Describe the range and the null space for the differentiation transformation of Example 2. Do the same for the integration transformation of Example 5.
Solution
For the differentiation map \(D\) on the space \(V\) of polynomial functions \(F\to F\), \[ \boxed{\operatorname{range}D=V, \qquad \ker D=\{\text{constant polynomial functions}\}.} \] For the integration map \(T\) on \(C(\mathbb R,\mathbb R)\), \[ \boxed{ \operatorname{range}T =\{g\in C^1(\mathbb R):g(0)=0\}, \qquad \ker T=\{0\}.} \]
1. Determine the range and null space of differentiation.
In Example 2, a polynomial function \[ p(x)=c_0+c_1x+\cdots+c_nx^n \] is sent to its derivative. Given any polynomial function \(p\) as above, define \[ q(x)=c_0x+\frac{c_1}{2}x^2+\cdots+\frac{c_n}{n+1}x^{n+1}. \] The setting in Hoffman–Kunze uses a field of characteristic zero here, so the displayed denominators are nonzero scalars. Then \(Dq=p\), proving \(\operatorname{range}D=V\) [@HK71].
Also \(Dp=0\) if and only if every coefficient of positive degree is zero, so \(p\) is constant. Thus the null space is exactly the subspace of constant polynomial functions.
2. Determine the range of the integration transformation.
Example 5 takes \[ V=C(\mathbb R,\mathbb R), \qquad (Tf)(x)=\int_0^x f(t)\,dt \] [@HK71]. If \(f\in V\), the fundamental theorem of calculus gives \[ (Tf)'=f, \] so \(Tf\) is continuously differentiable, and clearly \((Tf)(0)=0\). Hence \[ \operatorname{range}T \subseteq\{g\in C^1(\mathbb R):g(0)=0\}. \]
Conversely, if \(g\in C^1(\mathbb R)\) and \(g(0)=0\), then \(f=g'\) is continuous, so \(f\in V\), and the fundamental theorem gives \[ (Tf)(x)=\int_0^x g'(t)\,dt=g(x)-g(0)=g(x). \] Thus every such \(g\) lies in the range, proving equality.
3. Determine the null space of the integration transformation.
If \(Tf=0\), then the identically zero function \(Tf\) has derivative zero. But \((Tf)'=f\) by the fundamental theorem of calculus, so \(f=0\) identically. Conversely \(T0=0\). Therefore \(\ker T=\{0\}\).