Let \(V\) be a finite-dimensional complex linear subspace of \(C^\infty(\mathbb R)\) and suppose \(V\) is invariant under differentiation: \[ f\in V\implies Df\in V. \] Prove that there is a nonzero constant-coefficient differential operator \[ L=\sum_{k=0}^{n}a_kD^k \] such that \(V\) is exactly the space of solutions of \[ Lf=0. \]
Solution
Let \[ d\coloneqq\dim_{\CC}V. \]
1. If \(d=0\), the assertion holds with\[ \boxed{L=1}. \]
Then \[ V=\{0\}, \] and the equation \[ Lf=f=0 \] has exactly the zero function as its solution.
2. Assume \(d>0\). Differentiation restricts to a linear endomorphism\[ T\coloneqq D|_V:V\to V. \]
This is exactly the hypothesis that \(V\) is invariant under differentiation.
3. Let\[ P(\lambda) \coloneqq \det(\lambda I-T) = \lambda^d+a_{d-1}\lambda^{d-1}+\cdots+a_0 \]be the characteristic polynomial of \(T\). Then\[ P(T)=0 \]on \(V\).
This is the Cayley–Hamilton theorem applied to the endomorphism \(T\) of the \(d\)-dimensional complex vector space \(V\).
4. Define the nonzero constant-coefficient differential operator\[ L \coloneqq P(D) = D^d+a_{d-1}D^{d-1}+\cdots+a_0. \]Then\[ V\subseteq\ker L. \]
If \(f\in V\), every derivative of \(f\) that occurs in \(P(D)f\) is computed by repeated application of the restricted operator \(T\). Hence \[ Lf = P(D)f = P(T)f =0 \] by step 3.
5. A solution \(f\) of\[ Lf=0 \]is uniquely determined by the \(d\) initial values\[ f(0),f'(0),\ldots,f^{(d-1)}(0). \]
The equation is monic of order \(d\): \[ f^{(d)} = -a_{d-1}f^{(d-1)} -\cdots -a_0f. \] Thus it is the standard linear homogeneous initial-value problem with continuous, in fact constant, coefficients. The uniqueness theorem for linear ordinary differential equations says that prescribed values of \[ f(0),f'(0),\ldots,f^{(d-1)}(0) \] admit at most one solution. Equivalently, a solution whose first \(d\) initial derivatives vanish is identically zero.
6. The solution space\[ W\coloneqq\ker L \]has dimension at most \(d\).
Consider the linear map \[ J:W\to\CC^d, \qquad J(f) = \bigl(f(0),f'(0),\ldots,f^{(d-1)}(0)\bigr). \] By step 5, \(J\) is injective. Therefore \[ \dim_{\CC}W \leq \dim_{\CC}\CC^d =d. \]
7. One has\[ \boxed{V=\ker L}. \]
Step 4 gives \[ V\subseteq W. \] Since \[ \dim V=d, \] this inclusion gives \[ d\leq\dim W. \] Step 6 gives the reverse inequality \[ \dim W\leq d. \] Hence the dimensions are equal, and the inclusion of finite-dimensional spaces must be equality.
8. Q.E.D.
Step 1 settles the zero-dimensional case. For \(d>0\), step 7 gives a nonzero constant-coefficient operator whose solution space is exactly \(V\).