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Applying Hahn-Banach to space of polynomials?
Clash Royale CLAN TAG#URR8PPP
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Does $phi$ extend to a bounded linear functional on $L^2([0,1]),$ where $phi$ is defined on the space of polynomials, $mathcalP$, on $[0, 1]$ and $phi(f)=f(1/2)?$ Prove or disprove.
Few questions:
- Is there a sound strategy to approach such open-ended questions (i.e. provide a proof or disproof)?
- I know that $overlinemathcalP[0,1] = overlineC([0,1])$ under the supremum norm, and $overlineC([0,1])= L^2([0,1])$ under the $L^2$-norm, so I think the answer is affirmative (using Hahn-Banach). Am I on the right track?
Hope someone can shed insight.
functional-analysis
asked yesterday
Mathemagica
85922762
add a comment |Â
up vote
0
down vote
favorite
Does $phi$ extend to a bounded linear functional on $L^2([0,1]),$ where $phi$ is defined on the space of polynomials, $mathcalP$, on $[0, 1]$ and $phi(f)=f(1/2)?$ Prove or disprove.
Few questions:
- Is there a sound strategy to approach such open-ended questions (i.e. provide a proof or disproof)?
- I know that $overlinemathcalP[0,1] = overlineC([0,1])$ under the supremum norm, and $overlineC([0,1])= L^2([0,1])$ under the $L^2$-norm, so I think the answer is affirmative (using Hahn-Banach). Am I on the right track?
Hope someone can shed insight.
functional-analysis
asked yesterday
Mathemagica
85922762
$L^2([0,1])$ is not endowed with the supremum norm.
– Paul Frost
yesterday
The members of $L^2[0,1]$ are not functions but equivalence classes of functions. But if $P$ is a polynomial on $[0,1]$ then the equivalence class $[P]$ contains no polynomials other than $P,$ so you can make $phi$ well-defined on the equivalence classes of polynomials. However , as pointed out in the Answer below, $phi$ is not bounded with respect to the $L^2$ norm.
– DanielWainfleet
yesterday
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Does $phi$ extend to a bounded linear functional on $L^2([0,1]),$ where $phi$ is defined on the space of polynomials, $mathcalP$, on $[0, 1]$ and $phi(f)=f(1/2)?$ Prove or disprove.
Few questions:
- Is there a sound strategy to approach such open-ended questions (i.e. provide a proof or disproof)?
- I know that $overlinemathcalP[0,1] = overlineC([0,1])$ under the supremum norm, and $overlineC([0,1])= L^2([0,1])$ under the $L^2$-norm, so I think the answer is affirmative (using Hahn-Banach). Am I on the right track?
Hope someone can shed insight.
functional-analysis
asked yesterday
Mathemagica
85922762
Does $phi$ extend to a bounded linear functional on $L^2([0,1]),$ where $phi$ is defined on the space of polynomials, $mathcalP$, on $[0, 1]$ and $phi(f)=f(1/2)?$ Prove or disprove.
Few questions:
- Is there a sound strategy to approach such open-ended questions (i.e. provide a proof or disproof)?
- I know that $overlinemathcalP[0,1] = overlineC([0,1])$ under the supremum norm, and $overlineC([0,1])= L^2([0,1])$ under the $L^2$-norm, so I think the answer is affirmative (using Hahn-Banach). Am I on the right track?
Hope someone can shed insight.
functional-analysis
asked yesterday
Mathemagica
85922762
edited yesterday
asked yesterday
Mathemagica
85922762
asked yesterday
Mathemagica
85922762
asked yesterday
Mathemagica
85922762
85922762
$L^2([0,1])$ is not endowed with the supremum norm.
– Paul Frost
yesterday
The members of $L^2[0,1]$ are not functions but equivalence classes of functions. But if $P$ is a polynomial on $[0,1]$ then the equivalence class $[P]$ contains no polynomials other than $P,$ so you can make $phi$ well-defined on the equivalence classes of polynomials. However , as pointed out in the Answer below, $phi$ is not bounded with respect to the $L^2$ norm.
– DanielWainfleet
yesterday
add a comment |Â
$L^2([0,1])$ is not endowed with the supremum norm.
– Paul Frost
yesterday
The members of $L^2[0,1]$ are not functions but equivalence classes of functions. But if $P$ is a polynomial on $[0,1]$ then the equivalence class $[P]$ contains no polynomials other than $P,$ so you can make $phi$ well-defined on the equivalence classes of polynomials. However , as pointed out in the Answer below, $phi$ is not bounded with respect to the $L^2$ norm.
– DanielWainfleet
yesterday
$L^2([0,1])$ is not endowed with the supremum norm.
– Paul Frost
yesterday
$L^2([0,1])$ is not endowed with the supremum norm.
– Paul Frost
yesterday
The members of $L^2[0,1]$ are not functions but equivalence classes of functions. But if $P$ is a polynomial on $[0,1]$ then the equivalence class $[P]$ contains no polynomials other than $P,$ so you can make $phi$ well-defined on the equivalence classes of polynomials. However , as pointed out in the Answer below, $phi$ is not bounded with respect to the $L^2$ norm.
– DanielWainfleet
yesterday
The members of $L^2[0,1]$ are not functions but equivalence classes of functions. But if $P$ is a polynomial on $[0,1]$ then the equivalence class $[P]$ contains no polynomials other than $P,$ so you can make $phi$ well-defined on the equivalence classes of polynomials. However , as pointed out in the Answer below, $phi$ is not bounded with respect to the $L^2$ norm.
– DanielWainfleet
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
In order that you can extend $phi$ to a continuous linear functional on $L^2[0,1]$ it has to be a continuous linear functional on the space of polynomials with $L^2$ norm. If this is so then there is a finite constant $C$ such that $|f(frac 1 2 )| leq C|f|_2$. If this holds for polynomials it must hold for continuous functions (an easy consequence of Weierstrass approximation). However, it is easy to construct continuous functions $f$ with $f(frac 1 2 )=1$ and $|f|_2$ arbitrarily small by making $f$ small outside a neighborhood of $frac 1 2$ so $phi $ is not a bounded linear functional to begin with.
edited yesterday
Mathemagica
85922762
answered yesterday
Kavi Rama Murthy
19k2828
Could you clarify the "easy" part - an example of such a continuous function? In general, what is the procedure to construct a continuous function with arbitrarily small $L^p$ norm?
– Mathemagica
yesterday
1
If $f(frac 1 2)=1$, $f(x)=0$ for $|x-frac 1 2 | >c$ and $0leq f leq 1$ then $|f|_2 leq sqrt 2c$ and you can choose $c$ as small as you want. To construct such a function $f$ just take straight line graphs between $frac 1 2 -c$ and $frac 1 2 +c$. Drawing a picture helps.
– Kavi Rama Murthy
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
In order that you can extend $phi$ to a continuous linear functional on $L^2[0,1]$ it has to be a continuous linear functional on the space of polynomials with $L^2$ norm. If this is so then there is a finite constant $C$ such that $|f(frac 1 2 )| leq C|f|_2$. If this holds for polynomials it must hold for continuous functions (an easy consequence of Weierstrass approximation). However, it is easy to construct continuous functions $f$ with $f(frac 1 2 )=1$ and $|f|_2$ arbitrarily small by making $f$ small outside a neighborhood of $frac 1 2$ so $phi $ is not a bounded linear functional to begin with.
edited yesterday
Mathemagica
85922762
answered yesterday
Kavi Rama Murthy
19k2828
Could you clarify the "easy" part - an example of such a continuous function? In general, what is the procedure to construct a continuous function with arbitrarily small $L^p$ norm?
– Mathemagica
yesterday
1
If $f(frac 1 2)=1$, $f(x)=0$ for $|x-frac 1 2 | >c$ and $0leq f leq 1$ then $|f|_2 leq sqrt 2c$ and you can choose $c$ as small as you want. To construct such a function $f$ just take straight line graphs between $frac 1 2 -c$ and $frac 1 2 +c$. Drawing a picture helps.
– Kavi Rama Murthy
yesterday
add a comment |Â
up vote
3
down vote
accepted
In order that you can extend $phi$ to a continuous linear functional on $L^2[0,1]$ it has to be a continuous linear functional on the space of polynomials with $L^2$ norm. If this is so then there is a finite constant $C$ such that $|f(frac 1 2 )| leq C|f|_2$. If this holds for polynomials it must hold for continuous functions (an easy consequence of Weierstrass approximation). However, it is easy to construct continuous functions $f$ with $f(frac 1 2 )=1$ and $|f|_2$ arbitrarily small by making $f$ small outside a neighborhood of $frac 1 2$ so $phi $ is not a bounded linear functional to begin with.
edited yesterday
Mathemagica
85922762
answered yesterday
Kavi Rama Murthy
19k2828
Could you clarify the "easy" part - an example of such a continuous function? In general, what is the procedure to construct a continuous function with arbitrarily small $L^p$ norm?
– Mathemagica
yesterday
1
If $f(frac 1 2)=1$, $f(x)=0$ for $|x-frac 1 2 | >c$ and $0leq f leq 1$ then $|f|_2 leq sqrt 2c$ and you can choose $c$ as small as you want. To construct such a function $f$ just take straight line graphs between $frac 1 2 -c$ and $frac 1 2 +c$. Drawing a picture helps.
– Kavi Rama Murthy
yesterday
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
In order that you can extend $phi$ to a continuous linear functional on $L^2[0,1]$ it has to be a continuous linear functional on the space of polynomials with $L^2$ norm. If this is so then there is a finite constant $C$ such that $|f(frac 1 2 )| leq C|f|_2$. If this holds for polynomials it must hold for continuous functions (an easy consequence of Weierstrass approximation). However, it is easy to construct continuous functions $f$ with $f(frac 1 2 )=1$ and $|f|_2$ arbitrarily small by making $f$ small outside a neighborhood of $frac 1 2$ so $phi $ is not a bounded linear functional to begin with.
edited yesterday
Mathemagica
85922762
answered yesterday
Kavi Rama Murthy
19k2828
In order that you can extend $phi$ to a continuous linear functional on $L^2[0,1]$ it has to be a continuous linear functional on the space of polynomials with $L^2$ norm. If this is so then there is a finite constant $C$ such that $|f(frac 1 2 )| leq C|f|_2$. If this holds for polynomials it must hold for continuous functions (an easy consequence of Weierstrass approximation). However, it is easy to construct continuous functions $f$ with $f(frac 1 2 )=1$ and $|f|_2$ arbitrarily small by making $f$ small outside a neighborhood of $frac 1 2$ so $phi $ is not a bounded linear functional to begin with.
edited yesterday
Mathemagica
85922762
answered yesterday
Kavi Rama Murthy
19k2828
edited yesterday
Mathemagica
85922762
edited yesterday
Mathemagica
85922762
edited yesterday
Mathemagica
85922762
85922762
answered yesterday
Kavi Rama Murthy
19k2828
answered yesterday
Kavi Rama Murthy
19k2828
answered yesterday
Kavi Rama Murthy
19k2828
19k2828
Could you clarify the "easy" part - an example of such a continuous function? In general, what is the procedure to construct a continuous function with arbitrarily small $L^p$ norm?
– Mathemagica
yesterday
1
If $f(frac 1 2)=1$, $f(x)=0$ for $|x-frac 1 2 | >c$ and $0leq f leq 1$ then $|f|_2 leq sqrt 2c$ and you can choose $c$ as small as you want. To construct such a function $f$ just take straight line graphs between $frac 1 2 -c$ and $frac 1 2 +c$. Drawing a picture helps.
– Kavi Rama Murthy
yesterday
add a comment |Â
Could you clarify the "easy" part - an example of such a continuous function? In general, what is the procedure to construct a continuous function with arbitrarily small $L^p$ norm?
– Mathemagica
yesterday
1
If $f(frac 1 2)=1$, $f(x)=0$ for $|x-frac 1 2 | >c$ and $0leq f leq 1$ then $|f|_2 leq sqrt 2c$ and you can choose $c$ as small as you want. To construct such a function $f$ just take straight line graphs between $frac 1 2 -c$ and $frac 1 2 +c$. Drawing a picture helps.
– Kavi Rama Murthy
yesterday
Could you clarify the "easy" part - an example of such a continuous function? In general, what is the procedure to construct a continuous function with arbitrarily small $L^p$ norm?
– Mathemagica
yesterday
Could you clarify the "easy" part - an example of such a continuous function? In general, what is the procedure to construct a continuous function with arbitrarily small $L^p$ norm?
– Mathemagica
yesterday
1
1
If $f(frac 1 2)=1$, $f(x)=0$ for $|x-frac 1 2 | >c$ and $0leq f leq 1$ then $|f|_2 leq sqrt 2c$ and you can choose $c$ as small as you want. To construct such a function $f$ just take straight line graphs between $frac 1 2 -c$ and $frac 1 2 +c$. Drawing a picture helps.
– Kavi Rama Murthy
yesterday
If $f(frac 1 2)=1$, $f(x)=0$ for $|x-frac 1 2 | >c$ and $0leq f leq 1$ then $|f|_2 leq sqrt 2c$ and you can choose $c$ as small as you want. To construct such a function $f$ just take straight line graphs between $frac 1 2 -c$ and $frac 1 2 +c$. Drawing a picture helps.
– Kavi Rama Murthy
yesterday
add a comment |Â
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$L^2([0,1])$ is not endowed with the supremum norm.
– Paul Frost
yesterday
The members of $L^2[0,1]$ are not functions but equivalence classes of functions. But if $P$ is a polynomial on $[0,1]$ then the equivalence class $[P]$ contains no polynomials other than $P,$ so you can make $phi$ well-defined on the equivalence classes of polynomials. However , as pointed out in the Answer below, $phi$ is not bounded with respect to the $L^2$ norm.
– DanielWainfleet
yesterday