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relation between linear functional and kernel on nls [closed]
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Let $f$ be a linear functional on a normed linear space $X$. Prove that $f$ is discontinious iff $ker(f)$ is a proper dense subspace of $X$
I tried to get with discontinuity of $f$ but not getting the idea how to deal with kernel
functional-analysis
asked Jul 20 at 16:24
ravi yadav
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closed as off-topic by Shaun, Adrian Keister, Xander Henderson, John Ma, Shailesh Jul 21 at 9:29
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- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Adrian Keister, Xander Henderson, John Ma, Shailesh
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Let $f$ be a linear functional on a normed linear space $X$. Prove that $f$ is discontinious iff $ker(f)$ is a proper dense subspace of $X$
I tried to get with discontinuity of $f$ but not getting the idea how to deal with kernel
functional-analysis
asked Jul 20 at 16:24
ravi yadav
295
closed as off-topic by Shaun, Adrian Keister, Xander Henderson, John Ma, Shailesh Jul 21 at 9:29
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Adrian Keister, Xander Henderson, John Ma, Shailesh
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f$ be a linear functional on a normed linear space $X$. Prove that $f$ is discontinious iff $ker(f)$ is a proper dense subspace of $X$
I tried to get with discontinuity of $f$ but not getting the idea how to deal with kernel
functional-analysis
asked Jul 20 at 16:24
ravi yadav
295
Let $f$ be a linear functional on a normed linear space $X$. Prove that $f$ is discontinious iff $ker(f)$ is a proper dense subspace of $X$
I tried to get with discontinuity of $f$ but not getting the idea how to deal with kernel
functional-analysis
asked Jul 20 at 16:24
ravi yadav
295
asked Jul 20 at 16:24
ravi yadav
295
asked Jul 20 at 16:24
ravi yadav
295
asked Jul 20 at 16:24
ravi yadav
295
295
closed as off-topic by Shaun, Adrian Keister, Xander Henderson, John Ma, Shailesh Jul 21 at 9:29
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Adrian Keister, Xander Henderson, John Ma, Shailesh
closed as off-topic by Shaun, Adrian Keister, Xander Henderson, John Ma, Shailesh Jul 21 at 9:29
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Adrian Keister, Xander Henderson, John Ma, Shailesh
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1 Answer
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Suppose that $f$ is discontinuous. Then $fneq 0$, so $ker f$ is proper. To prove that $ker f$ is dense in $X$, let $x_0in X$, let $lambda:=f(x_0)$, and let $lefty_nrightsubset X$ be a sequence such that $|y_n|=1$ and $|f(y_n)|to +infty $ (it exists because $f$ is discontinuous).
Now consider
$$x_n:=x_0-fracy_nf(y_n)lambda $$
then $f(x_n)=0$, so that $x_nin ker f$ for all $n$, and
$$|x_n-x_0|=left|fraclambdaf(y_n)right|to 0 $$
Therefore, $x_nto x_0$, and $x_0in overlineker f$.
Conversely, if $ker f$ is proper and dense, it is non-closed, so since $ker f$ is the preimage of the singleton $left0right$ (which is closed in $mathbbR$), $f$ has to be discontinuous.
By the way, this is basically equivalent to the fact that a linear functional is continuous if and only if its kernel is closed.
answered Jul 20 at 16:43
Lorenzo Quarisa
2,319314
But this not true if we have any linear transformation ,Just take $X=C^1[a,b] and Y =C[a,b]$ and $T(f)=f'$ then its kernel is closed but still T is not continuous.
– ravi yadav
Jul 20 at 18:03
Indeed. It only holds if the codomain is finite dimensional.
– Lorenzo Quarisa
Jul 20 at 20:16
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Suppose that $f$ is discontinuous. Then $fneq 0$, so $ker f$ is proper. To prove that $ker f$ is dense in $X$, let $x_0in X$, let $lambda:=f(x_0)$, and let $lefty_nrightsubset X$ be a sequence such that $|y_n|=1$ and $|f(y_n)|to +infty $ (it exists because $f$ is discontinuous).
Now consider
$$x_n:=x_0-fracy_nf(y_n)lambda $$
then $f(x_n)=0$, so that $x_nin ker f$ for all $n$, and
$$|x_n-x_0|=left|fraclambdaf(y_n)right|to 0 $$
Therefore, $x_nto x_0$, and $x_0in overlineker f$.
Conversely, if $ker f$ is proper and dense, it is non-closed, so since $ker f$ is the preimage of the singleton $left0right$ (which is closed in $mathbbR$), $f$ has to be discontinuous.
By the way, this is basically equivalent to the fact that a linear functional is continuous if and only if its kernel is closed.
answered Jul 20 at 16:43
Lorenzo Quarisa
2,319314
But this not true if we have any linear transformation ,Just take $X=C^1[a,b] and Y =C[a,b]$ and $T(f)=f'$ then its kernel is closed but still T is not continuous.
– ravi yadav
Jul 20 at 18:03
Indeed. It only holds if the codomain is finite dimensional.
– Lorenzo Quarisa
Jul 20 at 20:16
add a comment |Â
up vote
3
down vote
accepted
Suppose that $f$ is discontinuous. Then $fneq 0$, so $ker f$ is proper. To prove that $ker f$ is dense in $X$, let $x_0in X$, let $lambda:=f(x_0)$, and let $lefty_nrightsubset X$ be a sequence such that $|y_n|=1$ and $|f(y_n)|to +infty $ (it exists because $f$ is discontinuous).
Now consider
$$x_n:=x_0-fracy_nf(y_n)lambda $$
then $f(x_n)=0$, so that $x_nin ker f$ for all $n$, and
$$|x_n-x_0|=left|fraclambdaf(y_n)right|to 0 $$
Therefore, $x_nto x_0$, and $x_0in overlineker f$.
Conversely, if $ker f$ is proper and dense, it is non-closed, so since $ker f$ is the preimage of the singleton $left0right$ (which is closed in $mathbbR$), $f$ has to be discontinuous.
By the way, this is basically equivalent to the fact that a linear functional is continuous if and only if its kernel is closed.
answered Jul 20 at 16:43
Lorenzo Quarisa
2,319314
But this not true if we have any linear transformation ,Just take $X=C^1[a,b] and Y =C[a,b]$ and $T(f)=f'$ then its kernel is closed but still T is not continuous.
– ravi yadav
Jul 20 at 18:03
Indeed. It only holds if the codomain is finite dimensional.
– Lorenzo Quarisa
Jul 20 at 20:16
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Suppose that $f$ is discontinuous. Then $fneq 0$, so $ker f$ is proper. To prove that $ker f$ is dense in $X$, let $x_0in X$, let $lambda:=f(x_0)$, and let $lefty_nrightsubset X$ be a sequence such that $|y_n|=1$ and $|f(y_n)|to +infty $ (it exists because $f$ is discontinuous).
Now consider
$$x_n:=x_0-fracy_nf(y_n)lambda $$
then $f(x_n)=0$, so that $x_nin ker f$ for all $n$, and
$$|x_n-x_0|=left|fraclambdaf(y_n)right|to 0 $$
Therefore, $x_nto x_0$, and $x_0in overlineker f$.
Conversely, if $ker f$ is proper and dense, it is non-closed, so since $ker f$ is the preimage of the singleton $left0right$ (which is closed in $mathbbR$), $f$ has to be discontinuous.
By the way, this is basically equivalent to the fact that a linear functional is continuous if and only if its kernel is closed.
answered Jul 20 at 16:43
Lorenzo Quarisa
2,319314
Suppose that $f$ is discontinuous. Then $fneq 0$, so $ker f$ is proper. To prove that $ker f$ is dense in $X$, let $x_0in X$, let $lambda:=f(x_0)$, and let $lefty_nrightsubset X$ be a sequence such that $|y_n|=1$ and $|f(y_n)|to +infty $ (it exists because $f$ is discontinuous).
Now consider
$$x_n:=x_0-fracy_nf(y_n)lambda $$
then $f(x_n)=0$, so that $x_nin ker f$ for all $n$, and
$$|x_n-x_0|=left|fraclambdaf(y_n)right|to 0 $$
Therefore, $x_nto x_0$, and $x_0in overlineker f$.
Conversely, if $ker f$ is proper and dense, it is non-closed, so since $ker f$ is the preimage of the singleton $left0right$ (which is closed in $mathbbR$), $f$ has to be discontinuous.
By the way, this is basically equivalent to the fact that a linear functional is continuous if and only if its kernel is closed.
answered Jul 20 at 16:43
Lorenzo Quarisa
2,319314
edited Jul 20 at 20:13
answered Jul 20 at 16:43
Lorenzo Quarisa
2,319314
answered Jul 20 at 16:43
Lorenzo Quarisa
2,319314
answered Jul 20 at 16:43
Lorenzo Quarisa
2,319314
2,319314
But this not true if we have any linear transformation ,Just take $X=C^1[a,b] and Y =C[a,b]$ and $T(f)=f'$ then its kernel is closed but still T is not continuous.
– ravi yadav
Jul 20 at 18:03
Indeed. It only holds if the codomain is finite dimensional.
– Lorenzo Quarisa
Jul 20 at 20:16
add a comment |Â
But this not true if we have any linear transformation ,Just take $X=C^1[a,b] and Y =C[a,b]$ and $T(f)=f'$ then its kernel is closed but still T is not continuous.
– ravi yadav
Jul 20 at 18:03
Indeed. It only holds if the codomain is finite dimensional.
– Lorenzo Quarisa
Jul 20 at 20:16
But this not true if we have any linear transformation ,Just take $X=C^1[a,b] and Y =C[a,b]$ and $T(f)=f'$ then its kernel is closed but still T is not continuous.
– ravi yadav
Jul 20 at 18:03
But this not true if we have any linear transformation ,Just take $X=C^1[a,b] and Y =C[a,b]$ and $T(f)=f'$ then its kernel is closed but still T is not continuous.
– ravi yadav
Jul 20 at 18:03
Indeed. It only holds if the codomain is finite dimensional.
– Lorenzo Quarisa
Jul 20 at 20:16
Indeed. It only holds if the codomain is finite dimensional.
– Lorenzo Quarisa
Jul 20 at 20:16
add a comment |Â