Let \(X\) be the Banach space of continuous real-valued functions on \([0,\pi]\) that vanish at \(0\) and \(\pi\), with the supremum norm. Suppose \(Y\subset X\) is a closed subspace and every element of \(Y\) is a trigonometric polynomial. Prove that \(Y\) is finite dimensional.
Solution
For each \(N\ge0\), let \[ V_N=\operatorname{span}\{1,\sin x,\cos x,\ldots,\sin Nx,\cos Nx\}, \] and set \[ Y_N=Y\cap V_N. \] Each \(V_N\) is finite dimensional, hence closed in \(X\), so \(Y_N\) is closed in \(Y\). By hypothesis, every \(y\in Y\) is a trigonometric polynomial of some finite degree, therefore \[ Y=\bigcup_{N=0}^\infty Y_N. \] Since \(Y\) is closed in the Banach space \(X\), it is itself Banach. By the Baire category theorem, some \(Y_N\) has nonempty interior relative to \(Y\). Because \(Y_N\) is a linear subspace of \(Y\), nonempty interior forces \[ Y_N=Y. \] Indeed, if a ball \(B_Y(y_0,r)\subset Y_N\), then \(y_0\in Y_N\) and hence \(B_Y(0,r)\subset Y_N\); by scalar multiplication every vector of \(Y\) then lies in \(Y_N\).
Thus \[ Y\subset V_N, \] and \(V_N\) is finite dimensional. Hence \(Y\) is finite dimensional.