Scalar product of complex valued square integrable functions












0














So I was told that if $p,q in mathbb{L}^2 _mathbb{C} [a,b]$ which is the set of all square integrable functions on the interval $[a,b]$ then:
$$(p,q) = int^b_a{dx p^*(x)q(x)}$$
is a scalar product.
However, for this to be the case $(p,p) = 0 leftrightarrow p =0$ should be satisfied. $(0,0) = 0$ is obvious but on $mathbb{L}^2 _mathbb{C}$ there can be other functions that can yield 0 upon integration (e.g. a function that takes a finite value at one point but is zero everywhere else), so how can this be true?










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  • Remember that $L^2$ only makes sense up to equality at almost every point. So a function which is zero except at a single point is equal to the zero function (or more precisely... lies in the same equivalence class, and these classes are the actual elements of $L^2$).
    – T. Bongers
    Jan 4 at 22:48










  • Are you sure that $L^2_C [a,b]$ does not only contain contineous functions, because otherwise you are right. the "C" in $L^2_C [a,b]$ makes me think so
    – A. P
    Jan 4 at 23:21










  • @A.P yes the $mathbb{C}$ stands for the complex valuedness of the functions.
    – daljit97
    Jan 4 at 23:33
















0














So I was told that if $p,q in mathbb{L}^2 _mathbb{C} [a,b]$ which is the set of all square integrable functions on the interval $[a,b]$ then:
$$(p,q) = int^b_a{dx p^*(x)q(x)}$$
is a scalar product.
However, for this to be the case $(p,p) = 0 leftrightarrow p =0$ should be satisfied. $(0,0) = 0$ is obvious but on $mathbb{L}^2 _mathbb{C}$ there can be other functions that can yield 0 upon integration (e.g. a function that takes a finite value at one point but is zero everywhere else), so how can this be true?










share|cite|improve this question






















  • Remember that $L^2$ only makes sense up to equality at almost every point. So a function which is zero except at a single point is equal to the zero function (or more precisely... lies in the same equivalence class, and these classes are the actual elements of $L^2$).
    – T. Bongers
    Jan 4 at 22:48










  • Are you sure that $L^2_C [a,b]$ does not only contain contineous functions, because otherwise you are right. the "C" in $L^2_C [a,b]$ makes me think so
    – A. P
    Jan 4 at 23:21










  • @A.P yes the $mathbb{C}$ stands for the complex valuedness of the functions.
    – daljit97
    Jan 4 at 23:33














0












0








0







So I was told that if $p,q in mathbb{L}^2 _mathbb{C} [a,b]$ which is the set of all square integrable functions on the interval $[a,b]$ then:
$$(p,q) = int^b_a{dx p^*(x)q(x)}$$
is a scalar product.
However, for this to be the case $(p,p) = 0 leftrightarrow p =0$ should be satisfied. $(0,0) = 0$ is obvious but on $mathbb{L}^2 _mathbb{C}$ there can be other functions that can yield 0 upon integration (e.g. a function that takes a finite value at one point but is zero everywhere else), so how can this be true?










share|cite|improve this question













So I was told that if $p,q in mathbb{L}^2 _mathbb{C} [a,b]$ which is the set of all square integrable functions on the interval $[a,b]$ then:
$$(p,q) = int^b_a{dx p^*(x)q(x)}$$
is a scalar product.
However, for this to be the case $(p,p) = 0 leftrightarrow p =0$ should be satisfied. $(0,0) = 0$ is obvious but on $mathbb{L}^2 _mathbb{C}$ there can be other functions that can yield 0 upon integration (e.g. a function that takes a finite value at one point but is zero everywhere else), so how can this be true?







complex-analysis vector-spaces inner-product-space






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asked Jan 4 at 22:46









daljit97daljit97

106110




106110












  • Remember that $L^2$ only makes sense up to equality at almost every point. So a function which is zero except at a single point is equal to the zero function (or more precisely... lies in the same equivalence class, and these classes are the actual elements of $L^2$).
    – T. Bongers
    Jan 4 at 22:48










  • Are you sure that $L^2_C [a,b]$ does not only contain contineous functions, because otherwise you are right. the "C" in $L^2_C [a,b]$ makes me think so
    – A. P
    Jan 4 at 23:21










  • @A.P yes the $mathbb{C}$ stands for the complex valuedness of the functions.
    – daljit97
    Jan 4 at 23:33


















  • Remember that $L^2$ only makes sense up to equality at almost every point. So a function which is zero except at a single point is equal to the zero function (or more precisely... lies in the same equivalence class, and these classes are the actual elements of $L^2$).
    – T. Bongers
    Jan 4 at 22:48










  • Are you sure that $L^2_C [a,b]$ does not only contain contineous functions, because otherwise you are right. the "C" in $L^2_C [a,b]$ makes me think so
    – A. P
    Jan 4 at 23:21










  • @A.P yes the $mathbb{C}$ stands for the complex valuedness of the functions.
    – daljit97
    Jan 4 at 23:33
















Remember that $L^2$ only makes sense up to equality at almost every point. So a function which is zero except at a single point is equal to the zero function (or more precisely... lies in the same equivalence class, and these classes are the actual elements of $L^2$).
– T. Bongers
Jan 4 at 22:48




Remember that $L^2$ only makes sense up to equality at almost every point. So a function which is zero except at a single point is equal to the zero function (or more precisely... lies in the same equivalence class, and these classes are the actual elements of $L^2$).
– T. Bongers
Jan 4 at 22:48












Are you sure that $L^2_C [a,b]$ does not only contain contineous functions, because otherwise you are right. the "C" in $L^2_C [a,b]$ makes me think so
– A. P
Jan 4 at 23:21




Are you sure that $L^2_C [a,b]$ does not only contain contineous functions, because otherwise you are right. the "C" in $L^2_C [a,b]$ makes me think so
– A. P
Jan 4 at 23:21












@A.P yes the $mathbb{C}$ stands for the complex valuedness of the functions.
– daljit97
Jan 4 at 23:33




@A.P yes the $mathbb{C}$ stands for the complex valuedness of the functions.
– daljit97
Jan 4 at 23:33










2 Answers
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You'll need some measure theory for this. $L^2$ actually isn't a space of functions but a space of equivalence classes of functions. Two square-integral functions $f$, $g$ are defined to be equivalent if they differ only on a set of measure zero (If you don't know measure theory, think of this as a set with zero length/volume/area).



The function in your example differs from zero in one point only and since the measure of a one-point set is zero, your function is actually identical to zero in $L^2$.



More generally, if a function $p$ is nonzero on a set $A$ of positive measure, then $f = |p|^2 = p^* p$ is also nonzero on $A$. If you divide $A$ as to $A = bigcup_{n geq 1} A_n = bigcup_{n geq 1} {x in A | f(x) > frac1n}$ you get that one of the $A_n$ must have nonzero measure and therefore $int_a^b f(x) ,mathrm{d}x > frac1n mu(A_n) > 0$. (here $mu$ denotes the measure). But this means that a function $p$ which is not equivalent to $0$ in $L^2$ also has a positive "scalar square" $(p, p)$, proving definiteness of the scalar product on $L^2$.






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    0














    The set $mathbb{L}^2[a, b]$ is defined as the quotient space $mathcal{L}^2[a, b]$ / $mathcal{N}$. Thus $mathbb{L}^2$ is actually a set of equivalence classes. Here $mathcal{N}$ denotes the set of all functions that are non-zero on a set with Lebesgue measure zero. $mathcal{L}^2[a, b]$ denotes the set of all square-integrable functions (i.e. all functions f such that $||f||_2^2 = langle f, frangle = int_a^b |f|^2(x) dx < infty$.) So $mathbb{L}^2$ is the set of all equivalence classes of functions that are equal almost everywhere (i.e. they differ only on a set with measure $0$).



    Note that the above 2-norm on $mathcal{L}^2[a, b]$ isn't an actual norm, as $||f||_2^2 = 0 iff f equiv 0$ is obviously not satisfied, similarly as how $langle f, f rangle = 0 iff f equiv 0$ is not satisfied on $mathcal{L}^2[a, b]$. However, writing $f = 0$ for an $f in mathbb{L}^2$ means that $f$ is the equivalence class of the functions $f in mathcal{L}^2$ that are $0$ almost everywhere.



    Note that, $||f||_2^2 = 0 iff f = 0$ and $langle f, f rangle = 0 iff f = 0$ are satisfied on $mathbb{L}^2$ (contrary to $mathcal{L}^2$), since the Integral of $|f|^2$ is $0$ if and only if $f$ is zero almost everywhere (i.e. if $f$ is the equivalence class of functions that are zero almost everywhere).



    EDIT: Some further reading (also for more general $mathbb{L}^2$ spaces than just $mathbb{L}^2[a,b]$): https://en.wikipedia.org/wiki/Lp_space#Lp_spaces






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      2 Answers
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      2 Answers
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      You'll need some measure theory for this. $L^2$ actually isn't a space of functions but a space of equivalence classes of functions. Two square-integral functions $f$, $g$ are defined to be equivalent if they differ only on a set of measure zero (If you don't know measure theory, think of this as a set with zero length/volume/area).



      The function in your example differs from zero in one point only and since the measure of a one-point set is zero, your function is actually identical to zero in $L^2$.



      More generally, if a function $p$ is nonzero on a set $A$ of positive measure, then $f = |p|^2 = p^* p$ is also nonzero on $A$. If you divide $A$ as to $A = bigcup_{n geq 1} A_n = bigcup_{n geq 1} {x in A | f(x) > frac1n}$ you get that one of the $A_n$ must have nonzero measure and therefore $int_a^b f(x) ,mathrm{d}x > frac1n mu(A_n) > 0$. (here $mu$ denotes the measure). But this means that a function $p$ which is not equivalent to $0$ in $L^2$ also has a positive "scalar square" $(p, p)$, proving definiteness of the scalar product on $L^2$.






      share|cite|improve this answer


























        1














        You'll need some measure theory for this. $L^2$ actually isn't a space of functions but a space of equivalence classes of functions. Two square-integral functions $f$, $g$ are defined to be equivalent if they differ only on a set of measure zero (If you don't know measure theory, think of this as a set with zero length/volume/area).



        The function in your example differs from zero in one point only and since the measure of a one-point set is zero, your function is actually identical to zero in $L^2$.



        More generally, if a function $p$ is nonzero on a set $A$ of positive measure, then $f = |p|^2 = p^* p$ is also nonzero on $A$. If you divide $A$ as to $A = bigcup_{n geq 1} A_n = bigcup_{n geq 1} {x in A | f(x) > frac1n}$ you get that one of the $A_n$ must have nonzero measure and therefore $int_a^b f(x) ,mathrm{d}x > frac1n mu(A_n) > 0$. (here $mu$ denotes the measure). But this means that a function $p$ which is not equivalent to $0$ in $L^2$ also has a positive "scalar square" $(p, p)$, proving definiteness of the scalar product on $L^2$.






        share|cite|improve this answer
























          1












          1








          1






          You'll need some measure theory for this. $L^2$ actually isn't a space of functions but a space of equivalence classes of functions. Two square-integral functions $f$, $g$ are defined to be equivalent if they differ only on a set of measure zero (If you don't know measure theory, think of this as a set with zero length/volume/area).



          The function in your example differs from zero in one point only and since the measure of a one-point set is zero, your function is actually identical to zero in $L^2$.



          More generally, if a function $p$ is nonzero on a set $A$ of positive measure, then $f = |p|^2 = p^* p$ is also nonzero on $A$. If you divide $A$ as to $A = bigcup_{n geq 1} A_n = bigcup_{n geq 1} {x in A | f(x) > frac1n}$ you get that one of the $A_n$ must have nonzero measure and therefore $int_a^b f(x) ,mathrm{d}x > frac1n mu(A_n) > 0$. (here $mu$ denotes the measure). But this means that a function $p$ which is not equivalent to $0$ in $L^2$ also has a positive "scalar square" $(p, p)$, proving definiteness of the scalar product on $L^2$.






          share|cite|improve this answer












          You'll need some measure theory for this. $L^2$ actually isn't a space of functions but a space of equivalence classes of functions. Two square-integral functions $f$, $g$ are defined to be equivalent if they differ only on a set of measure zero (If you don't know measure theory, think of this as a set with zero length/volume/area).



          The function in your example differs from zero in one point only and since the measure of a one-point set is zero, your function is actually identical to zero in $L^2$.



          More generally, if a function $p$ is nonzero on a set $A$ of positive measure, then $f = |p|^2 = p^* p$ is also nonzero on $A$. If you divide $A$ as to $A = bigcup_{n geq 1} A_n = bigcup_{n geq 1} {x in A | f(x) > frac1n}$ you get that one of the $A_n$ must have nonzero measure and therefore $int_a^b f(x) ,mathrm{d}x > frac1n mu(A_n) > 0$. (here $mu$ denotes the measure). But this means that a function $p$ which is not equivalent to $0$ in $L^2$ also has a positive "scalar square" $(p, p)$, proving definiteness of the scalar product on $L^2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 23:16









          0x5390x539

          1,042317




          1,042317























              0














              The set $mathbb{L}^2[a, b]$ is defined as the quotient space $mathcal{L}^2[a, b]$ / $mathcal{N}$. Thus $mathbb{L}^2$ is actually a set of equivalence classes. Here $mathcal{N}$ denotes the set of all functions that are non-zero on a set with Lebesgue measure zero. $mathcal{L}^2[a, b]$ denotes the set of all square-integrable functions (i.e. all functions f such that $||f||_2^2 = langle f, frangle = int_a^b |f|^2(x) dx < infty$.) So $mathbb{L}^2$ is the set of all equivalence classes of functions that are equal almost everywhere (i.e. they differ only on a set with measure $0$).



              Note that the above 2-norm on $mathcal{L}^2[a, b]$ isn't an actual norm, as $||f||_2^2 = 0 iff f equiv 0$ is obviously not satisfied, similarly as how $langle f, f rangle = 0 iff f equiv 0$ is not satisfied on $mathcal{L}^2[a, b]$. However, writing $f = 0$ for an $f in mathbb{L}^2$ means that $f$ is the equivalence class of the functions $f in mathcal{L}^2$ that are $0$ almost everywhere.



              Note that, $||f||_2^2 = 0 iff f = 0$ and $langle f, f rangle = 0 iff f = 0$ are satisfied on $mathbb{L}^2$ (contrary to $mathcal{L}^2$), since the Integral of $|f|^2$ is $0$ if and only if $f$ is zero almost everywhere (i.e. if $f$ is the equivalence class of functions that are zero almost everywhere).



              EDIT: Some further reading (also for more general $mathbb{L}^2$ spaces than just $mathbb{L}^2[a,b]$): https://en.wikipedia.org/wiki/Lp_space#Lp_spaces






              share|cite|improve this answer










              New contributor




              Maximilian Janisch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.























                0














                The set $mathbb{L}^2[a, b]$ is defined as the quotient space $mathcal{L}^2[a, b]$ / $mathcal{N}$. Thus $mathbb{L}^2$ is actually a set of equivalence classes. Here $mathcal{N}$ denotes the set of all functions that are non-zero on a set with Lebesgue measure zero. $mathcal{L}^2[a, b]$ denotes the set of all square-integrable functions (i.e. all functions f such that $||f||_2^2 = langle f, frangle = int_a^b |f|^2(x) dx < infty$.) So $mathbb{L}^2$ is the set of all equivalence classes of functions that are equal almost everywhere (i.e. they differ only on a set with measure $0$).



                Note that the above 2-norm on $mathcal{L}^2[a, b]$ isn't an actual norm, as $||f||_2^2 = 0 iff f equiv 0$ is obviously not satisfied, similarly as how $langle f, f rangle = 0 iff f equiv 0$ is not satisfied on $mathcal{L}^2[a, b]$. However, writing $f = 0$ for an $f in mathbb{L}^2$ means that $f$ is the equivalence class of the functions $f in mathcal{L}^2$ that are $0$ almost everywhere.



                Note that, $||f||_2^2 = 0 iff f = 0$ and $langle f, f rangle = 0 iff f = 0$ are satisfied on $mathbb{L}^2$ (contrary to $mathcal{L}^2$), since the Integral of $|f|^2$ is $0$ if and only if $f$ is zero almost everywhere (i.e. if $f$ is the equivalence class of functions that are zero almost everywhere).



                EDIT: Some further reading (also for more general $mathbb{L}^2$ spaces than just $mathbb{L}^2[a,b]$): https://en.wikipedia.org/wiki/Lp_space#Lp_spaces






                share|cite|improve this answer










                New contributor




                Maximilian Janisch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.





















                  0












                  0








                  0






                  The set $mathbb{L}^2[a, b]$ is defined as the quotient space $mathcal{L}^2[a, b]$ / $mathcal{N}$. Thus $mathbb{L}^2$ is actually a set of equivalence classes. Here $mathcal{N}$ denotes the set of all functions that are non-zero on a set with Lebesgue measure zero. $mathcal{L}^2[a, b]$ denotes the set of all square-integrable functions (i.e. all functions f such that $||f||_2^2 = langle f, frangle = int_a^b |f|^2(x) dx < infty$.) So $mathbb{L}^2$ is the set of all equivalence classes of functions that are equal almost everywhere (i.e. they differ only on a set with measure $0$).



                  Note that the above 2-norm on $mathcal{L}^2[a, b]$ isn't an actual norm, as $||f||_2^2 = 0 iff f equiv 0$ is obviously not satisfied, similarly as how $langle f, f rangle = 0 iff f equiv 0$ is not satisfied on $mathcal{L}^2[a, b]$. However, writing $f = 0$ for an $f in mathbb{L}^2$ means that $f$ is the equivalence class of the functions $f in mathcal{L}^2$ that are $0$ almost everywhere.



                  Note that, $||f||_2^2 = 0 iff f = 0$ and $langle f, f rangle = 0 iff f = 0$ are satisfied on $mathbb{L}^2$ (contrary to $mathcal{L}^2$), since the Integral of $|f|^2$ is $0$ if and only if $f$ is zero almost everywhere (i.e. if $f$ is the equivalence class of functions that are zero almost everywhere).



                  EDIT: Some further reading (also for more general $mathbb{L}^2$ spaces than just $mathbb{L}^2[a,b]$): https://en.wikipedia.org/wiki/Lp_space#Lp_spaces






                  share|cite|improve this answer










                  New contributor




                  Maximilian Janisch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  The set $mathbb{L}^2[a, b]$ is defined as the quotient space $mathcal{L}^2[a, b]$ / $mathcal{N}$. Thus $mathbb{L}^2$ is actually a set of equivalence classes. Here $mathcal{N}$ denotes the set of all functions that are non-zero on a set with Lebesgue measure zero. $mathcal{L}^2[a, b]$ denotes the set of all square-integrable functions (i.e. all functions f such that $||f||_2^2 = langle f, frangle = int_a^b |f|^2(x) dx < infty$.) So $mathbb{L}^2$ is the set of all equivalence classes of functions that are equal almost everywhere (i.e. they differ only on a set with measure $0$).



                  Note that the above 2-norm on $mathcal{L}^2[a, b]$ isn't an actual norm, as $||f||_2^2 = 0 iff f equiv 0$ is obviously not satisfied, similarly as how $langle f, f rangle = 0 iff f equiv 0$ is not satisfied on $mathcal{L}^2[a, b]$. However, writing $f = 0$ for an $f in mathbb{L}^2$ means that $f$ is the equivalence class of the functions $f in mathcal{L}^2$ that are $0$ almost everywhere.



                  Note that, $||f||_2^2 = 0 iff f = 0$ and $langle f, f rangle = 0 iff f = 0$ are satisfied on $mathbb{L}^2$ (contrary to $mathcal{L}^2$), since the Integral of $|f|^2$ is $0$ if and only if $f$ is zero almost everywhere (i.e. if $f$ is the equivalence class of functions that are zero almost everywhere).



                  EDIT: Some further reading (also for more general $mathbb{L}^2$ spaces than just $mathbb{L}^2[a,b]$): https://en.wikipedia.org/wiki/Lp_space#Lp_spaces







                  share|cite|improve this answer










                  New contributor




                  Maximilian Janisch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









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                  share|cite|improve this answer








                  edited Jan 4 at 23:32





















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                  answered Jan 4 at 23:16









                  Maximilian JanischMaximilian Janisch

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                  46110




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                  New contributor





                  Maximilian Janisch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  Maximilian Janisch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






























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