Let \(T\) be the indefinite integral operator \[ (Tf)(x) = \int_0^x f(t)\,dt \] on the space of continuous functions on the interval \([0, 1]\). Is the space of polynomial functions invariant under \(T\)? The space of differentiable functions? The space of functions which vanish at \(x = \frac{1}{2}\)?
Solution
The answers are, respectively, \[ \boxed{\text{yes},\qquad \text{yes},\qquad \text{no}.} \]
1. Polynomial functions form a \(T\)-invariant subspace.
If \[ f(x)=\sum_{k=0}^m a_kx^k, \] then direct integration gives \[ (Tf)(x)=\int_0^x f(t)\,dt =\sum_{k=0}^m\frac{a_k}{k+1}x^{k+1}, \] which is again a polynomial function. Thus \(T\) maps the polynomial subspace into itself, so that subspace is invariant [@HK71].
2. Differentiable functions form a \(T\)-invariant subspace.
Let \(f\) be differentiable on \([0,1]\). In particular \(f\) is continuous, so the fundamental theorem of calculus gives \[ (Tf)'(x)=f(x) \] for \(0<x<1\), with the corresponding one-sided derivatives at the endpoints. Hence \(Tf\) is differentiable on \([0,1]\). Therefore the differentiable functions are invariant under \(T\).
3. Functions vanishing at \(1/2\) do not form a \(T\)-invariant subspace.
Take \[ f(x)=x-\frac12. \] Then \(f(1/2)=0\), so \(f\) belongs to the stated subspace. But \[ (Tf)\left(\frac12\right) =\int_0^{1/2}\left(t-\frac12\right)dt =\left[\frac{t^2}{2}-\frac{t}{2}\right]_0^{1/2} =-\frac18\ne0. \] Thus \(Tf\) does not belong to the subspace. Hence that subspace is not \(T\)-invariant.