Let \(V\) be the vector space of continuous real-valued functions on the interval \([-1, 1]\) of the real line. Let \(W_e\) be the subspace of even functions, \(f(-x) = f(x)\), and let \(W_o\) be the subspace of odd functions, \(f(-x) = -f(x)\).
Show that \(V = W_e \oplus W_o\).
If \(T\) is the indefinite integral operator \[ (Tf)(x) = \int_0^x f(t)\,dt \] are \(W_e\) and \(W_o\) invariant under \(T\)?
Solution
1. Every continuous function is uniquely a sum of an even and an odd function.
For \(f\in V\), define \[ f_e(x)=\frac{f(x)+f(-x)}2, \qquad f_o(x)=\frac{f(x)-f(-x)}2. \] Both functions are continuous, and direct substitution gives \[ f_e(-x)=f_e(x), \qquad f_o(-x)=-f_o(x). \] Thus \(f_e\in W_e\), \(f_o\in W_o\), and \[ f=f_e+f_o. \] Hence \(V=W_e+W_o\). If \(h\in W_e\cap W_o\), then \[ h(x)=h(-x)=-h(x), \] so \(2h(x)=0\) for every \(x\). Over \(\mathbb R\), this implies \(h=0\). Therefore \[ \boxed{V=W_e\oplus W_o}. \]
2. Integration interchanges parity rather than preserving either summand.
Let \[ (Tf)(x)=\int_0^x f(t)\,dt. \] If \(f\) is even, then using \(t=-u\) gives \[ (Tf)(-x) =-\int_0^x f(-u)\,du =-\int_0^x f(u)\,du =-(Tf)(x). \] Thus \(T(W_e)\subseteq W_o\). If \(f\) is odd, the same substitution gives \[ (Tf)(-x) =-\int_0^x f(-u)\,du =\int_0^x f(u)\,du =(Tf)(x), \] so \(T(W_o)\subseteq W_e\) [@HK71].
Neither subspace is invariant. Indeed, \(f(x)=1\) lies in \(W_e\) but \[ Tf(x)=x\notin W_e, \] while \(f(x)=x\) lies in \(W_o\) but \[ Tf(x)=\frac{x^2}{2}\notin W_o. \] Therefore the answer to both invariance questions is \(\boxed{\text{no}}\).