Let \(V\) be a finite-dimensional complex vector space of \(C^\infty\) functions on \(\mathbb R\). Suppose \(V\) is closed under differentiation. Prove that \(V\) is closed under translations: if \(f\in V\) and \(a\in\mathbb R\), then \[ x\longmapsto f(x+a) \] also lies in \(V\).
Solution
If \(V=\{0\}\), the conclusion is immediate. Assume henceforth that \[ \dim_{\CC}V=d\geq1, \] and choose a basis \(f_1,\ldots,f_d\) of \(V\).
1. There is a matrix \(M=(M_{ij})\in M_d(\CC)\) such that\[ f_j' = \sum_{i=1}^d M_{ij}f_i \]for every \(1\leq j\leq d\).
By hypothesis, differentiation maps \(V\) into itself. Therefore each \(f_j'\) is a unique linear combination of the chosen basis. Put its coefficients in the \(j\)th column of \(M\).
2. For the row vector\[ F(x)\coloneqq \begin{pmatrix} f_1(x)&\cdots&f_d(x) \end{pmatrix}, \]one has\[ F'(x)=F(x)M. \]
The \(j\)th component of \(F'(x)\) is \(f_j'(x)\). By step 1, \[ f_j'(x) = \sum_{i=1}^d f_i(x)M_{ij}, \] which is exactly the \(j\)th component of the row vector \(F(x)M\).
3. For every \(x\in\RR\),\[ F(x)=F(0)e^{xM}. \]
Consider \[ G(x)\coloneqq F(x)e^{-xM}. \] Since \[ \frac{d}{dx}e^{-xM} = -Me^{-xM}, \] step 2 gives \[ \begin{aligned} G'(x) &= F'(x)e^{-xM} +F(x)\frac{d}{dx}e^{-xM}\\ &= F(x)Me^{-xM} -F(x)Me^{-xM}\\ &=0. \end{aligned} \] Thus \(G\) is constant, so \[ F(x)e^{-xM}=F(0). \] Right multiplication by \(e^{xM}\) yields the claim.
4. For every \(a,x\in\RR\),\[ F(x+a)=F(x)e^{aM}. \]
By step 3, \[ \begin{aligned} F(x+a) &= F(0)e^{(x+a)M}\\ &= F(0)e^{xM}e^{aM}\\ &= F(x)e^{aM}. \end{aligned} \] The middle equality holds because both matrix exponentials are functions of the same matrix \(M\).
5. If\[ f=\sum_{j=1}^d c_jf_j\in V, \]then for every \(a\in\RR\) the translate\[ x\longmapsto f(x+a) \]also belongs to \(V\).
Let \[ c= \begin{pmatrix} c_1\\ \vdots\\ c_d \end{pmatrix}. \] Then \[ f(x)=F(x)c. \] By step 4, \[ f(x+a) = F(x+a)c = F(x)e^{aM}c. \] The column vector \(e^{aM}c\) is independent of \(x\), so the last expression is a fixed complex linear combination of \(f_1(x),\ldots,f_d(x)\). Hence the translated function lies in \(V\).
6. Therefore\[ \boxed{V\text{ is closed under every real translation}}. \]
Step 5 applies to every \(f\in V\) and every \(a\in\RR\); the zero-space case was settled before step 1.
7. Q.E.D.
Step 6 is the required conclusion.