- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.8 Exercises
27.$\quad$ Let $V$ be the linear space of all real polynomials $p(x).$ Let $D$ denote the differentiation operator and let $T$ denote the integration operator which maps each polynomial $p$ onto the polynomial $q$ given by $q(x) = \int_0^x p(t) \, dt.$ Prove that $DT = I_V$ but that $TD \neq I_V.$ Describe the null space and range of $TD.$
Solution. $\quad$ Using the First Fundamental Theorem of Calculus (Volume 1, Section 5.1), we find that for every $p$ in $V,$ we have \begin{align*} DT[p(x)] &= D[q(x)] \\ &= D\left[\int_0^x p(t)\,dt\right] \\ &= p(x) \end{align*} Thus, $DT = I_V.$
$\quad$ A counterexample to show that $TD \neq I_V$ is when $p(x) = c$ (where $c$ is a nonzero scalar). If $p(x)$ is constant, $D(p) = 0$ for all $x,$ giving us \begin{align*} TD[p(x)] &= T(0) \\ &= \int_0^x 0\,dt \\ &= 0 \\ &\neq p(x) \end{align*} As demonstrated above, the null space of $TD$ is the set of all constant real polynomials (that is, real polynomials of degree $0$), and the range of $TD$ is the set of all real polynomials with no constant coefficient. That is, the set of real polynomials $p$ such that $p(0) = 0. \quad \blacksquare$