If $X$ has more than one element, then the norm $||f||=sup_xin X|f(x)|$ doesn't come from a inner product
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If $X$ has more than one element, then the norm $|f|=sup_xin X|f(x)|$ doesn't come from a inner product in $B(X;mathbbR) := f mathpunct: Xto mathbbR$ such that $f$ is bounded$$
That is, there is no inner product $langle-,-rangle$ such that $langle f,frangle = |f|^2$.
I thought of doing this in two ways: Set $langle f,f rangle = |f|^2$ and then getting a contradiction to the definition of inner product. Or showing a counterexample of the parallelogram law. In both cases I'm stuck on how to use that $X$ has more than one element. Any hint or idea will be appreciated.
linear-algebra metric-spaces
edited Aug 17 at 7:51
Henno Brandsma
92k342100
asked Aug 16 at 19:46
AnalyticHarmony
569213
add a comment |Â
up vote
1
down vote
favorite
If $X$ has more than one element, then the norm $|f|=sup_xin X|f(x)|$ doesn't come from a inner product in $B(X;mathbbR) := f mathpunct: Xto mathbbR$ such that $f$ is bounded$$
That is, there is no inner product $langle-,-rangle$ such that $langle f,frangle = |f|^2$.
I thought of doing this in two ways: Set $langle f,f rangle = |f|^2$ and then getting a contradiction to the definition of inner product. Or showing a counterexample of the parallelogram law. In both cases I'm stuck on how to use that $X$ has more than one element. Any hint or idea will be appreciated.
linear-algebra metric-spaces
edited Aug 17 at 7:51
Henno Brandsma
92k342100
asked Aug 16 at 19:46
AnalyticHarmony
569213
1
A norm comes from an inner product if and only if it satisfies the parallelogram law (en.wikipedia.org/wiki/Parallelogram_law) which this norm does not.
– User8128
Aug 16 at 20:02
1
Parallelogram law is the way to go. Say $a,bin X$, $ane b$. Use $a$ and $b$ to define two functions $f$ and $g$...
– David C. Ullrich
Aug 16 at 20:04
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $X$ has more than one element, then the norm $|f|=sup_xin X|f(x)|$ doesn't come from a inner product in $B(X;mathbbR) := f mathpunct: Xto mathbbR$ such that $f$ is bounded$$
That is, there is no inner product $langle-,-rangle$ such that $langle f,frangle = |f|^2$.
I thought of doing this in two ways: Set $langle f,f rangle = |f|^2$ and then getting a contradiction to the definition of inner product. Or showing a counterexample of the parallelogram law. In both cases I'm stuck on how to use that $X$ has more than one element. Any hint or idea will be appreciated.
linear-algebra metric-spaces
edited Aug 17 at 7:51
Henno Brandsma
92k342100
asked Aug 16 at 19:46
AnalyticHarmony
569213
If $X$ has more than one element, then the norm $|f|=sup_xin X|f(x)|$ doesn't come from a inner product in $B(X;mathbbR) := f mathpunct: Xto mathbbR$ such that $f$ is bounded$$
That is, there is no inner product $langle-,-rangle$ such that $langle f,frangle = |f|^2$.
I thought of doing this in two ways: Set $langle f,f rangle = |f|^2$ and then getting a contradiction to the definition of inner product. Or showing a counterexample of the parallelogram law. In both cases I'm stuck on how to use that $X$ has more than one element. Any hint or idea will be appreciated.
linear-algebra metric-spaces
edited Aug 17 at 7:51
Henno Brandsma
92k342100
asked Aug 16 at 19:46
AnalyticHarmony
569213
edited Aug 17 at 7:51
Henno Brandsma
92k342100
edited Aug 17 at 7:51
Henno Brandsma
92k342100
edited Aug 17 at 7:51
Henno Brandsma
92k342100
92k342100
asked Aug 16 at 19:46
AnalyticHarmony
569213
asked Aug 16 at 19:46
AnalyticHarmony
569213
asked Aug 16 at 19:46
AnalyticHarmony
569213
569213
1
A norm comes from an inner product if and only if it satisfies the parallelogram law (en.wikipedia.org/wiki/Parallelogram_law) which this norm does not.
– User8128
Aug 16 at 20:02
1
Parallelogram law is the way to go. Say $a,bin X$, $ane b$. Use $a$ and $b$ to define two functions $f$ and $g$...
– David C. Ullrich
Aug 16 at 20:04
add a comment |Â
1
A norm comes from an inner product if and only if it satisfies the parallelogram law (en.wikipedia.org/wiki/Parallelogram_law) which this norm does not.
– User8128
Aug 16 at 20:02
1
Parallelogram law is the way to go. Say $a,bin X$, $ane b$. Use $a$ and $b$ to define two functions $f$ and $g$...
– David C. Ullrich
Aug 16 at 20:04
1
1
A norm comes from an inner product if and only if it satisfies the parallelogram law (en.wikipedia.org/wiki/Parallelogram_law) which this norm does not.
– User8128
Aug 16 at 20:02
A norm comes from an inner product if and only if it satisfies the parallelogram law (en.wikipedia.org/wiki/Parallelogram_law) which this norm does not.
– User8128
Aug 16 at 20:02
1
1
Parallelogram law is the way to go. Say $a,bin X$, $ane b$. Use $a$ and $b$ to define two functions $f$ and $g$...
– David C. Ullrich
Aug 16 at 20:04
Parallelogram law is the way to go. Say $a,bin X$, $ane b$. Use $a$ and $b$ to define two functions $f$ and $g$...
– David C. Ullrich
Aug 16 at 20:04
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
If $X$ contains elements $aneq b$, define $f(x) = delta_xa$ and $g(x) = delta_xa + delta_ba$. Then
$$
2(|f|_infty^2 + |g|_infty^2) = 2 neq 5 = |f+g|_infty^2 + |f-g|_infty^2
$$
so $|cdot|_infty$ doesn't satisfy the parallelogram law and hence isn't generated by an inner product.
answered Aug 16 at 20:10
cdipaolo
677311
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
If $X$ contains elements $aneq b$, define $f(x) = delta_xa$ and $g(x) = delta_xa + delta_ba$. Then
$$
2(|f|_infty^2 + |g|_infty^2) = 2 neq 5 = |f+g|_infty^2 + |f-g|_infty^2
$$
so $|cdot|_infty$ doesn't satisfy the parallelogram law and hence isn't generated by an inner product.
answered Aug 16 at 20:10
cdipaolo
677311
add a comment |Â
up vote
3
down vote
accepted
If $X$ contains elements $aneq b$, define $f(x) = delta_xa$ and $g(x) = delta_xa + delta_ba$. Then
$$
2(|f|_infty^2 + |g|_infty^2) = 2 neq 5 = |f+g|_infty^2 + |f-g|_infty^2
$$
so $|cdot|_infty$ doesn't satisfy the parallelogram law and hence isn't generated by an inner product.
answered Aug 16 at 20:10
cdipaolo
677311
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
If $X$ contains elements $aneq b$, define $f(x) = delta_xa$ and $g(x) = delta_xa + delta_ba$. Then
$$
2(|f|_infty^2 + |g|_infty^2) = 2 neq 5 = |f+g|_infty^2 + |f-g|_infty^2
$$
so $|cdot|_infty$ doesn't satisfy the parallelogram law and hence isn't generated by an inner product.
answered Aug 16 at 20:10
cdipaolo
677311
If $X$ contains elements $aneq b$, define $f(x) = delta_xa$ and $g(x) = delta_xa + delta_ba$. Then
$$
2(|f|_infty^2 + |g|_infty^2) = 2 neq 5 = |f+g|_infty^2 + |f-g|_infty^2
$$
so $|cdot|_infty$ doesn't satisfy the parallelogram law and hence isn't generated by an inner product.
answered Aug 16 at 20:10
cdipaolo
677311
answered Aug 16 at 20:10
cdipaolo
677311
answered Aug 16 at 20:10
cdipaolo
677311
answered Aug 16 at 20:10
cdipaolo
677311
677311
add a comment |Â
add a comment |Â
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1
A norm comes from an inner product if and only if it satisfies the parallelogram law (en.wikipedia.org/wiki/Parallelogram_law) which this norm does not.
– User8128
Aug 16 at 20:02
1
Parallelogram law is the way to go. Say $a,bin X$, $ane b$. Use $a$ and $b$ to define two functions $f$ and $g$...
– David C. Ullrich
Aug 16 at 20:04