Sequence of increasing null spaces and range Find a vector space W and T ∈ L(W)  such that null T^k ⊊ null T^(k+1) and range T^k ⊋ range T^(k+1) for every positive integer k.
 A: Define
$\tau(x) = \max(0,\min(1, 2(x-{1 \over 2})+{1 \over 2}))$.
We have $\tau^k (x) = \max(0,\min(1, 2^k(x-{1 \over 2})+{1 \over 2})))$, where $\tau^k$ means the $k$-fold composition of $\tau$.
Let $W=C[0,1]$ and define $Tf = f \circ \tau$. It is straightforward to see that $T^k f = f \circ \tau^k$.
We see that $\ker T^k = \{ f | f(x)=0\  \forall x \in [{1 \over 2}-{1 \over 2^{k+1}}, {1 \over 2}+{1 \over 2^{k+1}} ] \}$.
We see that ${\cal R}T^k = \{ f | f(x) = f(0) \  \forall x \in [0,{1 \over 2}-{1 \over 2^{k+1}}] \ \land \ f(x)=f(1) \ \forall x \in [{1 \over 2}+{1 \over 2^{k+1}},1] \}$.
It is straightforward to verify that $\ker T^k \subset \ker T^{k+1}$ and 
${\cal R}T^{k+1} \subset {\cal R}T^k$.
By taking any function that is zero on $[0, {1 \over 2}+{1 \over 2^{k+2}}]$ and $f({1 \over 2}+{1 \over 2^{k+1}}) \neq 0$, we see that the kernel containment is strict.
By taking the function that is zero on $[0, {1 \over 2}-{1 \over 2^{k+1}}]$,
one on $[{1 \over 2}+{1 \over 2^{k+1}},1]$ and linearly interpolated between,
we see that the range containment is strict.
A: Let $E$ an infinite dimensional vector space, with a basis $\{e_0,e_1, \cdots \}$. Define $T$ by $T(e_0)=0$, $T(e_{2k})=e_{2(k-1)}$ for $k\geq 1$, and $T(e_{2m+1})=e_{2m+3}$ if $m\geq 0$. We have ${\rm Ker}(T)={\rm Span}\{e_0\}=<e_0>$, and ${\rm Range}(T)=< e_{2k}, k\geq 0, e_{2l+1}, l\geq 1>$. We see ${\rm Ker}(T^2)=<e_0, e_2>$ and ${\rm Range}(T^2)=< e_{2k}, k\geq 0, e_{2l+1}, l\geq 2>$. Now generalise this and find ${\rm Ker}(T^k)$ and ${\rm Range}(T^k)$ for $k\geq 2$.  
