Invariant subspace

[Tsakanikas Nickos]

Let \( \displaystyle T : C \left( \left[ 0 , 1 \right] \right) \longrightarrow C \left( \left[ 0 , 1 \right] \right) \, , \, \left( Tf \right)(x) = \int_{0}^{x} f(t) \mathrm{d}t \) be an integral operator.

• Is the space of polynomial functions invariant under \( T \)?
• Is the space of differentiable functions invariant under \( T \)?
• Is the space of functions that vanish at \( x = { 1 \over 2 } \) invariant under \( T \)?

[Papapetros Vaggelis]

Hellos Nickos.

Firstly, the integral operator \(\displaystyle{T}\) is well defined since for every continuous function \(\displaystyle{f:\left[0,1\right]\longrightarrow \mathbb{R}}\)

the function \(\displaystyle{T(f):\left[0,1\right]\longrightarrow \mathbb{R}\,,T(f)(x)=\int_{0}^{x}f(t)\,\mathrm{d}t}\) is differentiable.

So, \(\displaystyle{T(f)}\) is continuous.

Also, the integral operator \(\displaystyle{T}\) is \(\displaystyle{\mathbb{R}}\) - linear.

Let \(\displaystyle{W}\) be the \(\displaystyle{\mathbb{R}}\) -subspace of polynomial functions on \(\displaystyle{\left[0,1\right]}\) .

If \(\displaystyle{f(x)=a_0+a_1\,x+...+a_{n}\,x^{n}\,,x\in\left[0,1\right]}\) is a typical element of \(\displaystyle{W}\), then :

\(\displaystyle{\begin{aligned} T(f)(x)&=\int_{0}^{x}f(t)\,\mathrm{d}t\\&=\int_{0}^{x}\sum_{i=0}^{n}a_{i}\,t^{i}\,\mathrm{d}t\\&=\sum_{i=0}^{n}\int_{0}^{x}a_{i}\,t^{i}\,\mathrm{d}t\\&=\sum_{i=0}^{n}\left[\dfrac{a_{i}}{i+1}\,t^{i+1}\right]_{0}^{x}\\&=\sum_{i=0}^{n}\dfrac{a_{i}}{i+1}\,x^{i+1}\,,\forall\,x\in\left[0,1\right]\end{aligned}}\)

so: \(\displaystyle{T(f)\in W}\) and thus: \(\displaystyle{T(W)\subseteq W}\) .

The set \(\displaystyle{D}\) of differentiable functions is \(\displaystyle{\mathbb{R}}\) - subspace of \(\displaystyle{\left(C(\left[0,1\right]),+,\cdot\right)}\) .

Let \(\displaystyle{f\in D}\) . Then, the function \(\displaystyle{T(f)}\) is differentiable and continuous and additionally,

\(\displaystyle{T(f)^\prime(x)=f(x)\,,\forall\,x\in\left[0,1\right]}\) . Therefore, \(\displaystyle{T(D)\subseteq D}\) .

Now, the last subspace is \(\displaystyle{V=\left\{f\in C(\left[0,1\right]): f(1/2)=0\right\}\leq C(\left[0,1\right])}\) .

We define \(\displaystyle{f:\left[0,1\right]\longrightarrow \mathbb{R}}\) by \(\displaystyle{f(x)=2\,x-1}\) .

Obviously, \(\displaystyle{f\in V}\) but :

\(\displaystyle{\begin{aligned} T(f)(x)&=\int_{0}^{x}f(t)\mathrm{d}t\\&=\int_{0}^{x}\left(2\,t-1\right)\,\mathrm{d}t\\&=\left[t^2-t\right]_{0}^{x}\\&=x^2-x\,\,\forall\,x\in\left[0,1\right]\end{aligned}}\)

and \(\displaystyle{T(f)\,\left(\dfrac{1}{2}\right)=\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{4}\neq 0}\) .

So, \(\displaystyle{T(f)\notin V}\) and \(\displaystyle{V}\) is not invariant under \(\displaystyle{T}\) .

Additional question

Find \(\displaystyle{\rm{Ker}(T)}\) and \(\displaystyle{Im(T)}\) .