} \end{align*} Multiplying both sides by $c,$ we find that $c_1c = 2\pi c_1,$ giving us two cases:

(1) $\quad$ $c_1 = 0,$ in which case $f(x) = \frac{c_2}{c}\cos x + \frac{c_3}{c}\sin x.$ Plugging $f(x)$ into the equation for $c_2$ gives us \begin{align*} c_2 &= \int_{-\pi}^{\pi} \frac{c_2}{c}\cos^2 x + \frac{c_3}{c}\sin x\cos x\,dx \\ &= \int_{-\pi}^{\pi} \frac{c_2}{c}\cos^2 x \,dx \\ &= \int_{-\pi}^{\pi} \frac{c_2}{c}\cos 2x +\sin^2 x\,dx \\ &= \int_{-\pi}^{\pi} \frac{c_2}{c} \sin^2 x\,dx \end{align*} But, if $c_2 = \int_{-\pi}^{\pi} \frac{c_2}{c} \sin^2 x\,dx = \int_{-\pi}^{\pi} \frac{c_2}{c} \cos^2 x\,dx,$ then \begin{align*} 2c_2 &= \int_{-\pi}^{\pi} \frac{c_2}{c} \sin^2 x\,dx + \int_{-\pi}^{\pi} \frac{c_2}{c} \cos^2 x\,dx \\ &= 2\pi\frac{c_2}{c} \end{align*} Multiplying both sides by $c/2,$ we get $c_2c = c_2\pi,$ leading to another two cases:
$\quad$ (1.1) $\quad c_2=0,$ in which case $f(x) = \frac{c_3}{c}\sin x,$ which, when plugged into the equation for $c_3,$ gives us: \begin{align*} c_3 &= \int_{-\pi}^{\pi}\frac{c_3}{c}\sin^2 x\,dx \\ &= \frac{c_3}{c}\pi \end{align*} Multiplying both sides by $c,$ we have $ck = c_3\pi.$ By hypothesis, $f \neq O,$ thus $c_3 \neq 0,$ giving us $c = \pi,$ and \begin{align*} \\ f(x) &= \frac{c_3}{\pi}\sin x \end{align*} for some arbitrary $c_3 \neq 0.$
$\quad$ (1.2) $\quad c_2 \neq 0.$ Then, $c = \pi,$ and \begin{align*} f(x) &= \frac{c_2}{\pi}\cos x + \frac{c_3}{\pi}\sin x \end{align*} where $c_2$ and $c_3$ are arbitrary scalars, not both zero.

(2) $\quad c_1 \neq 0,$ in which case $c = 2\pi,$ and $f(x) = c_1/2\pi,$ where $c_1$ is an arbitrary nonzero scalar. $\quad \blacksquare$