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$X,Y$ Banach, $V subset X$ a linear subspace, $T:Vto Y$ closed, then $T$ bounded $iff$ $V$ closed
When is the image of a linear operator closed?Closed range for maps between banach spaces?Closed linear operator on Banach spaces.Unbounded linear operator with bounded restrictionProve that if a functional is in the dual space then the null space of the functional is closed.Closed linear operators have closed kernelsCan every closed subspace be realized as kernel of a bounded linear operator from a Banach space to itself?Image of a closed set is closed under bounded linear transformation between Banach spaces?Bounded linear operator (between Banach spaces) with second category range has closed rangeShow that every closed linear subspace $ Y $ of a Banach space $ X $ is weakly sequentially closed
$begingroup$
Let $X$ and $Y$ be Banach spaces and let $V subset X$ be a linear subspace. Let $T: V to Y$ be a closed linear operator. Show that $T$ is bounded if and only if $V$ is closed.
For the direction $V$ closed $Rightarrow$ $T$ bounded this is simply the closed graph theorem. However I am stuck for the direction $T$ bounded $Rightarrow$ $V$ closed.
I tried to prove $V$ not closed $Rightarrow$ $T$ unbounded, by something along the lines of
let $v_n$ be a sequence in $V$ converging to a limit $x in Xsetminus V$ then $(v_n,Tv_n)$ has limit $(x,y)$ for $y=lim Tv_n$ and then I guess we seek to show something like that $fracy$ is unbounded if we choose $v_n$ in the right way. Can anyone help?
functional-analysis
asked Mar 22 at 9:15
DanDan
998
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be Banach spaces and let $V subset X$ be a linear subspace. Let $T: V to Y$ be a closed linear operator. Show that $T$ is bounded if and only if $V$ is closed.
For the direction $V$ closed $Rightarrow$ $T$ bounded this is simply the closed graph theorem. However I am stuck for the direction $T$ bounded $Rightarrow$ $V$ closed.
I tried to prove $V$ not closed $Rightarrow$ $T$ unbounded, by something along the lines of
let $v_n$ be a sequence in $V$ converging to a limit $x in Xsetminus V$ then $(v_n,Tv_n)$ has limit $(x,y)$ for $y=lim Tv_n$ and then I guess we seek to show something like that $fracy$ is unbounded if we choose $v_n$ in the right way. Can anyone help?
functional-analysis
asked Mar 22 at 9:15
DanDan
998
$endgroup$
$begingroup$
Probably, you mean be the closedness of $T$ that its graph $(x,T(x)):xin V$ is not only closed in $Vtimes Y$ but even in $Xtimes Y$ (this is not the standard terminology).
$endgroup$
– Jochen
Mar 22 at 12:53
$begingroup$
That's indeed what I meant, thanks for pointing that out.
$endgroup$
– Dan
Mar 22 at 17:41
add a comment |
$begingroup$
Let $X$ and $Y$ be Banach spaces and let $V subset X$ be a linear subspace. Let $T: V to Y$ be a closed linear operator. Show that $T$ is bounded if and only if $V$ is closed.
For the direction $V$ closed $Rightarrow$ $T$ bounded this is simply the closed graph theorem. However I am stuck for the direction $T$ bounded $Rightarrow$ $V$ closed.
I tried to prove $V$ not closed $Rightarrow$ $T$ unbounded, by something along the lines of
let $v_n$ be a sequence in $V$ converging to a limit $x in Xsetminus V$ then $(v_n,Tv_n)$ has limit $(x,y)$ for $y=lim Tv_n$ and then I guess we seek to show something like that $fracy$ is unbounded if we choose $v_n$ in the right way. Can anyone help?
functional-analysis
asked Mar 22 at 9:15
DanDan
998
$endgroup$
Let $X$ and $Y$ be Banach spaces and let $V subset X$ be a linear subspace. Let $T: V to Y$ be a closed linear operator. Show that $T$ is bounded if and only if $V$ is closed.
For the direction $V$ closed $Rightarrow$ $T$ bounded this is simply the closed graph theorem. However I am stuck for the direction $T$ bounded $Rightarrow$ $V$ closed.
I tried to prove $V$ not closed $Rightarrow$ $T$ unbounded, by something along the lines of
let $v_n$ be a sequence in $V$ converging to a limit $x in Xsetminus V$ then $(v_n,Tv_n)$ has limit $(x,y)$ for $y=lim Tv_n$ and then I guess we seek to show something like that $fracy$ is unbounded if we choose $v_n$ in the right way. Can anyone help?
functional-analysis
functional-analysis
asked Mar 22 at 9:15
DanDan
998
asked Mar 22 at 9:15
DanDan
998
asked Mar 22 at 9:15
DanDan
998
asked Mar 22 at 9:15
DanDan
998
asked Mar 22 at 9:15
DanDan
998
998
$begingroup$
Probably, you mean be the closedness of $T$ that its graph $(x,T(x)):xin V$ is not only closed in $Vtimes Y$ but even in $Xtimes Y$ (this is not the standard terminology).
$endgroup$
– Jochen
Mar 22 at 12:53
$begingroup$
That's indeed what I meant, thanks for pointing that out.
$endgroup$
– Dan
Mar 22 at 17:41
add a comment |
$begingroup$
Probably, you mean be the closedness of $T$ that its graph $(x,T(x)):xin V$ is not only closed in $Vtimes Y$ but even in $Xtimes Y$ (this is not the standard terminology).
$endgroup$
– Jochen
Mar 22 at 12:53
$begingroup$
That's indeed what I meant, thanks for pointing that out.
$endgroup$
– Dan
Mar 22 at 17:41
$begingroup$
Probably, you mean be the closedness of $T$ that its graph $(x,T(x)):xin V$ is not only closed in $Vtimes Y$ but even in $Xtimes Y$ (this is not the standard terminology).
$endgroup$
– Jochen
Mar 22 at 12:53
$begingroup$
Probably, you mean be the closedness of $T$ that its graph $(x,T(x)):xin V$ is not only closed in $Vtimes Y$ but even in $Xtimes Y$ (this is not the standard terminology).
$endgroup$
– Jochen
Mar 22 at 12:53
$begingroup$
That's indeed what I meant, thanks for pointing that out.
$endgroup$
– Dan
Mar 22 at 17:41
$begingroup$
That's indeed what I meant, thanks for pointing that out.
$endgroup$
– Dan
Mar 22 at 17:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The way I'd do it: Assume $T$ is a bounded linear operator. Let $x$ be a point in the closure of $V$, and let $v_n$ be a Cauchy sequence of points in $V$ converging to $x$. Then, since $T$ is uniformly continuous ($|Tx-Ty|le |T|cdot|x-y|$), $Tv_n$ is a Cauchy sequence in $Y$. Let its limit be $y$.
Then the points $(v_n,Tv_n)$ converge to $(x,y)$ in $Xtimes Y$. Since the graph of $T$ is closed in $Xtimes Y$ by hypothesis, $(x,y)$ is in the graph. In particular, $xin V$. This works for any $x$ in the closure, so $overlineVsubset V$ and $V$ is closed in $X$.
answered Mar 22 at 9:52
jmerryjmerry
17k11633
$endgroup$
$begingroup$
thanks jmerry that's great.
$endgroup$
– Dan
Mar 22 at 17:44
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The way I'd do it: Assume $T$ is a bounded linear operator. Let $x$ be a point in the closure of $V$, and let $v_n$ be a Cauchy sequence of points in $V$ converging to $x$. Then, since $T$ is uniformly continuous ($|Tx-Ty|le |T|cdot|x-y|$), $Tv_n$ is a Cauchy sequence in $Y$. Let its limit be $y$.
Then the points $(v_n,Tv_n)$ converge to $(x,y)$ in $Xtimes Y$. Since the graph of $T$ is closed in $Xtimes Y$ by hypothesis, $(x,y)$ is in the graph. In particular, $xin V$. This works for any $x$ in the closure, so $overlineVsubset V$ and $V$ is closed in $X$.
answered Mar 22 at 9:52
jmerryjmerry
17k11633
$endgroup$
$begingroup$
thanks jmerry that's great.
$endgroup$
– Dan
Mar 22 at 17:44
add a comment |
$begingroup$
The way I'd do it: Assume $T$ is a bounded linear operator. Let $x$ be a point in the closure of $V$, and let $v_n$ be a Cauchy sequence of points in $V$ converging to $x$. Then, since $T$ is uniformly continuous ($|Tx-Ty|le |T|cdot|x-y|$), $Tv_n$ is a Cauchy sequence in $Y$. Let its limit be $y$.
Then the points $(v_n,Tv_n)$ converge to $(x,y)$ in $Xtimes Y$. Since the graph of $T$ is closed in $Xtimes Y$ by hypothesis, $(x,y)$ is in the graph. In particular, $xin V$. This works for any $x$ in the closure, so $overlineVsubset V$ and $V$ is closed in $X$.
answered Mar 22 at 9:52
jmerryjmerry
17k11633
$endgroup$
$begingroup$
thanks jmerry that's great.
$endgroup$
– Dan
Mar 22 at 17:44
add a comment |
$begingroup$
The way I'd do it: Assume $T$ is a bounded linear operator. Let $x$ be a point in the closure of $V$, and let $v_n$ be a Cauchy sequence of points in $V$ converging to $x$. Then, since $T$ is uniformly continuous ($|Tx-Ty|le |T|cdot|x-y|$), $Tv_n$ is a Cauchy sequence in $Y$. Let its limit be $y$.
Then the points $(v_n,Tv_n)$ converge to $(x,y)$ in $Xtimes Y$. Since the graph of $T$ is closed in $Xtimes Y$ by hypothesis, $(x,y)$ is in the graph. In particular, $xin V$. This works for any $x$ in the closure, so $overlineVsubset V$ and $V$ is closed in $X$.
answered Mar 22 at 9:52
jmerryjmerry
17k11633
$endgroup$
The way I'd do it: Assume $T$ is a bounded linear operator. Let $x$ be a point in the closure of $V$, and let $v_n$ be a Cauchy sequence of points in $V$ converging to $x$. Then, since $T$ is uniformly continuous ($|Tx-Ty|le |T|cdot|x-y|$), $Tv_n$ is a Cauchy sequence in $Y$. Let its limit be $y$.
Then the points $(v_n,Tv_n)$ converge to $(x,y)$ in $Xtimes Y$. Since the graph of $T$ is closed in $Xtimes Y$ by hypothesis, $(x,y)$ is in the graph. In particular, $xin V$. This works for any $x$ in the closure, so $overlineVsubset V$ and $V$ is closed in $X$.
answered Mar 22 at 9:52
jmerryjmerry
17k11633
answered Mar 22 at 9:52
jmerryjmerry
17k11633
answered Mar 22 at 9:52
jmerryjmerry
17k11633
answered Mar 22 at 9:52
jmerryjmerry
17k11633
17k11633
$begingroup$
thanks jmerry that's great.
$endgroup$
– Dan
Mar 22 at 17:44
add a comment |
$begingroup$
thanks jmerry that's great.
$endgroup$
– Dan
Mar 22 at 17:44
$begingroup$
thanks jmerry that's great.
$endgroup$
– Dan
Mar 22 at 17:44
$begingroup$
thanks jmerry that's great.
$endgroup$
– Dan
Mar 22 at 17:44
add a comment |
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$begingroup$
Probably, you mean be the closedness of $T$ that its graph $(x,T(x)):xin V$ is not only closed in $Vtimes Y$ but even in $Xtimes Y$ (this is not the standard terminology).
$endgroup$
– Jochen
Mar 22 at 12:53
$begingroup$
That's indeed what I meant, thanks for pointing that out.
$endgroup$
– Dan
Mar 22 at 17:41