linear map of bounded sets into bounded sets implies a bounded operator
I was watching a video lecture on bounded linear operators from one normed linear space to another. It was stated that if $T$ sends bounded sets in $X$ to bounded sets in $Y$ then $T$ is a bounded operator. I found this to be very hard to prove. The complication, as I understand it, arises for sequences $x_n$ converging to zero in $X$ where we might need an arbitrary large constant $K$ to get $|Tx_n|leq K|x_n|$ even though there exists an constant $C$ such that $|Tx_n|leq C$. Another approach would be to prove that mapping bounded sets to bounded implies continuity at some point, for example zero, but I have not succeeding in proving this either. Also in the video lecture, it had not yet been proven that linear bounded and linear continuous operators are the same, whereas I suspect this is not the standard approach.
functional-analysis
edited Dec 1 '18 at 11:08
Martin Sleziak
44.7k8115271
asked Jan 13 '14 at 19:27
harajmharajm
7901819
add a comment |
I was watching a video lecture on bounded linear operators from one normed linear space to another. It was stated that if $T$ sends bounded sets in $X$ to bounded sets in $Y$ then $T$ is a bounded operator. I found this to be very hard to prove. The complication, as I understand it, arises for sequences $x_n$ converging to zero in $X$ where we might need an arbitrary large constant $K$ to get $|Tx_n|leq K|x_n|$ even though there exists an constant $C$ such that $|Tx_n|leq C$. Another approach would be to prove that mapping bounded sets to bounded implies continuity at some point, for example zero, but I have not succeeding in proving this either. Also in the video lecture, it had not yet been proven that linear bounded and linear continuous operators are the same, whereas I suspect this is not the standard approach.
functional-analysis
edited Dec 1 '18 at 11:08
Martin Sleziak
44.7k8115271
asked Jan 13 '14 at 19:27
harajmharajm
7901819
5
If $Vert TxVertle C$ for all $x$ of norm $1$, then for any non-zero $x$, $Vert T(x/Vert xVert)Vert le C$. But this implies $Vert TxVertle CVert xVert$.
– David Mitra
Jan 13 '14 at 19:30
1
@DavidMitra: that seems like it should be an answer, to me
– Ben Millwood
Jan 13 '14 at 20:12
add a comment |
I was watching a video lecture on bounded linear operators from one normed linear space to another. It was stated that if $T$ sends bounded sets in $X$ to bounded sets in $Y$ then $T$ is a bounded operator. I found this to be very hard to prove. The complication, as I understand it, arises for sequences $x_n$ converging to zero in $X$ where we might need an arbitrary large constant $K$ to get $|Tx_n|leq K|x_n|$ even though there exists an constant $C$ such that $|Tx_n|leq C$. Another approach would be to prove that mapping bounded sets to bounded implies continuity at some point, for example zero, but I have not succeeding in proving this either. Also in the video lecture, it had not yet been proven that linear bounded and linear continuous operators are the same, whereas I suspect this is not the standard approach.
functional-analysis
edited Dec 1 '18 at 11:08
Martin Sleziak
44.7k8115271
asked Jan 13 '14 at 19:27
harajmharajm
7901819
I was watching a video lecture on bounded linear operators from one normed linear space to another. It was stated that if $T$ sends bounded sets in $X$ to bounded sets in $Y$ then $T$ is a bounded operator. I found this to be very hard to prove. The complication, as I understand it, arises for sequences $x_n$ converging to zero in $X$ where we might need an arbitrary large constant $K$ to get $|Tx_n|leq K|x_n|$ even though there exists an constant $C$ such that $|Tx_n|leq C$. Another approach would be to prove that mapping bounded sets to bounded implies continuity at some point, for example zero, but I have not succeeding in proving this either. Also in the video lecture, it had not yet been proven that linear bounded and linear continuous operators are the same, whereas I suspect this is not the standard approach.
functional-analysis
functional-analysis
edited Dec 1 '18 at 11:08
Martin Sleziak
44.7k8115271
asked Jan 13 '14 at 19:27
harajmharajm
7901819
edited Dec 1 '18 at 11:08
Martin Sleziak
44.7k8115271
asked Jan 13 '14 at 19:27
harajmharajm
7901819
edited Dec 1 '18 at 11:08
Martin Sleziak
44.7k8115271
edited Dec 1 '18 at 11:08
Martin Sleziak
44.7k8115271
edited Dec 1 '18 at 11:08
Martin Sleziak
44.7k8115271
44.7k8115271
asked Jan 13 '14 at 19:27
harajmharajm
7901819
asked Jan 13 '14 at 19:27
harajmharajm
7901819
asked Jan 13 '14 at 19:27
harajmharajm
7901819
7901819
5
If $Vert TxVertle C$ for all $x$ of norm $1$, then for any non-zero $x$, $Vert T(x/Vert xVert)Vert le C$. But this implies $Vert TxVertle CVert xVert$.
– David Mitra
Jan 13 '14 at 19:30
1
@DavidMitra: that seems like it should be an answer, to me
– Ben Millwood
Jan 13 '14 at 20:12
add a comment |
5
If $Vert TxVertle C$ for all $x$ of norm $1$, then for any non-zero $x$, $Vert T(x/Vert xVert)Vert le C$. But this implies $Vert TxVertle CVert xVert$.
– David Mitra
Jan 13 '14 at 19:30
1
@DavidMitra: that seems like it should be an answer, to me
– Ben Millwood
Jan 13 '14 at 20:12
5
5
If $Vert TxVertle C$ for all $x$ of norm $1$, then for any non-zero $x$, $Vert T(x/Vert xVert)Vert le C$. But this implies $Vert TxVertle CVert xVert$.
– David Mitra
Jan 13 '14 at 19:30
If $Vert TxVertle C$ for all $x$ of norm $1$, then for any non-zero $x$, $Vert T(x/Vert xVert)Vert le C$. But this implies $Vert TxVertle CVert xVert$.
– David Mitra
Jan 13 '14 at 19:30
1
1
@DavidMitra: that seems like it should be an answer, to me
– Ben Millwood
Jan 13 '14 at 20:12
@DavidMitra: that seems like it should be an answer, to me
– Ben Millwood
Jan 13 '14 at 20:12
add a comment |
1 Answer
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It follows from the homogeneity of the norm (that is $Vert alpha xVert=|alpha|Vert xVert$):
If $Vert TxVert≤C$ for all $x$ of norm $1$, then for any non-zero $x$,
$$biggllVert Tleft(frac{x}{lVert xrVert}right)biggrrVert≤C.$$
But this implies $Vert TxVert≤CVert xVert$ for all $x$.
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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It follows from the homogeneity of the norm (that is $Vert alpha xVert=|alpha|Vert xVert$):
If $Vert TxVert≤C$ for all $x$ of norm $1$, then for any non-zero $x$,
$$biggllVert Tleft(frac{x}{lVert xrVert}right)biggrrVert≤C.$$
But this implies $Vert TxVert≤CVert xVert$ for all $x$.
add a comment |
It follows from the homogeneity of the norm (that is $Vert alpha xVert=|alpha|Vert xVert$):
If $Vert TxVert≤C$ for all $x$ of norm $1$, then for any non-zero $x$,
$$biggllVert Tleft(frac{x}{lVert xrVert}right)biggrrVert≤C.$$
But this implies $Vert TxVert≤CVert xVert$ for all $x$.
add a comment |
It follows from the homogeneity of the norm (that is $Vert alpha xVert=|alpha|Vert xVert$):
If $Vert TxVert≤C$ for all $x$ of norm $1$, then for any non-zero $x$,
$$biggllVert Tleft(frac{x}{lVert xrVert}right)biggrrVert≤C.$$
But this implies $Vert TxVert≤CVert xVert$ for all $x$.
It follows from the homogeneity of the norm (that is $Vert alpha xVert=|alpha|Vert xVert$):
If $Vert TxVert≤C$ for all $x$ of norm $1$, then for any non-zero $x$,
$$biggllVert Tleft(frac{x}{lVert xrVert}right)biggrrVert≤C.$$
But this implies $Vert TxVert≤CVert xVert$ for all $x$.
edited Dec 1 '18 at 3:27
community wiki
2 revs, 2 users 67%
David Mitra
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5
If $Vert TxVertle C$ for all $x$ of norm $1$, then for any non-zero $x$, $Vert T(x/Vert xVert)Vert le C$. But this implies $Vert TxVertle CVert xVert$.
– David Mitra
Jan 13 '14 at 19:30
1
@DavidMitra: that seems like it should be an answer, to me
– Ben Millwood
Jan 13 '14 at 20:12