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Consider the space $C_0 := \{f \in C[0,1], f(0)=0\}$ and the shift semigroup $(T^t, t \in [0,1])$, defined by

$T^t f(x) := \begin{cases} f(x-t), & t\le x\\ 0, & \text{otherwise} \end{cases}$

Clearly, for every $t \in [0,1]$ there is a $T$-invariant subspace $T^t C_0$ that consists of functions that vanish on $[0,t]$. Are there other closed $T$-invariant subspaces? I suspect not, but I failed to prove this...

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How about taking $f(x)=x^2$. The (non-closed) invariant subspace it generates, $\mbox{span}\{T^tf: 0\le t\le 1\}$, contains only differentiable functions, hence its (uniform) closure in $C_0$ also has this property. But not every function in $C_0$ is differentiable, so this seems to be a closed invariant subspace which isn't of the form you described since $f$ doesn't vanish away from $0$. – user108903 Jan 8 '13 at 8:59
@user108903: I believe this is wrong. That the non-closed invariant subspace contains only differentiable functions does not imply that their closure does. – Alexander Shamov Jan 8 '13 at 9:02
A uniform limit of differentiable functions is differentiable, and I suppose you want to give $C_0$ the uniform norm. So the set $D$ of differentiable functions in $C_0$ seems to be a closed subset of $C_0$, and the closure of any subset of $D$ is again a subset of $D$. Or maybe I'm doing something dumb... – user108903 Jan 8 '13 at 9:08
@user108903: "A uniform limit of differentiable functions is differentiable" - this is just plain wrong. See…. – Alexander Shamov Jan 8 '13 at 9:13
@user108903: Or even better, – Alexander Shamov Jan 8 '13 at 9:20

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