1

Suppose X be a Banach space, T be a bounded operator on X and Y be a T-invariant subspace of X (not necessary closed subspacace). If T is injective on Y, can we say T will be also injective on closure of Y?

Ben Grossmann
  • 225,327
Ali Farzaneh
  • 329

1 Answers1

4

Let $X = \ell^2(\mathbb{N})$, and $Y = c_{00}$ be the subspace of sequences with only finitely many nonzero terms. We know $\overline{Y} = X$. Then let $S$ be the left shift,

$$(S(x))_n = x_{n+1}$$

and $T = I - 2S$. Since $S$ is bounded and $Y$ is $S$-invariant, the same holds for $T$. Furthermore, $T$ is injective on $Y$: Let $0 \neq y\in Y$ and $n_y\in \mathbb{N}$ such that $y_{n_y} \neq 0$ and $y_n = 0$ for all $n > n_y$. Then

$$(T(y))_{n_y} = y_{n_y} - 2y_{n_y+1} = y_{n_y} \neq 0,$$

so $T(y) \neq 0$. But $x = \sum 2^{-n}e_n \in X\setminus Y$ satisfies $T(x) = 0$, so $T$ is not injective on $X$.

Daniel Fischer
  • 206,697
  • Tanks a lot Mr Fischer – Ali Farzaneh Aug 06 '15 at 15:22
  • Hi. I think that answers the issue from the comments in Resolvent Set. Your example is a closable densely-defined operator. It has densely-defined bounded inverse. According to the 'equivalent' definition zero would be part of the resolvent set. But the operator is not closed. So according to the 'correct' definition necessarily zero is not part of the resolvent. And indeed considering its closure one encounters that zero still does not belong to the resolvent set for either definition. ... – C-star-W-star Aug 07 '15 at 01:05
  • ... This counterexample is maximal in the following sense. Suppose there were a closable densely-defined operator with everywhere defined bounded inverse. Then it were already closed by the closed graph theorem. Besides, this counterexample is even a bounded one! – C-star-W-star Aug 07 '15 at 01:07
  • Besides nice answer :) – C-star-W-star Aug 07 '15 at 01:08
  • @Freeze_S The inverse of $T\lvert_Y$ is not bounded, $T(\sum_{n=0}^k 2^{-n}e_n) = 2^{-k}e_k$, so "$\lVert (T\lvert_Y)^{-1}\rVert \geqslant 2^k$". – Daniel Fischer Aug 07 '15 at 06:15
  • @DanielFischer: Right. Ok but at least the other way around: $T^{-1}$ is densely defined and has bounded inverse $T$. Unfortunately, $T^{-1}$ is not closable as the closure of its inverse is not injective, i.e. $\overline{T}(\tfrac{1}{2^n})_n=0$. Maybe one can modify it however. I'm thinking of something like $1-\frac12 L$. What do you think? – C-star-W-star Aug 07 '15 at 06:47
  • 1
    @Freeze_S If you have a densely defined injective operator $U\colon X\to Y$ with dense range, and its inverse $V$, then $\overline{\Gamma(U)} = \sigma(\overline{\Gamma(V)})$, where $\sigma(y,x) = (x,y)$. But if $V$ is bounded, then $\overline{\Gamma(V)}$ is the graph of the continuous extension $W$ of $V$ to $Y$, so $U$ is closable if and only if $W$ is injective. – Daniel Fischer Aug 07 '15 at 06:52
  • ...and that proves there is no intersting counterexample. heh-heh ^^ Thanks!!!! – C-star-W-star Aug 07 '15 at 06:56