Is every normable topological vector space “inner productable”?
$begingroup$
Not every norm on a vector space is induced by an inner product on that vector space. But suppose hat $X$ is a topological vector space that is normable, i.e. its topology is induced by some norm on the vector space. Then my question is, does that imply that $X$ is "inner product-able", i.e. that its topology is induced by some inner product on the vector space?
If not, then what is an example of a topological vector space which is normable but not "inner product-able"? And what properties must a topological vector space satisfy to be "inner product-able"?
general-topology examples-counterexamples normed-spaces inner-product-space topological-vector-spaces
asked Dec 18 '18 at 7:11
Keshav SrinivasanKeshav Srinivasan
2,31721445
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add a comment |
$begingroup$
Not every norm on a vector space is induced by an inner product on that vector space. But suppose hat $X$ is a topological vector space that is normable, i.e. its topology is induced by some norm on the vector space. Then my question is, does that imply that $X$ is "inner product-able", i.e. that its topology is induced by some inner product on the vector space?
If not, then what is an example of a topological vector space which is normable but not "inner product-able"? And what properties must a topological vector space satisfy to be "inner product-able"?
general-topology examples-counterexamples normed-spaces inner-product-space topological-vector-spaces
asked Dec 18 '18 at 7:11
Keshav SrinivasanKeshav Srinivasan
2,31721445
$endgroup$
add a comment |
$begingroup$
Not every norm on a vector space is induced by an inner product on that vector space. But suppose hat $X$ is a topological vector space that is normable, i.e. its topology is induced by some norm on the vector space. Then my question is, does that imply that $X$ is "inner product-able", i.e. that its topology is induced by some inner product on the vector space?
If not, then what is an example of a topological vector space which is normable but not "inner product-able"? And what properties must a topological vector space satisfy to be "inner product-able"?
general-topology examples-counterexamples normed-spaces inner-product-space topological-vector-spaces
asked Dec 18 '18 at 7:11
Keshav SrinivasanKeshav Srinivasan
2,31721445
$endgroup$
Not every norm on a vector space is induced by an inner product on that vector space. But suppose hat $X$ is a topological vector space that is normable, i.e. its topology is induced by some norm on the vector space. Then my question is, does that imply that $X$ is "inner product-able", i.e. that its topology is induced by some inner product on the vector space?
If not, then what is an example of a topological vector space which is normable but not "inner product-able"? And what properties must a topological vector space satisfy to be "inner product-able"?
general-topology examples-counterexamples normed-spaces inner-product-space topological-vector-spaces
general-topology examples-counterexamples normed-spaces inner-product-space topological-vector-spaces
asked Dec 18 '18 at 7:11
Keshav SrinivasanKeshav Srinivasan
2,31721445
asked Dec 18 '18 at 7:11
Keshav SrinivasanKeshav Srinivasan
2,31721445
asked Dec 18 '18 at 7:11
Keshav SrinivasanKeshav Srinivasan
2,31721445
asked Dec 18 '18 at 7:11
Keshav SrinivasanKeshav Srinivasan
2,31721445
asked Dec 18 '18 at 7:11
Keshav SrinivasanKeshav Srinivasan
2,31721445
2,31721445
add a comment |
add a comment |
1 Answer
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$C[0,1]$ is normable with the sup norm. If the topology of $C[0,1]$ is induced by an inner product then the norm corresponding to the inner product is equivalent to the sup norm because these norms induce the
same topology. But there is no norm equivalent to the sup norm which is given by an inner product. [ The existence of such a norm would make $C[0,1]$ reflexive].
answered Dec 18 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
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$begingroup$
Can you elaborate on the proof that no norm equivalent to the sup norm is induced by an inner product?
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– Keshav Srinivasan
Dec 18 '18 at 7:32
1
$begingroup$
The dual space and thee second dual are same for equivalent norms. Every Hilbert space is reflexive but $C[0,1]$ is not reflexive.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 7:34
$begingroup$
OK thanks for clarifying.
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 7:39
$begingroup$
I just posted a follow-up question: math.stackexchange.com/q/3045323/71829
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 16:00
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
$C[0,1]$ is normable with the sup norm. If the topology of $C[0,1]$ is induced by an inner product then the norm corresponding to the inner product is equivalent to the sup norm because these norms induce the
same topology. But there is no norm equivalent to the sup norm which is given by an inner product. [ The existence of such a norm would make $C[0,1]$ reflexive].
answered Dec 18 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
Can you elaborate on the proof that no norm equivalent to the sup norm is induced by an inner product?
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 7:32
1
$begingroup$
The dual space and thee second dual are same for equivalent norms. Every Hilbert space is reflexive but $C[0,1]$ is not reflexive.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 7:34
$begingroup$
OK thanks for clarifying.
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 7:39
$begingroup$
I just posted a follow-up question: math.stackexchange.com/q/3045323/71829
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 16:00
add a comment |
$begingroup$
$C[0,1]$ is normable with the sup norm. If the topology of $C[0,1]$ is induced by an inner product then the norm corresponding to the inner product is equivalent to the sup norm because these norms induce the
same topology. But there is no norm equivalent to the sup norm which is given by an inner product. [ The existence of such a norm would make $C[0,1]$ reflexive].
answered Dec 18 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
Can you elaborate on the proof that no norm equivalent to the sup norm is induced by an inner product?
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 7:32
1
$begingroup$
The dual space and thee second dual are same for equivalent norms. Every Hilbert space is reflexive but $C[0,1]$ is not reflexive.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 7:34
$begingroup$
OK thanks for clarifying.
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 7:39
$begingroup$
I just posted a follow-up question: math.stackexchange.com/q/3045323/71829
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 16:00
add a comment |
$begingroup$
$C[0,1]$ is normable with the sup norm. If the topology of $C[0,1]$ is induced by an inner product then the norm corresponding to the inner product is equivalent to the sup norm because these norms induce the
same topology. But there is no norm equivalent to the sup norm which is given by an inner product. [ The existence of such a norm would make $C[0,1]$ reflexive].
answered Dec 18 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$C[0,1]$ is normable with the sup norm. If the topology of $C[0,1]$ is induced by an inner product then the norm corresponding to the inner product is equivalent to the sup norm because these norms induce the
same topology. But there is no norm equivalent to the sup norm which is given by an inner product. [ The existence of such a norm would make $C[0,1]$ reflexive].
answered Dec 18 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Dec 18 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Dec 18 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Dec 18 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
62.9k42362
$begingroup$
Can you elaborate on the proof that no norm equivalent to the sup norm is induced by an inner product?
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 7:32
1
$begingroup$
The dual space and thee second dual are same for equivalent norms. Every Hilbert space is reflexive but $C[0,1]$ is not reflexive.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 7:34
$begingroup$
OK thanks for clarifying.
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 7:39
$begingroup$
I just posted a follow-up question: math.stackexchange.com/q/3045323/71829
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 16:00
add a comment |
$begingroup$
Can you elaborate on the proof that no norm equivalent to the sup norm is induced by an inner product?
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 7:32
1
$begingroup$
The dual space and thee second dual are same for equivalent norms. Every Hilbert space is reflexive but $C[0,1]$ is not reflexive.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 7:34
$begingroup$
OK thanks for clarifying.
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 7:39
$begingroup$
I just posted a follow-up question: math.stackexchange.com/q/3045323/71829
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 16:00
$begingroup$
Can you elaborate on the proof that no norm equivalent to the sup norm is induced by an inner product?
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 7:32
$begingroup$
Can you elaborate on the proof that no norm equivalent to the sup norm is induced by an inner product?
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 7:32
1
1
$begingroup$
The dual space and thee second dual are same for equivalent norms. Every Hilbert space is reflexive but $C[0,1]$ is not reflexive.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 7:34
$begingroup$
The dual space and thee second dual are same for equivalent norms. Every Hilbert space is reflexive but $C[0,1]$ is not reflexive.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 7:34
$begingroup$
OK thanks for clarifying.
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 7:39
$begingroup$
OK thanks for clarifying.
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 7:39
$begingroup$
I just posted a follow-up question: math.stackexchange.com/q/3045323/71829
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 16:00
$begingroup$
I just posted a follow-up question: math.stackexchange.com/q/3045323/71829
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 16:00
add a comment |
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