C-S and dependences with the scalar product

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Let's say I have a scalar product : $v$ on $C^0(mathbbR, mathbbR)$, then by Cauchy-Scwartz inequality I have :



$$forall f,g in(C^0(mathbbR, mathbbR),v)quad v(f,g) leq | f|_v |g|_v$$



Now let's say I have an other scalar product : $w$ on $C^0(mathbbR, mathbbR)$, then do I have $forall f, g in (C^0(mathbbR, mathbbR),colorredv)$ :



$$ w(f,g) leq | f|_w |g|_w$$



(so is this inequality true, even-if $f,g$ "lives" in a different inner-product space)




calculus real-analysis cauchy-schwarz-inequality




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    A particular space $X$ can be endowed with different scalar products, and the Cauchy-Schwarz inequality holds for each scalar product. For your last sentence, the other scalar product $v$ is irrelevant for the Cauchy-Schwarz inequality in $w$.
    – angryavian
    Jun 25 at 21:29















up vote
0
down vote

favorite












Let's say I have a scalar product : $v$ on $C^0(mathbbR, mathbbR)$, then by Cauchy-Scwartz inequality I have :



$$forall f,g in(C^0(mathbbR, mathbbR),v)quad v(f,g) leq | f|_v |g|_v$$



Now let's say I have an other scalar product : $w$ on $C^0(mathbbR, mathbbR)$, then do I have $forall f, g in (C^0(mathbbR, mathbbR),colorredv)$ :



$$ w(f,g) leq | f|_w |g|_w$$



(so is this inequality true, even-if $f,g$ "lives" in a different inner-product space)




calculus real-analysis cauchy-schwarz-inequality




share|cite|improve this question


















  • 2




    A particular space $X$ can be endowed with different scalar products, and the Cauchy-Schwarz inequality holds for each scalar product. For your last sentence, the other scalar product $v$ is irrelevant for the Cauchy-Schwarz inequality in $w$.
    – angryavian
    Jun 25 at 21:29













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let's say I have a scalar product : $v$ on $C^0(mathbbR, mathbbR)$, then by Cauchy-Scwartz inequality I have :



$$forall f,g in(C^0(mathbbR, mathbbR),v)quad v(f,g) leq | f|_v |g|_v$$



Now let's say I have an other scalar product : $w$ on $C^0(mathbbR, mathbbR)$, then do I have $forall f, g in (C^0(mathbbR, mathbbR),colorredv)$ :



$$ w(f,g) leq | f|_w |g|_w$$



(so is this inequality true, even-if $f,g$ "lives" in a different inner-product space)




calculus real-analysis cauchy-schwarz-inequality




share|cite|improve this question














Let's say I have a scalar product : $v$ on $C^0(mathbbR, mathbbR)$, then by Cauchy-Scwartz inequality I have :



$$forall f,g in(C^0(mathbbR, mathbbR),v)quad v(f,g) leq | f|_v |g|_v$$



Now let's say I have an other scalar product : $w$ on $C^0(mathbbR, mathbbR)$, then do I have $forall f, g in (C^0(mathbbR, mathbbR),colorredv)$ :



$$ w(f,g) leq | f|_w |g|_w$$



(so is this inequality true, even-if $f,g$ "lives" in a different inner-product space)





calculus real-analysis cauchy-schwarz-inequality





share|cite|improve this question













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edited Aug 24 at 8:43









Martin Sleziak

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asked Jun 25 at 21:21









auhasard

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  • 2




    A particular space $X$ can be endowed with different scalar products, and the Cauchy-Schwarz inequality holds for each scalar product. For your last sentence, the other scalar product $v$ is irrelevant for the Cauchy-Schwarz inequality in $w$.
    – angryavian
    Jun 25 at 21:29













  • 2




    A particular space $X$ can be endowed with different scalar products, and the Cauchy-Schwarz inequality holds for each scalar product. For your last sentence, the other scalar product $v$ is irrelevant for the Cauchy-Schwarz inequality in $w$.
    – angryavian
    Jun 25 at 21:29








2




2




A particular space $X$ can be endowed with different scalar products, and the Cauchy-Schwarz inequality holds for each scalar product. For your last sentence, the other scalar product $v$ is irrelevant for the Cauchy-Schwarz inequality in $w$.
– angryavian
Jun 25 at 21:29





A particular space $X$ can be endowed with different scalar products, and the Cauchy-Schwarz inequality holds for each scalar product. For your last sentence, the other scalar product $v$ is irrelevant for the Cauchy-Schwarz inequality in $w$.
– angryavian
Jun 25 at 21:29











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It seems like you are confused by notation. Suppose $X$ is a real vector space and $v,w$ are two inner products on $X;$ that is $(X,v)$ and $(X,w)$ are inner product spaces.



The notation "let $x in (X,v)$" just says "let $x in X$." Sometimes it's convenient to remind the reader that we are considering $X$ as an inner product space (with inner product $v$), but technically it is not giving any additional information. Hence the spaces $(X,v)$ and $(X,w)$ may not be "the same" as inner product spaces, but any $x in X$ "lives" in both spaces nevertheless.



As a result the Cauchy-Schwarz inequality holds for any $x,y in X,$ with respect to any choice of scalar product on $X.$






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    accepted










    It seems like you are confused by notation. Suppose $X$ is a real vector space and $v,w$ are two inner products on $X;$ that is $(X,v)$ and $(X,w)$ are inner product spaces.



    The notation "let $x in (X,v)$" just says "let $x in X$." Sometimes it's convenient to remind the reader that we are considering $X$ as an inner product space (with inner product $v$), but technically it is not giving any additional information. Hence the spaces $(X,v)$ and $(X,w)$ may not be "the same" as inner product spaces, but any $x in X$ "lives" in both spaces nevertheless.



    As a result the Cauchy-Schwarz inequality holds for any $x,y in X,$ with respect to any choice of scalar product on $X.$






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      It seems like you are confused by notation. Suppose $X$ is a real vector space and $v,w$ are two inner products on $X;$ that is $(X,v)$ and $(X,w)$ are inner product spaces.



      The notation "let $x in (X,v)$" just says "let $x in X$." Sometimes it's convenient to remind the reader that we are considering $X$ as an inner product space (with inner product $v$), but technically it is not giving any additional information. Hence the spaces $(X,v)$ and $(X,w)$ may not be "the same" as inner product spaces, but any $x in X$ "lives" in both spaces nevertheless.



      As a result the Cauchy-Schwarz inequality holds for any $x,y in X,$ with respect to any choice of scalar product on $X.$






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        It seems like you are confused by notation. Suppose $X$ is a real vector space and $v,w$ are two inner products on $X;$ that is $(X,v)$ and $(X,w)$ are inner product spaces.



        The notation "let $x in (X,v)$" just says "let $x in X$." Sometimes it's convenient to remind the reader that we are considering $X$ as an inner product space (with inner product $v$), but technically it is not giving any additional information. Hence the spaces $(X,v)$ and $(X,w)$ may not be "the same" as inner product spaces, but any $x in X$ "lives" in both spaces nevertheless.



        As a result the Cauchy-Schwarz inequality holds for any $x,y in X,$ with respect to any choice of scalar product on $X.$






        share|cite|improve this answer












        It seems like you are confused by notation. Suppose $X$ is a real vector space and $v,w$ are two inner products on $X;$ that is $(X,v)$ and $(X,w)$ are inner product spaces.



        The notation "let $x in (X,v)$" just says "let $x in X$." Sometimes it's convenient to remind the reader that we are considering $X$ as an inner product space (with inner product $v$), but technically it is not giving any additional information. Hence the spaces $(X,v)$ and $(X,w)$ may not be "the same" as inner product spaces, but any $x in X$ "lives" in both spaces nevertheless.



        As a result the Cauchy-Schwarz inequality holds for any $x,y in X,$ with respect to any choice of scalar product on $X.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jun 25 at 21:51









        ktoi

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