Normed spaces continuity at 0 proof
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It says for V and V' under two different normed spaces, let T:V->V' be a linear map. Then T is continuous at 0 implies T is continuous. But then they say
''Since $left | Tx-Ty right |' = left | T(x-y) right |'<epsilon$
, T is continuous at all y in V ''.
So I dont understand where they have used the fact that T is continuous at 0 as if it were wouldnt we use $left | Tx right |'<epsilon$ only to deduce that its continuous at any y.
functional-analysis
asked Dec 10 '18 at 13:55
NoteBookNoteBook
1197
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add a comment |
$begingroup$
It says for V and V' under two different normed spaces, let T:V->V' be a linear map. Then T is continuous at 0 implies T is continuous. But then they say
''Since $left | Tx-Ty right |' = left | T(x-y) right |'<epsilon$
, T is continuous at all y in V ''.
So I dont understand where they have used the fact that T is continuous at 0 as if it were wouldnt we use $left | Tx right |'<epsilon$ only to deduce that its continuous at any y.
functional-analysis
asked Dec 10 '18 at 13:55
NoteBookNoteBook
1197
$endgroup$
add a comment |
$begingroup$
It says for V and V' under two different normed spaces, let T:V->V' be a linear map. Then T is continuous at 0 implies T is continuous. But then they say
''Since $left | Tx-Ty right |' = left | T(x-y) right |'<epsilon$
, T is continuous at all y in V ''.
So I dont understand where they have used the fact that T is continuous at 0 as if it were wouldnt we use $left | Tx right |'<epsilon$ only to deduce that its continuous at any y.
functional-analysis
asked Dec 10 '18 at 13:55
NoteBookNoteBook
1197
$endgroup$
It says for V and V' under two different normed spaces, let T:V->V' be a linear map. Then T is continuous at 0 implies T is continuous. But then they say
''Since $left | Tx-Ty right |' = left | T(x-y) right |'<epsilon$
, T is continuous at all y in V ''.
So I dont understand where they have used the fact that T is continuous at 0 as if it were wouldnt we use $left | Tx right |'<epsilon$ only to deduce that its continuous at any y.
functional-analysis
functional-analysis
asked Dec 10 '18 at 13:55
NoteBookNoteBook
1197
asked Dec 10 '18 at 13:55
NoteBookNoteBook
1197
asked Dec 10 '18 at 13:55
NoteBookNoteBook
1197
asked Dec 10 '18 at 13:55
NoteBookNoteBook
1197
asked Dec 10 '18 at 13:55
NoteBookNoteBook
1197
1197
add a comment |
add a comment |
2 Answers
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Assume $T$ is continuous at $0$. It means given $epsilon >0$ there exist $delta>0$ such that for all $|z|<delta$ one has $|T(z)||'<epsilon$.
Now suppose we want to show that $T$ is continuous at $vin V$. Given $epsilon >0$, take the $delta >0$ above in continuity at $0$. Now for every $xin V$ such that $|x-v|<delta$ we have
$$|T(x)-T(v)|'=|T(x-v)|'=|T(z)|'<epsilon$$
where $z=x-v$.
answered Dec 10 '18 at 14:03
user9077user9077
1,239612
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If you use the sequence definition of continuity, you take an arbitrary sequence $x_nxrightarrow{to infty } x$ and want to prove that the image also converges.
Observe that any converging sequence $x_nxrightarrow{to infty } x$ defines a sequence converging to zero of the form $x-x_n$, but now the above equality shows that the converging of $T(x_n) to T(x)$ is equivalent to $T(x_n -x) xrightarrow{to infty} 0$. Which is guaranteed by continuity at 0.
answered Dec 10 '18 at 14:01
EnkiduEnkidu
1,30119
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
Assume $T$ is continuous at $0$. It means given $epsilon >0$ there exist $delta>0$ such that for all $|z|<delta$ one has $|T(z)||'<epsilon$.
Now suppose we want to show that $T$ is continuous at $vin V$. Given $epsilon >0$, take the $delta >0$ above in continuity at $0$. Now for every $xin V$ such that $|x-v|<delta$ we have
$$|T(x)-T(v)|'=|T(x-v)|'=|T(z)|'<epsilon$$
where $z=x-v$.
answered Dec 10 '18 at 14:03
user9077user9077
1,239612
$endgroup$
add a comment |
$begingroup$
Assume $T$ is continuous at $0$. It means given $epsilon >0$ there exist $delta>0$ such that for all $|z|<delta$ one has $|T(z)||'<epsilon$.
Now suppose we want to show that $T$ is continuous at $vin V$. Given $epsilon >0$, take the $delta >0$ above in continuity at $0$. Now for every $xin V$ such that $|x-v|<delta$ we have
$$|T(x)-T(v)|'=|T(x-v)|'=|T(z)|'<epsilon$$
where $z=x-v$.
answered Dec 10 '18 at 14:03
user9077user9077
1,239612
$endgroup$
add a comment |
$begingroup$
Assume $T$ is continuous at $0$. It means given $epsilon >0$ there exist $delta>0$ such that for all $|z|<delta$ one has $|T(z)||'<epsilon$.
Now suppose we want to show that $T$ is continuous at $vin V$. Given $epsilon >0$, take the $delta >0$ above in continuity at $0$. Now for every $xin V$ such that $|x-v|<delta$ we have
$$|T(x)-T(v)|'=|T(x-v)|'=|T(z)|'<epsilon$$
where $z=x-v$.
answered Dec 10 '18 at 14:03
user9077user9077
1,239612
$endgroup$
Assume $T$ is continuous at $0$. It means given $epsilon >0$ there exist $delta>0$ such that for all $|z|<delta$ one has $|T(z)||'<epsilon$.
Now suppose we want to show that $T$ is continuous at $vin V$. Given $epsilon >0$, take the $delta >0$ above in continuity at $0$. Now for every $xin V$ such that $|x-v|<delta$ we have
$$|T(x)-T(v)|'=|T(x-v)|'=|T(z)|'<epsilon$$
where $z=x-v$.
answered Dec 10 '18 at 14:03
user9077user9077
1,239612
answered Dec 10 '18 at 14:03
user9077user9077
1,239612
answered Dec 10 '18 at 14:03
user9077user9077
1,239612
answered Dec 10 '18 at 14:03
user9077user9077
1,239612
1,239612
add a comment |
add a comment |
$begingroup$
If you use the sequence definition of continuity, you take an arbitrary sequence $x_nxrightarrow{to infty } x$ and want to prove that the image also converges.
Observe that any converging sequence $x_nxrightarrow{to infty } x$ defines a sequence converging to zero of the form $x-x_n$, but now the above equality shows that the converging of $T(x_n) to T(x)$ is equivalent to $T(x_n -x) xrightarrow{to infty} 0$. Which is guaranteed by continuity at 0.
answered Dec 10 '18 at 14:01
EnkiduEnkidu
1,30119
$endgroup$
add a comment |
$begingroup$
If you use the sequence definition of continuity, you take an arbitrary sequence $x_nxrightarrow{to infty } x$ and want to prove that the image also converges.
Observe that any converging sequence $x_nxrightarrow{to infty } x$ defines a sequence converging to zero of the form $x-x_n$, but now the above equality shows that the converging of $T(x_n) to T(x)$ is equivalent to $T(x_n -x) xrightarrow{to infty} 0$. Which is guaranteed by continuity at 0.
answered Dec 10 '18 at 14:01
EnkiduEnkidu
1,30119
$endgroup$
add a comment |
$begingroup$
If you use the sequence definition of continuity, you take an arbitrary sequence $x_nxrightarrow{to infty } x$ and want to prove that the image also converges.
Observe that any converging sequence $x_nxrightarrow{to infty } x$ defines a sequence converging to zero of the form $x-x_n$, but now the above equality shows that the converging of $T(x_n) to T(x)$ is equivalent to $T(x_n -x) xrightarrow{to infty} 0$. Which is guaranteed by continuity at 0.
answered Dec 10 '18 at 14:01
EnkiduEnkidu
1,30119
$endgroup$
If you use the sequence definition of continuity, you take an arbitrary sequence $x_nxrightarrow{to infty } x$ and want to prove that the image also converges.
Observe that any converging sequence $x_nxrightarrow{to infty } x$ defines a sequence converging to zero of the form $x-x_n$, but now the above equality shows that the converging of $T(x_n) to T(x)$ is equivalent to $T(x_n -x) xrightarrow{to infty} 0$. Which is guaranteed by continuity at 0.
answered Dec 10 '18 at 14:01
EnkiduEnkidu
1,30119
answered Dec 10 '18 at 14:01
EnkiduEnkidu
1,30119
answered Dec 10 '18 at 14:01
EnkiduEnkidu
1,30119
answered Dec 10 '18 at 14:01
EnkiduEnkidu
1,30119
1,30119
add a comment |
add a comment |
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