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edited Jul 5 2010 at 8:05 |
For a positive answer to Q1, I came to the same assumptions stated by Bill Johnson in his answer, so I'll adopt his notations.
Let $X$ be a dense, countably generated linear subspace of the separable Banach space $Y$ with unit ball $B_Y.$ Let $T\in L(Y)$ and $\epsilon>0$. Write $X$ as increasing union of a sequence $0=X_0\subset X_1\subset \dots$ of finite dimensional subspaces, with linear projectors $P_n:Y\to X_n$ (in particular, $P_0=0.$)
I think we can choose the projectors $P_n$ (depending on $T$) in such a way that, for every $k$ we have $(I-P_n)T_{|X_k}\to 0$ in the operator norm, as $n\to\infty$. As a consequence, there exists a natural number $n_k$ such that $$\| (I-P_{n_k})\, T\, (P_k-P_{k-1})\|\leq \| (I-P_{n_k})\, T_{|X_k}\|\, \|P_k-P_{k-1}\| \leq \epsilon\ 2^{-k}.$$
The sum
$$\sum_{k=1}^\infty\ P_{n_k}\, T\, (P_k-P_{k-1})$$
is punctually finite on $X$, therefore it defines a linear map $T_{\epsilon}:X\to X$ (indeed, it takes $X_k$ into $X_{n_k}$ for every $k$). On the subspace $X$, the operator $T$ also writes in the form
$$\sum_{k=1}^\infty\ T\, (P_k-P_{k-1})$$
and one has, on the subspace $X$
$$T-T_{\epsilon}=\sum_{k=1}^\infty\ (I-P_{n_k})\, T\, (P_k-P_{k-1}).$$
By the choice of the sequence $n_k$ the latter series is normally convergent to an operator of norm less than $\epsilon$. Therefore $T_{\epsilon}$ extends to a bounded operator on $Y$ with a distance less than or equal to $\epsilon$ from $T$ such that $T_{\epsilon}(X)\subset\,X.$
The claim should be proved as suggested below by Bill Johnson. Also, a suitable lemma for proving the claim could be stated as follows:
Given the subspaces $\{X_n\}_n$ as above and a countable subset $A\subset Y,$ there are linear projectors $P_n:X\to X_n$ such that $P_na\to a$ as $n\to\infty,$ for all $a\in A.$
Applying this to $A$ equals to the image of a Hamel basis of $X$ via $T$, one has $\|(I-P_n)T_{|X_k}\|=o(1)$ as $n\to\infty$ as we wanted.
Rmk. It seems to me that the statement gains something in generality and semplicity if one considers a different Banach space as codomain: if $T:F\to F'$ is a bounded linear operator; $D\subset F$ is a countable subset; $D'\subset F'$ is dense subset linear subspace; then $T$ can be approximated in operator norm by operators that map $D$ into $D'$ -hence of course $\mathrm{span}(D)$ into $\mathrm{span}(D')$. D'$. This way one sees where the assumptions are needed: countability is only relevant for $D$, density is only relevant for $D'$, and linearity is not really, for both. D'$.
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edited Jul 5 2010 at 7:17 |
For a positive answer to Q1, I came to the same assumptions stated by Bill Johnson in his answer, so I'll adopt his notations.
Let $X$ be a dense, countably generated linear subspace of the separable Banach space $Y$ with unit ball $B_Y.$ Let $T\in L(Y)$ and $\epsilon>0$. Write $X$ as increasing union of a sequence $0=X_0\subset X_1\subset \dots$ of finite dimensional subspaces, with linear projectors $P_n:Y\to X_n$ (in particular, $P_0=0.$)
I think we can choose the projectors $P_n$ (depending on $T$) in such a way that, for every $k$ we have $(I-P_n)T_{|X_k}\to 0$ in the operator norm, as $n\to\infty$. As a consequence, there exists a natural number $n_k$ such that $$\| (I-P_{n_k})\, T\, (P_k-P_{k-1})\|\leq \| (I-P_{n_k})\, T_{|X_k}\|\, \|P_k-P_{k-1}\| \leq \epsilon\ 2^{-k}.$$
The sum
$$\sum_{k=1}^\infty\ P_{n_k}\, T\, (P_k-P_{k-1})$$
is punctually finite on $X$, therefore it defines a linear map $T_{\epsilon}:X\to X$ (indeed, it takes $X_k$ into $X_{n_k}$ for every $k$). On the subspace $X$, the operator $T$ also writes in the form
$$\sum_{k=1}^\infty\ T\, (P_k-P_{k-1})$$
and one has, on the subspace $X$
$$T-T_{\epsilon}=\sum_{k=1}^\infty\ (I-P_{n_k})\, T\, (P_k-P_{k-1}).$$
By the choice of the sequence $n_k$ the latter series is normally convergent to an operator of norm less than $\epsilon$. Therefore $T_{\epsilon}$ extends to a bounded operator on $Y$ with a distance less than or equal to $\epsilon$ from $T$ such that $T_{\epsilon}(X)\subset\,X.$
The claim should be proved as suggested below by Bill Johnson. Also, a suitable lemma for proving the claim could be stated as follows:
Given the subspaces $\{X_n\}_n$ as above and a countable subset $A\subset Y,$ there are linear projectors $P_n:X\to X_n$ such that $P_na\to a$ as $n\to\infty,$ for all $a\in A.$
Applying this to the $A$ equals to the image of a Hamel basis of $X$ via $T$, one has $\|(I-P_n)T_{|X_k}\|=o(1)$ as $n\to\infty$ as we wanted.
Rmk. It seems to me that the statement gains something in generality and semplicity if one considers a different Banach space as codomain: if $T:F\to F'$ is a bounded linear operator; $D\subset F$ is a countable subset; $D'\subset F'$ is dense subset; then $T$ can be approximated in operator norm by operators that map $D$ into $D'$ -hence of course $\mathrm{span}(D)$ into $\mathrm{span}(D')$. This way one sees where the assumptions are needed: countability is only relevant for $D$, density is only relevant for $D'$, and linearity is not really, for both.
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edited Jul 4 2010 at 18:35 |
For a positive answer to Q1, I came to the same assumptions stated by Bill Johnson in his answer, so I'll adopt his notations.
Let $X$ be a dense, countably generated linear subspace of the separable Banach space $Y$ with unit ball $B_Y.$ Let $T\in L(Y)$ and $\epsilon>0$. Write $X$ as increasing union of a sequence $0=X_0\subset X_1\subset \dots$ of finite dimensional subspaces, with linear projectors $P_n:Y\to X_n$ (in particular, $P_0=0.$)
I think we can choose the projectors $P_n$ (depending on $T$) in such a way that, for every $k$ we have $(I-P_n)T_{|X_k}\to 0$ in the operator norm, as $n\to\infty$. As a consequence, there exists a natural number $n_k$ such that $$\| (I-P_{n_k})\, T\, (P_k-P_{k-1})\|\leq \| (I-P_{n_k})\, T_{|X_k}\|\, \|P_k-P_{k-1}\| \leq \epsilon\ 2^{-k}.$$
The sum
$$\sum_{k=1}^\infty\ P_{n_k}\, T\, (P_k-P_{k-1})$$
is punctually finite on $X$, therefore it defines a linear map $T_{\epsilon}:X\to X$ (indeed, it takes $X_k$ into $X_{n_k}$ for every $k$). On the subspace $X$, the operator $T$ also writes in the form
$$\sum_{k=1}^\infty\ T\, (P_k-P_{k-1})$$
and one has, on the subspace $X$
$$T-T_{\epsilon}=\sum_{k=1}^\infty\ (I-P_{n_k})\, T\, (P_k-P_{k-1}).$$
By the choice of the sequence $n_k$ the latter series is normally convergent to an operator of norm less than $\epsilon$. Therefore $T_{\epsilon}$ extends to a bounded operator on $Y$ with a distance less than or equal to $\epsilon$ from $T$ such that $T_{\epsilon}(X)\subset\,X.$
The claim should be proved as suggested below by Bill Johnson. Also, a suitable lemma for proving the claim could be stated:
Given the subspaces $\{X_n\}_n$ as above and a countable subset $A\subset Y,$ there are linear projectors $P_n:X\to X_n$ such that $P_na\to a$ as $n\to\infty,$ for all $a\in A.$
Applying this to the $A$ image of a Hamel basis of $X$ one has $\|(I-P_n)T_{|X_k}\|=o(1)$ as $n\to\infty$ as we wanted.
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edited Jul 4 2010 at 10:40 |
For a positive answer to Q1, I came to the same assumptions stated by Bill Johnson in his answer, so I'll adopt his notations.
Let $X$ be a dense, countably generated linear subspace of the separable Banach space $Y$ with unit ball $B_Y.$ Let $T\in L(Y)$ and $\epsilon>0$. Write $X$ as increasing union of a sequence $0=X_0\subset X_1\subset \dots$ of finite dimensional subspaces, with linear projectors $P_k:Y\to X_k$ P_n:Y\to X_n$ (in particular, $P_0=0.$)
Let
I think we can choose the projectors $T\in L(Y)$ and P_n$ (depending on $\epsilon>0$.
For T$) in such a way that, for every $k$ we have $(I-P_n)T_{|X_k}\to 0$ in the operator norm, therefore as $n\to\infty$. As a consequence, there exists a natural number $n_k$ such that $$\| (I-P_{n_k})\, T\, (P_k-P_{k-1})\|\leq \| (I-P_{n_k})\, T_{|X_k}\|\, \|P_k-P_{k-1}\| \leq \epsilon\ 2^{-k}.$$
The sum
$$\sum_{k=1}^\infty\ P_{n_k}\, T\, (P_k-P_{k-1})$$
is punctually finite on $X$, therefore it defines a linear map $T_{\epsilon}:X\to X$ (indeed, it takes $X_k$ into $X_{n_k}$ for every $k$). On the subspace $X$, the operator $T$ also writes in the form
$$\sum_{k=1}^\infty\ T\, (P_k-P_{k-1})$$
and one has, on the subspace $X$
$$T-T_{\epsilon}=\sum_{k=1}^\infty\ (I-P_{n_k})\, T\, (P_k-P_{k-1}).$$
By the choice of the sequence $n_k$ the latter series is normally convergent to an operator of norm less than $\epsilon$. Therefore $T_{\epsilon}$ extends to a bounded operator on $Y$ with a distance less than or equal to $\epsilon$ from $T$ such that $T_{\epsilon}(X)\subset\,X .$T_{\epsilon}(X)\subset\,X.$
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edited Jul 4 2010 at 10:23 |
Let $X$ be a dense, countably generated linear subspace of the separable Banach space $Y$ with unit ball $B_Y.$ Write $X$ as increasing union of a sequence $0=X_0\subset X_1\subset \dots$ of finite dimensional subspaces, with linear projectors $P_k:Y\to X_k.$ X_k$ (in particular, $P_0=0.$) For any positive integer every $k,$ let k$ we have $n_k$ be so large that the image of (I-P_n)T_{|X_k}\to 0$ in the unit ball of Y is inside a uniform $\epsilon\ 2^{-k}$ neighborhood of $X_{n_k},$ that is $T(P_k-P_{k-1})B_Y\subset X_{n_k}+ \epsilon\ 2^{-k} B_Y.$ There exists such $n_k$ since $T(P_k-P_{k-1})B_Y$ is a relatively compact set. Thenoperator norm, let $Q_k$ therefore there exists a linear projector onto natural number $X_{n_k}$ n_k$ such that $\|(I-Q_k) T(P_k-P_{k-1})||\leq $\| (I-P_{n_k})\, T\, (P_k-P_{k-1})\|\leq \| (I-P_{n_k})\, T_{|X_k}\|\, \|P_k-P_{k-1}\| \leq \epsilon\ 2^{-k}$ . Now define$$T_\epsilon:=\sum_{k=1}^\infty\ Q_k\$\sum_{k=1}^\infty\ P_{n_k}\, T\, (P_k-P_{k-1}).$$ The series converges to an operator $T_\epsilon$ which P_k-P_{k-1})$$is punctually finite on $\epsilon$ apart from X$, therefore it defines a linear map $T$ and T_{\epsilon}:X\to X$ (indeed, it takes $X_k$ into $X_{n_k}$, so that takes X_{n_k}$ for every $X$ into k$). On the subspace $X$ X$, the operator $T$ also writes in the form $$\sum_{k=1}^\infty\ T\, (note thatP_k-P_{k-1})$$and one has, of courseon the subspace $X$ $$T-T_{\epsilon}=\sum_{k=1}^\infty\ (I-P_{n_k})\, T\, (P_k-P_{k-1}).$$ By the norms choice of the sequence $P_k$ and of n_k$ the latter series is normally convergent to an operator of norm less than $Q_k$ may be unbounded)\epsilon$. Therefore $T_{\epsilon}$ extends to a bounded operator on $Y$ with a distance less than or equal to $\epsilon$ from $T$ such that $T_{\epsilon}(X)\subset\,X .$
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edited Jul 4 2010 at 1:23 |
For a positive answer to Q1, I came to the same assumptions stated by Bill Johnson in his answer, so I'll adopt his notations.
Let $X$ be a dense, countably generated linear subspace of the separable Banach space $Y$ with unit ball $B_Y.$ Write $X$ as increasing union of a sequence $0=X_0\subset X_1\subset \dots$ of finite dimensional subspaces, with linear projectors $P_k:Y\to X_k.$
Let $T\in L(Y)$ and $\epsilon>0$. For any positive integer $k,$ let $n_k$ be so large that the image of the unit ball of Y is inside a uniform $\epsilon\ 2^{-k}$ neighborhood of $X_{n_k},$ that is $T(P_k-P_{k-1})B_Y\subset X_{n_k}+ \epsilon\ 2^{-k} B_Y.$ There exists such $n_k$ since $T(P_k-P_{k-1})B_Y$ is a relatively compact set. Then, let $Q_k$ a linear projector onto $X_{n_k}$ such that $\|(I-Q_k) T(P_k-P_{k-1})||\leq \epsilon\ 2^{-k}$ . Now define
$$T_\epsilon:=\sum_{k=1}^\infty\ Q_k\, T\, (P_k-P_{k-1}).$$
The series converges to an operator $T_\epsilon$ which is $\epsilon$ apart from $T$ and takes $X_k$ into $X_{n_k}$, so that has invariant subspace takes $X$. X$ into $X$ (note that, of course, the norms of the $P_k$ and of the $Q_k$ may be unbounded).
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edited Jul 4 2010 at 0:48 |
For a positive answer to Q1, I came to the same assumptions stated by Bill Johnson in his answer, so I'll adopt his notations.
Let $X$ be a dense, countably generated linear subspace of the separable Banach space $Y$ with unit ball $B_Y.$ Write $X$ as increasing union of a sequence $0=X_0\subset X_1\subset \dots$ of finite dimensional subspaces, with linear projectors $P_k:Y\to X_k.$
Let $T\in L(Y)$ and $\epsilon>0$. For any positive integer $k,$ let $n_k$ be so large that the image of the unit ball of Y is inside a uniform $\epsilon\ 2^{-k}$ neighborhood of $X_{n_k},$ that is $T(P_k-P_{k-1})B_Y\subset X_{n_k}+ \epsilon\ 2^{-k} B_Y.$ There exists such $n_k$ since $T(P_k-P_{k-1})B_Y$ is a relatively compact set. Then, let $Q_k$ a linear projector onto $X_{n_k}$ such that $\|Q_k \|(I-Q_k) T(P_k-P_{k-1})||\leq \epsilon\ 2^{-k}$ . Now define
$$T_\epsilon:=\sum_{k=1}^\infty\ Q_k\, T\, (P_k-P_{k-1}).$$
The series normally converges to an operator $T_\epsilon$ which is $\epsilon$ apart from $T$ and takes $X_k$ into $X_{n_k}$, so that has with invariant subspace $X$. (note that, of course, the norms of the $P_k$ and of the $Q_k$ may be unbounded).
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Post Undeleted by Pietro Majer
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occurred Jul 4 2010 at 0:44
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Post Deleted by Pietro Majer
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occurred Jul 4 2010 at 0:44
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edited Jul 4 2010 at 0:37 |
For a positive answer to Q1, I came to the same assumptions stated by Bill Johnson in his answer, so I'll adopt his notations.
Let $X$ be a dense, countably generated linear subspace of the separable Banach space $Y$ with unit ball $B_Y.$ Write $X$ as increasing union of a sequence $0=X_0\subset X_1\subset \dots$ of finite dimensional subspaces, with linear projectors $P_k:Y\to X_k.$
Let $T\in L(Y)$ and $\epsilon>0$. For any positive integer $k,$ let $n_k$ be so large that the image of the unit ball of Y is inside a uniform $\epsilon\ 2^{-k}$ neighborhood of $X_{n_k},$ that is $T(P_k-P_{k-1})B_Y\subset X_{n_k}+ \epsilon\ 2^{-k} B_Y.$ There exists such $n_k$ since $T(P_k-P_{k-1})B_Y$ is a relatively compact set. Then, let $Q_k$ a linear projector onto $X_{n_k}$ such that $\|Q_k T(P_k-P_{k-1}||\leq T(P_k-P_{k-1})||\leq \epsilon\ 2^{-k}.$ 2^{-k}$ . Now define
$$T_\epsilon:=\sum_{k=1}^\infty\ Q_k\, T\, (P_k-P_{k-1}).$$
The series normally converges to an operator $T_\epsilon$ which is $\epsilon$ apart from $T$ and takes $X_k$ into $X_{n_k}$, so that has with invariant subspace $X$. (note that, of course, the norms of the $P_k$ and of the $Q_k$ may be unbounded).
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answered Jul 4 2010 at 0:31 |
For a positive answer to Q1, I came to the same assumptions stated by Bill Johnson in his answer, so I'll adopt his notations.
Let $X$ be a dense, countably generated linear subspace of the separable Banach space $Y$ with unit ball $B_Y.$ Write $X$ as increasing union of a sequence $0=X_0\subset X_1\subset \dots$ of finite dimensional subspaces, with linear projectors $P_k:Y\to X_k.$
Let $T\in L(Y)$ and $\epsilon>0$. For any positive integer $k,$ let $n_k$ be so large that the image of the unit ball of Y is inside a uniform $\epsilon\ 2^{-k}$ neighborhood of $X_{n_k},$ that is $T(P_k-P_{k-1})B_Y\subset X_{n_k}+ \epsilon\ 2^{-k} B_Y.$ There exists such $n_k$ since $T(P_k-P_{k-1})B_Y$ is a relatively compact set. Then, let $Q_k$ a linear projector onto $X_{n_k}$ such that $\|Q_k T(P_k-P_{k-1}||\leq \epsilon\ 2^{-k}.$ Now define
$$T_\epsilon:=\sum_{k=1}^\infty\ Q_k\, T\, (P_k-P_{k-1}).$$
The series normally converges to an operator $T_\epsilon$ which is $\epsilon$ apart from $T$ and takes $X_k$ into $X_{n_k}$, so that has with invariant subspace $X$.
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