How understand that the $|f|_{infty}=max(|f(x)|)$ norm on $C[0,1]$ is not equivalent to another one,...
How to prove it using parallelogram law?
functional-analysis hilbert-spaces
edited Dec 13 '18 at 6:04
Martin Sleziak
44.6k8115271
asked Dec 12 '18 at 22:31
PechoraPechora
62
closed as unclear what you're asking by José Carlos Santos, KReiser, Leucippus, Giuseppe Negro, Cesareo Dec 13 '18 at 8:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
How to prove it using parallelogram law?
functional-analysis hilbert-spaces
edited Dec 13 '18 at 6:04
Martin Sleziak
44.6k8115271
asked Dec 12 '18 at 22:31
PechoraPechora
62
closed as unclear what you're asking by José Carlos Santos, KReiser, Leucippus, Giuseppe Negro, Cesareo Dec 13 '18 at 8:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
Could you please elaborate?
– Mindlack
Dec 12 '18 at 22:37
1
I believe the norm is usually called the supremum norm, and denoted by $|_|_sup$ or $|_|_infty$.
– Batominovski
Dec 12 '18 at 22:45
1
I think "is not equivalent to another one, constructed from a scalar product" may be better rephrased as "is not equivalent to a norm $|_|$ induced by a scalar product on $C[0,1]$." And I don't know how the Parallelogram Law helps. I can prove the claim without this restriction.
– Batominovski
Dec 12 '18 at 22:49
You could have a look at: Is the norm on $ell^infty$ induced by an inner product? and Show that the sup-norm is not derived from an inner product
– Martin Sleziak
Dec 13 '18 at 6:03
add a comment |
How to prove it using parallelogram law?
functional-analysis hilbert-spaces
edited Dec 13 '18 at 6:04
Martin Sleziak
44.6k8115271
asked Dec 12 '18 at 22:31
PechoraPechora
62
How to prove it using parallelogram law?
functional-analysis hilbert-spaces
functional-analysis hilbert-spaces
edited Dec 13 '18 at 6:04
Martin Sleziak
44.6k8115271
asked Dec 12 '18 at 22:31
PechoraPechora
62
edited Dec 13 '18 at 6:04
Martin Sleziak
44.6k8115271
asked Dec 12 '18 at 22:31
PechoraPechora
62
edited Dec 13 '18 at 6:04
Martin Sleziak
44.6k8115271
edited Dec 13 '18 at 6:04
Martin Sleziak
44.6k8115271
edited Dec 13 '18 at 6:04
Martin Sleziak
44.6k8115271
44.6k8115271
asked Dec 12 '18 at 22:31
PechoraPechora
62
asked Dec 12 '18 at 22:31
PechoraPechora
62
asked Dec 12 '18 at 22:31
PechoraPechora
62
62
closed as unclear what you're asking by José Carlos Santos, KReiser, Leucippus, Giuseppe Negro, Cesareo Dec 13 '18 at 8:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by José Carlos Santos, KReiser, Leucippus, Giuseppe Negro, Cesareo Dec 13 '18 at 8:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
Could you please elaborate?
– Mindlack
Dec 12 '18 at 22:37
1
I believe the norm is usually called the supremum norm, and denoted by $|_|_sup$ or $|_|_infty$.
– Batominovski
Dec 12 '18 at 22:45
1
I think "is not equivalent to another one, constructed from a scalar product" may be better rephrased as "is not equivalent to a norm $|_|$ induced by a scalar product on $C[0,1]$." And I don't know how the Parallelogram Law helps. I can prove the claim without this restriction.
– Batominovski
Dec 12 '18 at 22:49
You could have a look at: Is the norm on $ell^infty$ induced by an inner product? and Show that the sup-norm is not derived from an inner product
– Martin Sleziak
Dec 13 '18 at 6:03
add a comment |
1
Could you please elaborate?
– Mindlack
Dec 12 '18 at 22:37
1
I believe the norm is usually called the supremum norm, and denoted by $|_|_sup$ or $|_|_infty$.
– Batominovski
Dec 12 '18 at 22:45
1
I think "is not equivalent to another one, constructed from a scalar product" may be better rephrased as "is not equivalent to a norm $|_|$ induced by a scalar product on $C[0,1]$." And I don't know how the Parallelogram Law helps. I can prove the claim without this restriction.
– Batominovski
Dec 12 '18 at 22:49
You could have a look at: Is the norm on $ell^infty$ induced by an inner product? and Show that the sup-norm is not derived from an inner product
– Martin Sleziak
Dec 13 '18 at 6:03
1
1
Could you please elaborate?
– Mindlack
Dec 12 '18 at 22:37
Could you please elaborate?
– Mindlack
Dec 12 '18 at 22:37
1
1
I believe the norm is usually called the supremum norm, and denoted by $|_|_sup$ or $|_|_infty$.
– Batominovski
Dec 12 '18 at 22:45
I believe the norm is usually called the supremum norm, and denoted by $|_|_sup$ or $|_|_infty$.
– Batominovski
Dec 12 '18 at 22:45
1
1
I think "is not equivalent to another one, constructed from a scalar product" may be better rephrased as "is not equivalent to a norm $|_|$ induced by a scalar product on $C[0,1]$." And I don't know how the Parallelogram Law helps. I can prove the claim without this restriction.
– Batominovski
Dec 12 '18 at 22:49
I think "is not equivalent to another one, constructed from a scalar product" may be better rephrased as "is not equivalent to a norm $|_|$ induced by a scalar product on $C[0,1]$." And I don't know how the Parallelogram Law helps. I can prove the claim without this restriction.
– Batominovski
Dec 12 '18 at 22:49
You could have a look at: Is the norm on $ell^infty$ induced by an inner product? and Show that the sup-norm is not derived from an inner product
– Martin Sleziak
Dec 13 '18 at 6:03
You could have a look at: Is the norm on $ell^infty$ induced by an inner product? and Show that the sup-norm is not derived from an inner product
– Martin Sleziak
Dec 13 '18 at 6:03
add a comment |
2 Answers
2
active
oldest
votes
If you are asking whether $|f|_{infty}$ is induced by an inner product here is my answer: let $f(x)=1-2x$ for $0leq x leq frac 1 2$ and $0$ for $x >frac 1 2$. Let $g(x)=2x$ for $0leq x leq frac 1 2$ and $0$ for $x >frac 1 2$. Verify that $|f|=|g|=|f+g|=|f-g|=1$. So the prallelogram identity $|f+g|^{2}+|f-g|^{2}=2|f|^{2}+2|g|^{2}$ does not hold. Hence there cannot be any inner product that induces $|f|_{infty}$.
If you are asking us to show that there is no norm equivalent to $|f|_{infty}$ which is induced by an inner product here is the argument: if $|f|_{infty}$ is induced by an inner product then the unit ball of $C[0,1]$ would be weakly compact which also implies that $C[0,1]$ is reflexive. It is well known that this is not the case. In fact any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$ which makes any separable Banach space reflexive. $ell^{1}$ gives a counterexample to this.
answered Dec 12 '18 at 23:29
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
1
I don't think that this argument works. It seems the OP wants to show that $|_|_infty$ is not equivalent to any norm induced by an inner product. $$phantom{aa}$$ However, given an odd request regarding the Parallelogram Law, you might be right. The question may be just that $|_|_infty$ is not a norn induced by an inner product. The OP's wording is awful.
– Batominovski
Dec 13 '18 at 0:24
@Batominovski I have edited my answer. Thanks for your comment.
– Kavi Rama Murthy
Dec 13 '18 at 0:44
add a comment |
We know that $(C[0,1],||cdot||_infty)$ is a separable Banach space. If the norm $||cdot||_infty$ can be induced by an inner product then $X=(C[0,1],||cdot||_infty)$ would be a Hilbert space, which means that $X^*$ is separable.
However, the evaluation functional $delta_x in X^*$ defined by
$$
delta_x(f) := f(x)
$$
satisfies $||delta_x - delta_y||ge 1$ for distinct $x,yin [0,1]$. Indeed, assuming $x<y$ we can test $delta_x - delta_y$ on a continuous function $f$ defined by
$$
f(t)=begin{cases}1 &;tle x \
1-frac{(t-x)}{y-x}&; tin [x,y]\
0 &; yle t
end{cases}
$$
to deduce the inequality. This means that each element in the (uncountable) set $D:={delta_x:xin[0,1]}subset X^*$ is at least of distance $1$ apart from each other. Hence any dense subset of $X^*$ must be uncountable. This is a contradiction to the separability of $X^*$.
Edit: Sorry I missed the part where you specified that you want to use the parallelogram law.
answered Dec 13 '18 at 4:45
BigbearZzzBigbearZzz
7,55821650
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you are asking whether $|f|_{infty}$ is induced by an inner product here is my answer: let $f(x)=1-2x$ for $0leq x leq frac 1 2$ and $0$ for $x >frac 1 2$. Let $g(x)=2x$ for $0leq x leq frac 1 2$ and $0$ for $x >frac 1 2$. Verify that $|f|=|g|=|f+g|=|f-g|=1$. So the prallelogram identity $|f+g|^{2}+|f-g|^{2}=2|f|^{2}+2|g|^{2}$ does not hold. Hence there cannot be any inner product that induces $|f|_{infty}$.
If you are asking us to show that there is no norm equivalent to $|f|_{infty}$ which is induced by an inner product here is the argument: if $|f|_{infty}$ is induced by an inner product then the unit ball of $C[0,1]$ would be weakly compact which also implies that $C[0,1]$ is reflexive. It is well known that this is not the case. In fact any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$ which makes any separable Banach space reflexive. $ell^{1}$ gives a counterexample to this.
answered Dec 12 '18 at 23:29
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
1
I don't think that this argument works. It seems the OP wants to show that $|_|_infty$ is not equivalent to any norm induced by an inner product. $$phantom{aa}$$ However, given an odd request regarding the Parallelogram Law, you might be right. The question may be just that $|_|_infty$ is not a norn induced by an inner product. The OP's wording is awful.
– Batominovski
Dec 13 '18 at 0:24
@Batominovski I have edited my answer. Thanks for your comment.
– Kavi Rama Murthy
Dec 13 '18 at 0:44
add a comment |
If you are asking whether $|f|_{infty}$ is induced by an inner product here is my answer: let $f(x)=1-2x$ for $0leq x leq frac 1 2$ and $0$ for $x >frac 1 2$. Let $g(x)=2x$ for $0leq x leq frac 1 2$ and $0$ for $x >frac 1 2$. Verify that $|f|=|g|=|f+g|=|f-g|=1$. So the prallelogram identity $|f+g|^{2}+|f-g|^{2}=2|f|^{2}+2|g|^{2}$ does not hold. Hence there cannot be any inner product that induces $|f|_{infty}$.
If you are asking us to show that there is no norm equivalent to $|f|_{infty}$ which is induced by an inner product here is the argument: if $|f|_{infty}$ is induced by an inner product then the unit ball of $C[0,1]$ would be weakly compact which also implies that $C[0,1]$ is reflexive. It is well known that this is not the case. In fact any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$ which makes any separable Banach space reflexive. $ell^{1}$ gives a counterexample to this.
answered Dec 12 '18 at 23:29
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
1
I don't think that this argument works. It seems the OP wants to show that $|_|_infty$ is not equivalent to any norm induced by an inner product. $$phantom{aa}$$ However, given an odd request regarding the Parallelogram Law, you might be right. The question may be just that $|_|_infty$ is not a norn induced by an inner product. The OP's wording is awful.
– Batominovski
Dec 13 '18 at 0:24
@Batominovski I have edited my answer. Thanks for your comment.
– Kavi Rama Murthy
Dec 13 '18 at 0:44
add a comment |
If you are asking whether $|f|_{infty}$ is induced by an inner product here is my answer: let $f(x)=1-2x$ for $0leq x leq frac 1 2$ and $0$ for $x >frac 1 2$. Let $g(x)=2x$ for $0leq x leq frac 1 2$ and $0$ for $x >frac 1 2$. Verify that $|f|=|g|=|f+g|=|f-g|=1$. So the prallelogram identity $|f+g|^{2}+|f-g|^{2}=2|f|^{2}+2|g|^{2}$ does not hold. Hence there cannot be any inner product that induces $|f|_{infty}$.
If you are asking us to show that there is no norm equivalent to $|f|_{infty}$ which is induced by an inner product here is the argument: if $|f|_{infty}$ is induced by an inner product then the unit ball of $C[0,1]$ would be weakly compact which also implies that $C[0,1]$ is reflexive. It is well known that this is not the case. In fact any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$ which makes any separable Banach space reflexive. $ell^{1}$ gives a counterexample to this.
answered Dec 12 '18 at 23:29
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
If you are asking whether $|f|_{infty}$ is induced by an inner product here is my answer: let $f(x)=1-2x$ for $0leq x leq frac 1 2$ and $0$ for $x >frac 1 2$. Let $g(x)=2x$ for $0leq x leq frac 1 2$ and $0$ for $x >frac 1 2$. Verify that $|f|=|g|=|f+g|=|f-g|=1$. So the prallelogram identity $|f+g|^{2}+|f-g|^{2}=2|f|^{2}+2|g|^{2}$ does not hold. Hence there cannot be any inner product that induces $|f|_{infty}$.
If you are asking us to show that there is no norm equivalent to $|f|_{infty}$ which is induced by an inner product here is the argument: if $|f|_{infty}$ is induced by an inner product then the unit ball of $C[0,1]$ would be weakly compact which also implies that $C[0,1]$ is reflexive. It is well known that this is not the case. In fact any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$ which makes any separable Banach space reflexive. $ell^{1}$ gives a counterexample to this.
answered Dec 12 '18 at 23:29
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
edited Dec 13 '18 at 0:40
answered Dec 12 '18 at 23:29
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
answered Dec 12 '18 at 23:29
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
answered Dec 12 '18 at 23:29
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
52.1k32055
1
I don't think that this argument works. It seems the OP wants to show that $|_|_infty$ is not equivalent to any norm induced by an inner product. $$phantom{aa}$$ However, given an odd request regarding the Parallelogram Law, you might be right. The question may be just that $|_|_infty$ is not a norn induced by an inner product. The OP's wording is awful.
– Batominovski
Dec 13 '18 at 0:24
@Batominovski I have edited my answer. Thanks for your comment.
– Kavi Rama Murthy
Dec 13 '18 at 0:44
add a comment |
1
I don't think that this argument works. It seems the OP wants to show that $|_|_infty$ is not equivalent to any norm induced by an inner product. $$phantom{aa}$$ However, given an odd request regarding the Parallelogram Law, you might be right. The question may be just that $|_|_infty$ is not a norn induced by an inner product. The OP's wording is awful.
– Batominovski
Dec 13 '18 at 0:24
@Batominovski I have edited my answer. Thanks for your comment.
– Kavi Rama Murthy
Dec 13 '18 at 0:44
1
1
I don't think that this argument works. It seems the OP wants to show that $|_|_infty$ is not equivalent to any norm induced by an inner product. $$phantom{aa}$$ However, given an odd request regarding the Parallelogram Law, you might be right. The question may be just that $|_|_infty$ is not a norn induced by an inner product. The OP's wording is awful.
– Batominovski
Dec 13 '18 at 0:24
I don't think that this argument works. It seems the OP wants to show that $|_|_infty$ is not equivalent to any norm induced by an inner product. $$phantom{aa}$$ However, given an odd request regarding the Parallelogram Law, you might be right. The question may be just that $|_|_infty$ is not a norn induced by an inner product. The OP's wording is awful.
– Batominovski
Dec 13 '18 at 0:24
@Batominovski I have edited my answer. Thanks for your comment.
– Kavi Rama Murthy
Dec 13 '18 at 0:44
@Batominovski I have edited my answer. Thanks for your comment.
– Kavi Rama Murthy
Dec 13 '18 at 0:44
add a comment |
We know that $(C[0,1],||cdot||_infty)$ is a separable Banach space. If the norm $||cdot||_infty$ can be induced by an inner product then $X=(C[0,1],||cdot||_infty)$ would be a Hilbert space, which means that $X^*$ is separable.
However, the evaluation functional $delta_x in X^*$ defined by
$$
delta_x(f) := f(x)
$$
satisfies $||delta_x - delta_y||ge 1$ for distinct $x,yin [0,1]$. Indeed, assuming $x<y$ we can test $delta_x - delta_y$ on a continuous function $f$ defined by
$$
f(t)=begin{cases}1 &;tle x \
1-frac{(t-x)}{y-x}&; tin [x,y]\
0 &; yle t
end{cases}
$$
to deduce the inequality. This means that each element in the (uncountable) set $D:={delta_x:xin[0,1]}subset X^*$ is at least of distance $1$ apart from each other. Hence any dense subset of $X^*$ must be uncountable. This is a contradiction to the separability of $X^*$.
Edit: Sorry I missed the part where you specified that you want to use the parallelogram law.
answered Dec 13 '18 at 4:45
BigbearZzzBigbearZzz
7,55821650
add a comment |
We know that $(C[0,1],||cdot||_infty)$ is a separable Banach space. If the norm $||cdot||_infty$ can be induced by an inner product then $X=(C[0,1],||cdot||_infty)$ would be a Hilbert space, which means that $X^*$ is separable.
However, the evaluation functional $delta_x in X^*$ defined by
$$
delta_x(f) := f(x)
$$
satisfies $||delta_x - delta_y||ge 1$ for distinct $x,yin [0,1]$. Indeed, assuming $x<y$ we can test $delta_x - delta_y$ on a continuous function $f$ defined by
$$
f(t)=begin{cases}1 &;tle x \
1-frac{(t-x)}{y-x}&; tin [x,y]\
0 &; yle t
end{cases}
$$
to deduce the inequality. This means that each element in the (uncountable) set $D:={delta_x:xin[0,1]}subset X^*$ is at least of distance $1$ apart from each other. Hence any dense subset of $X^*$ must be uncountable. This is a contradiction to the separability of $X^*$.
Edit: Sorry I missed the part where you specified that you want to use the parallelogram law.
answered Dec 13 '18 at 4:45
BigbearZzzBigbearZzz
7,55821650
add a comment |
We know that $(C[0,1],||cdot||_infty)$ is a separable Banach space. If the norm $||cdot||_infty$ can be induced by an inner product then $X=(C[0,1],||cdot||_infty)$ would be a Hilbert space, which means that $X^*$ is separable.
However, the evaluation functional $delta_x in X^*$ defined by
$$
delta_x(f) := f(x)
$$
satisfies $||delta_x - delta_y||ge 1$ for distinct $x,yin [0,1]$. Indeed, assuming $x<y$ we can test $delta_x - delta_y$ on a continuous function $f$ defined by
$$
f(t)=begin{cases}1 &;tle x \
1-frac{(t-x)}{y-x}&; tin [x,y]\
0 &; yle t
end{cases}
$$
to deduce the inequality. This means that each element in the (uncountable) set $D:={delta_x:xin[0,1]}subset X^*$ is at least of distance $1$ apart from each other. Hence any dense subset of $X^*$ must be uncountable. This is a contradiction to the separability of $X^*$.
Edit: Sorry I missed the part where you specified that you want to use the parallelogram law.
answered Dec 13 '18 at 4:45
BigbearZzzBigbearZzz
7,55821650
We know that $(C[0,1],||cdot||_infty)$ is a separable Banach space. If the norm $||cdot||_infty$ can be induced by an inner product then $X=(C[0,1],||cdot||_infty)$ would be a Hilbert space, which means that $X^*$ is separable.
However, the evaluation functional $delta_x in X^*$ defined by
$$
delta_x(f) := f(x)
$$
satisfies $||delta_x - delta_y||ge 1$ for distinct $x,yin [0,1]$. Indeed, assuming $x<y$ we can test $delta_x - delta_y$ on a continuous function $f$ defined by
$$
f(t)=begin{cases}1 &;tle x \
1-frac{(t-x)}{y-x}&; tin [x,y]\
0 &; yle t
end{cases}
$$
to deduce the inequality. This means that each element in the (uncountable) set $D:={delta_x:xin[0,1]}subset X^*$ is at least of distance $1$ apart from each other. Hence any dense subset of $X^*$ must be uncountable. This is a contradiction to the separability of $X^*$.
Edit: Sorry I missed the part where you specified that you want to use the parallelogram law.
answered Dec 13 '18 at 4:45
BigbearZzzBigbearZzz
7,55821650
edited Dec 13 '18 at 6:58
answered Dec 13 '18 at 4:45
BigbearZzzBigbearZzz
7,55821650
answered Dec 13 '18 at 4:45
BigbearZzzBigbearZzz
7,55821650
answered Dec 13 '18 at 4:45
BigbearZzzBigbearZzz
7,55821650
7,55821650
add a comment |
add a comment |
1
Could you please elaborate?
– Mindlack
Dec 12 '18 at 22:37
1
I believe the norm is usually called the supremum norm, and denoted by $|_|_sup$ or $|_|_infty$.
– Batominovski
Dec 12 '18 at 22:45
1
I think "is not equivalent to another one, constructed from a scalar product" may be better rephrased as "is not equivalent to a norm $|_|$ induced by a scalar product on $C[0,1]$." And I don't know how the Parallelogram Law helps. I can prove the claim without this restriction.
– Batominovski
Dec 12 '18 at 22:49
You could have a look at: Is the norm on $ell^infty$ induced by an inner product? and Show that the sup-norm is not derived from an inner product
– Martin Sleziak
Dec 13 '18 at 6:03