Let \(X\) be a topological space and let \(V\) be an \(n\)-dimensional subspace of the real vector space of continuous functions on \(X\). Prove that there are a basis \[ \{f_1,\dots,f_n\} \] of \(V\) and points \(x_1,\dots,x_n\in X\) such that \[ f_i(x_j)=\delta_{ij}. \]
Solution
For each \(x\in X\), let \[ \operatorname{ev}_x:V\longrightarrow\RR, \qquad \operatorname{ev}_x(f)=f(x) \] be the evaluation functional.
1. The evaluation functionals span \(V^*\):\[ \spanof\{\operatorname{ev}_x:x\in X\}=V^*. \]
Let \[ W\coloneqq\spanof\{\operatorname{ev}_x:x\in X\}\subseteq V^*. \] Suppose \(W\neq V^*\). Since \(V\) is finite-dimensional, the annihilator \[ W^\perp = \{f\in V:\lambda(f)=0\text{ for every }\lambda\in W\} \] is then nonzero. Choose \(0\neq f\in W^\perp\). For every \(x\in X\), \[ f(x)=\operatorname{ev}_x(f)=0, \] because \(\operatorname{ev}_x\in W\). Hence \(f\) is the zero function, a contradiction. Thus \(W=V^*\).
2. There are points \(x_1,\ldots,x_n\in X\) such that\[ \operatorname{ev}_{x_1},\ldots,\operatorname{ev}_{x_n} \]is a basis of \(V^*\).
By step 1, the family of all evaluation functionals spans the \(n\)-dimensional vector space \(V^*\). A spanning family in a finite-dimensional vector space contains a basis, so choose \(n\) evaluations forming one.
3. There is a basis \(f_1,\ldots,f_n\) of \(V\) satisfying\[ f_i(x_j)=\delta_{ij}. \]
Define \[ T:V\longrightarrow\RR^n, \qquad T(f)=\bigl(f(x_1),\ldots,f(x_n)\bigr). \] If \(T(f)=0\), then \[ \operatorname{ev}_{x_j}(f)=0 \] for every \(j\). Since the functionals in step 2 form a basis of \(V^*\), every element of \(V^*\) vanishes on \(f\), hence \(f=0\). Thus \(T\) is injective. Both \(V\) and \(\RR^n\) have dimension \(n\), so \(T\) is an isomorphism.
For the standard basis \(e_1,\ldots,e_n\) of \(\RR^n\), put \[ f_i\coloneqq T^{-1}(e_i). \] Then \(f_1,\ldots,f_n\) is a basis of \(V\), and the \(j\)-th coordinate of \(T(f_i)=e_i\) gives \[ f_i(x_j)=\delta_{ij}. \]
4. Q.E.D.
Step 3 constructs the required basis and points.