Scalar product of complex valued square integrable functions
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
add a comment |
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
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
add a comment |
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
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
complex-analysis vector-spaces inner-product-space
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
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 $L^2$.
add a comment |
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
New contributor
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062204%2fscalar-product-of-complex-valued-square-integrable-functions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $L^2$.
add a comment |
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 $L^2$.
add a comment |
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 $L^2$.
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 $L^2$.
answered Jan 4 at 23:16
0x5390x539
1,042317
1,042317
add a comment |
add a comment |
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
New contributor
add a comment |
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
New contributor
add a comment |
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
New contributor
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
New contributor
edited Jan 4 at 23:32
New contributor
answered Jan 4 at 23:16
Maximilian JanischMaximilian Janisch
46110
46110
New contributor
New contributor
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062204%2fscalar-product-of-complex-valued-square-integrable-functions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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