Mixing
Proving that a given function $f^*$ is the best least square approximation
$begingroup$
In De Boor (1972) it is stated that
Let be $$ $ a finite dimensional linear space of functions defined on the interval $[a,b]$. We are searching for the best approximation from $$$ to $g$.
The function $f^*$ is a best approximation from $$$ to $g$ with respect to the $mathcal{L}_2$ norm if and only if the function $f^*$ is in $$$ and the error term $g-f^*$ is orthogonal to $$$.
To show that the double implication holds the author provides the following statement (I add my own procedure because it's not given in the book).
For any function $fin $ $ we have that
$$|g-f|_2^2=|g-f^*+f^*-f|_2^2=|g-f^*|_2^2+2langle f^*-f,g-f^*rangle+|f^*-f|_2^2$$
If condition is satisfied we have
$$|g-f|_2^2=|g-f^*|+|f^*-f|_2^2geq|g-f^*|_2^2$$
Which proves that $f^*$ is the best least squares approximation in $$$
If $langle f,g-f^*rangle neq 0$
we have that by letting $tf:=f^*-f$
$$|g-f|_2^2=|g-(f^*+tf)+(f^*+tf)-f|_2^2$$
$$=|g-(f^*+tf)|_2^2+2langle 2tf,g-f^*-tfrangle+|2tf|_2^2$$
$$=|g-(f^*+tf)|_2^2+4langle tf,g-f^*rangle$$
Given that for all nonzero $t$ of the same sign as $langle f,g-f^*rangle$ and sufficiently close to $0$.
$$2tlangle f,g-f^*rangle>|tf|_2^2$$
This completes the proof since it implies that
begin{equation}
|g-(f^*+tf)|_2^2<|g-f^*|_2^2
end{equation}
and hence $f^*$ is not the best approximation.
I am not sure to understand the reason why it holds that $$2tlangle f,g-f^*rangle>|tf|_2^2$$
Is it because $t^2$ goes to zero faster than $t$.
Is there some way to show it more formally?
Thanks in advance.
approximation regression least-squares
asked Jan 23 at 10:23
RScrlliRScrlli
649114
$endgroup$
|
show 2 more comments
$begingroup$
In De Boor (1972) it is stated that
Let be $$ $ a finite dimensional linear space of functions defined on the interval $[a,b]$. We are searching for the best approximation from $$$ to $g$.
The function $f^*$ is a best approximation from $$$ to $g$ with respect to the $mathcal{L}_2$ norm if and only if the function $f^*$ is in $$$ and the error term $g-f^*$ is orthogonal to $$$.
To show that the double implication holds the author provides the following statement (I add my own procedure because it's not given in the book).
For any function $fin $ $ we have that
$$|g-f|_2^2=|g-f^*+f^*-f|_2^2=|g-f^*|_2^2+2langle f^*-f,g-f^*rangle+|f^*-f|_2^2$$
If condition is satisfied we have
$$|g-f|_2^2=|g-f^*|+|f^*-f|_2^2geq|g-f^*|_2^2$$
Which proves that $f^*$ is the best least squares approximation in $$$
If $langle f,g-f^*rangle neq 0$
we have that by letting $tf:=f^*-f$
$$|g-f|_2^2=|g-(f^*+tf)+(f^*+tf)-f|_2^2$$
$$=|g-(f^*+tf)|_2^2+2langle 2tf,g-f^*-tfrangle+|2tf|_2^2$$
$$=|g-(f^*+tf)|_2^2+4langle tf,g-f^*rangle$$
Given that for all nonzero $t$ of the same sign as $langle f,g-f^*rangle$ and sufficiently close to $0$.
$$2tlangle f,g-f^*rangle>|tf|_2^2$$
This completes the proof since it implies that
begin{equation}
|g-(f^*+tf)|_2^2<|g-f^*|_2^2
end{equation}
and hence $f^*$ is not the best approximation.
I am not sure to understand the reason why it holds that $$2tlangle f,g-f^*rangle>|tf|_2^2$$
Is it because $t^2$ goes to zero faster than $t$.
Is there some way to show it more formally?
Thanks in advance.
approximation regression least-squares
asked Jan 23 at 10:23
RScrlliRScrlli
649114
$endgroup$
$begingroup$
I suspect there are some typos, e.g. $langle f^*-f,g-f^*rangle$ in the first line after "we have that" and $||g-f||_2^2=||g-f^*||_2^2+||f^*-f||_2^2geq ||g-f^*||_2^2$
$endgroup$
– Peter Melech
Jan 23 at 16:33
$begingroup$
@PeterMelech I agree with you about the second typo, but the first one,( the inner product) looks okey to me, what do you think?
$endgroup$
– RScrlli
Jan 23 at 16:37
1
$begingroup$
I'm quite sure it has to be $f^*$ instead of $f$ in the right argument of the inner product, similarly: If $langle f,g-f^*rangleneq 0$ some lines under, when showing the other direction
$endgroup$
– Peter Melech
Jan 23 at 16:46
$begingroup$
Besides: wonder how" by letting $tf:=f^*-f$" is to be understood
$endgroup$
– Peter Melech
Jan 23 at 16:57
$begingroup$
@PeterMelech you are right, I've done a copy and paste from another document and I didn't realize that. Thanks!
$endgroup$
– RScrlli
Jan 23 at 16:57
|
show 2 more comments
$begingroup$
In De Boor (1972) it is stated that
Let be $$ $ a finite dimensional linear space of functions defined on the interval $[a,b]$. We are searching for the best approximation from $$$ to $g$.
The function $f^*$ is a best approximation from $$$ to $g$ with respect to the $mathcal{L}_2$ norm if and only if the function $f^*$ is in $$$ and the error term $g-f^*$ is orthogonal to $$$.
To show that the double implication holds the author provides the following statement (I add my own procedure because it's not given in the book).
For any function $fin $ $ we have that
$$|g-f|_2^2=|g-f^*+f^*-f|_2^2=|g-f^*|_2^2+2langle f^*-f,g-f^*rangle+|f^*-f|_2^2$$
If condition is satisfied we have
$$|g-f|_2^2=|g-f^*|+|f^*-f|_2^2geq|g-f^*|_2^2$$
Which proves that $f^*$ is the best least squares approximation in $$$
If $langle f,g-f^*rangle neq 0$
we have that by letting $tf:=f^*-f$
$$|g-f|_2^2=|g-(f^*+tf)+(f^*+tf)-f|_2^2$$
$$=|g-(f^*+tf)|_2^2+2langle 2tf,g-f^*-tfrangle+|2tf|_2^2$$
$$=|g-(f^*+tf)|_2^2+4langle tf,g-f^*rangle$$
Given that for all nonzero $t$ of the same sign as $langle f,g-f^*rangle$ and sufficiently close to $0$.
$$2tlangle f,g-f^*rangle>|tf|_2^2$$
This completes the proof since it implies that
begin{equation}
|g-(f^*+tf)|_2^2<|g-f^*|_2^2
end{equation}
and hence $f^*$ is not the best approximation.
I am not sure to understand the reason why it holds that $$2tlangle f,g-f^*rangle>|tf|_2^2$$
Is it because $t^2$ goes to zero faster than $t$.
Is there some way to show it more formally?
Thanks in advance.
approximation regression least-squares
asked Jan 23 at 10:23
RScrlliRScrlli
649114
$endgroup$
In De Boor (1972) it is stated that
Let be $$ $ a finite dimensional linear space of functions defined on the interval $[a,b]$. We are searching for the best approximation from $$$ to $g$.
The function $f^*$ is a best approximation from $$$ to $g$ with respect to the $mathcal{L}_2$ norm if and only if the function $f^*$ is in $$$ and the error term $g-f^*$ is orthogonal to $$$.
To show that the double implication holds the author provides the following statement (I add my own procedure because it's not given in the book).
For any function $fin $ $ we have that
$$|g-f|_2^2=|g-f^*+f^*-f|_2^2=|g-f^*|_2^2+2langle f^*-f,g-f^*rangle+|f^*-f|_2^2$$
If condition is satisfied we have
$$|g-f|_2^2=|g-f^*|+|f^*-f|_2^2geq|g-f^*|_2^2$$
Which proves that $f^*$ is the best least squares approximation in $$$
If $langle f,g-f^*rangle neq 0$
we have that by letting $tf:=f^*-f$
$$|g-f|_2^2=|g-(f^*+tf)+(f^*+tf)-f|_2^2$$
$$=|g-(f^*+tf)|_2^2+2langle 2tf,g-f^*-tfrangle+|2tf|_2^2$$
$$=|g-(f^*+tf)|_2^2+4langle tf,g-f^*rangle$$
Given that for all nonzero $t$ of the same sign as $langle f,g-f^*rangle$ and sufficiently close to $0$.
$$2tlangle f,g-f^*rangle>|tf|_2^2$$
This completes the proof since it implies that
begin{equation}
|g-(f^*+tf)|_2^2<|g-f^*|_2^2
end{equation}
and hence $f^*$ is not the best approximation.
I am not sure to understand the reason why it holds that $$2tlangle f,g-f^*rangle>|tf|_2^2$$
Is it because $t^2$ goes to zero faster than $t$.
Is there some way to show it more formally?
Thanks in advance.
approximation regression least-squares
approximation regression least-squares
asked Jan 23 at 10:23
RScrlliRScrlli
649114
asked Jan 23 at 10:23
RScrlliRScrlli
649114
edited Jan 23 at 17:02
RScrlli
asked Jan 23 at 10:23
RScrlliRScrlli
649114
asked Jan 23 at 10:23
RScrlliRScrlli
649114
asked Jan 23 at 10:23
RScrlliRScrlli
649114
649114
$begingroup$
I suspect there are some typos, e.g. $langle f^*-f,g-f^*rangle$ in the first line after "we have that" and $||g-f||_2^2=||g-f^*||_2^2+||f^*-f||_2^2geq ||g-f^*||_2^2$
$endgroup$
– Peter Melech
Jan 23 at 16:33
$begingroup$
@PeterMelech I agree with you about the second typo, but the first one,( the inner product) looks okey to me, what do you think?
$endgroup$
– RScrlli
Jan 23 at 16:37
1
$begingroup$
I'm quite sure it has to be $f^*$ instead of $f$ in the right argument of the inner product, similarly: If $langle f,g-f^*rangleneq 0$ some lines under, when showing the other direction
$endgroup$
– Peter Melech
Jan 23 at 16:46
$begingroup$
Besides: wonder how" by letting $tf:=f^*-f$" is to be understood
$endgroup$
– Peter Melech
Jan 23 at 16:57
$begingroup$
@PeterMelech you are right, I've done a copy and paste from another document and I didn't realize that. Thanks!
$endgroup$
– RScrlli
Jan 23 at 16:57
|
show 2 more comments
$begingroup$
I suspect there are some typos, e.g. $langle f^*-f,g-f^*rangle$ in the first line after "we have that" and $||g-f||_2^2=||g-f^*||_2^2+||f^*-f||_2^2geq ||g-f^*||_2^2$
$endgroup$
– Peter Melech
Jan 23 at 16:33
$begingroup$
@PeterMelech I agree with you about the second typo, but the first one,( the inner product) looks okey to me, what do you think?
$endgroup$
– RScrlli
Jan 23 at 16:37
1
$begingroup$
I'm quite sure it has to be $f^*$ instead of $f$ in the right argument of the inner product, similarly: If $langle f,g-f^*rangleneq 0$ some lines under, when showing the other direction
$endgroup$
– Peter Melech
Jan 23 at 16:46
$begingroup$
Besides: wonder how" by letting $tf:=f^*-f$" is to be understood
$endgroup$
– Peter Melech
Jan 23 at 16:57
$begingroup$
@PeterMelech you are right, I've done a copy and paste from another document and I didn't realize that. Thanks!
$endgroup$
– RScrlli
Jan 23 at 16:57
$begingroup$
I suspect there are some typos, e.g. $langle f^*-f,g-f^*rangle$ in the first line after "we have that" and $||g-f||_2^2=||g-f^*||_2^2+||f^*-f||_2^2geq ||g-f^*||_2^2$
$endgroup$
– Peter Melech
Jan 23 at 16:33
$begingroup$
I suspect there are some typos, e.g. $langle f^*-f,g-f^*rangle$ in the first line after "we have that" and $||g-f||_2^2=||g-f^*||_2^2+||f^*-f||_2^2geq ||g-f^*||_2^2$
$endgroup$
– Peter Melech
Jan 23 at 16:33
$begingroup$
@PeterMelech I agree with you about the second typo, but the first one,( the inner product) looks okey to me, what do you think?
$endgroup$
– RScrlli
Jan 23 at 16:37
$begingroup$
@PeterMelech I agree with you about the second typo, but the first one,( the inner product) looks okey to me, what do you think?
$endgroup$
– RScrlli
Jan 23 at 16:37
1
1
$begingroup$
I'm quite sure it has to be $f^*$ instead of $f$ in the right argument of the inner product, similarly: If $langle f,g-f^*rangleneq 0$ some lines under, when showing the other direction
$endgroup$
– Peter Melech
Jan 23 at 16:46
$begingroup$
I'm quite sure it has to be $f^*$ instead of $f$ in the right argument of the inner product, similarly: If $langle f,g-f^*rangleneq 0$ some lines under, when showing the other direction
$endgroup$
– Peter Melech
Jan 23 at 16:46
$begingroup$
Besides: wonder how" by letting $tf:=f^*-f$" is to be understood
$endgroup$
– Peter Melech
Jan 23 at 16:57
$begingroup$
Besides: wonder how" by letting $tf:=f^*-f$" is to be understood
$endgroup$
– Peter Melech
Jan 23 at 16:57
$begingroup$
@PeterMelech you are right, I've done a copy and paste from another document and I didn't realize that. Thanks!
$endgroup$
– RScrlli
Jan 23 at 16:57
$begingroup$
@PeterMelech you are right, I've done a copy and paste from another document and I didn't realize that. Thanks!
$endgroup$
– RScrlli
Jan 23 at 16:57
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
I would like to suggest a slightly different proof, because I don't feel well with the argument " by letting $tf:=f^*-f$".
Assume $langle f,g-f^*rangleneq 0$ for some $fin U$ (I call the subspace $U$), then for all $tinmathbb{R}$ one has:
$$||g-f^*||_2^2=||g-f^*+tf-tf||_2^2=||g-(f^*+tf)||_2^2+2langle g-(f^*+tf),tfrangle +||tf||_2^2$$
$$=||g-(f^*+tf)||_2^2+2tlangle g-f^*,frangle -2t^2langle f,frangle+||tf||_2 ^2$$
$$=|||g-(f^*+tf)||_2^2+2tlangle g-f^*,frangle-t^2||f||_2^2.$$
And now choose $t$ so that
$$2tlangle g-f^*,frangle-t^2||f||_2^2>0$$
which in case that $t$ has the same sign as $langle g-f^*,frangle$ is equivalent to
$$|t|<frac{2|langle g-f^*,frangle|}{||f||_2^2}.$$
(Note that $langle g-f^*,frangleneq 0$ is essential to that.)
Then:
$$|||g-(f^*+tf)||_2^2<||g-f^*||_2^2$$
which contradicts the best-approximation-property.
answered Jan 24 at 13:04
Peter MelechPeter Melech
2,687813
$endgroup$
$begingroup$
It's much more clear and general as a proof, for sure I will include it in my thesis with a footnote thanking you for the hint
$endgroup$
– RScrlli
Jan 24 at 15:27
$begingroup$
Thank You very much!
$endgroup$
– Peter Melech
Jan 25 at 7:49
add a comment |
$begingroup$
If $t$ has same sign as $langle f, g-f^{*}rangle$ the inequality you want is $2|t||langle f, g-f^{*}rangle| >|t|^{2} |f|_2^{2}$. This is true whenever $|t| <frac {2|langle f, g-f^{*}rangle|} { |f|_2^{2}}$.
answered Jan 23 at 10:29
Kavi Rama MurthyKavi Rama Murthy
66.8k53067
$endgroup$
$begingroup$
by taking the absolute value everything makes sense, but the problem is that what I want to show is $2langle f^*-f,g-frangle+|f^*-f|_2^2>4langle tf,g-f^*rangle$.
$endgroup$
– RScrlli
Jan 23 at 10:40
$begingroup$
I've edited the question, showing the first part of the proof.
$endgroup$
– RScrlli
Jan 23 at 10:40
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%2f3084293%2fproving-that-a-given-function-f-is-the-best-least-square-approximation%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
$begingroup$
I would like to suggest a slightly different proof, because I don't feel well with the argument " by letting $tf:=f^*-f$".
Assume $langle f,g-f^*rangleneq 0$ for some $fin U$ (I call the subspace $U$), then for all $tinmathbb{R}$ one has:
$$||g-f^*||_2^2=||g-f^*+tf-tf||_2^2=||g-(f^*+tf)||_2^2+2langle g-(f^*+tf),tfrangle +||tf||_2^2$$
$$=||g-(f^*+tf)||_2^2+2tlangle g-f^*,frangle -2t^2langle f,frangle+||tf||_2 ^2$$
$$=|||g-(f^*+tf)||_2^2+2tlangle g-f^*,frangle-t^2||f||_2^2.$$
And now choose $t$ so that
$$2tlangle g-f^*,frangle-t^2||f||_2^2>0$$
which in case that $t$ has the same sign as $langle g-f^*,frangle$ is equivalent to
$$|t|<frac{2|langle g-f^*,frangle|}{||f||_2^2}.$$
(Note that $langle g-f^*,frangleneq 0$ is essential to that.)
Then:
$$|||g-(f^*+tf)||_2^2<||g-f^*||_2^2$$
which contradicts the best-approximation-property.
answered Jan 24 at 13:04
Peter MelechPeter Melech
2,687813
$endgroup$
$begingroup$
It's much more clear and general as a proof, for sure I will include it in my thesis with a footnote thanking you for the hint
$endgroup$
– RScrlli
Jan 24 at 15:27
$begingroup$
Thank You very much!
$endgroup$
– Peter Melech
Jan 25 at 7:49
add a comment |
$begingroup$
I would like to suggest a slightly different proof, because I don't feel well with the argument " by letting $tf:=f^*-f$".
Assume $langle f,g-f^*rangleneq 0$ for some $fin U$ (I call the subspace $U$), then for all $tinmathbb{R}$ one has:
$$||g-f^*||_2^2=||g-f^*+tf-tf||_2^2=||g-(f^*+tf)||_2^2+2langle g-(f^*+tf),tfrangle +||tf||_2^2$$
$$=||g-(f^*+tf)||_2^2+2tlangle g-f^*,frangle -2t^2langle f,frangle+||tf||_2 ^2$$
$$=|||g-(f^*+tf)||_2^2+2tlangle g-f^*,frangle-t^2||f||_2^2.$$
And now choose $t$ so that
$$2tlangle g-f^*,frangle-t^2||f||_2^2>0$$
which in case that $t$ has the same sign as $langle g-f^*,frangle$ is equivalent to
$$|t|<frac{2|langle g-f^*,frangle|}{||f||_2^2}.$$
(Note that $langle g-f^*,frangleneq 0$ is essential to that.)
Then:
$$|||g-(f^*+tf)||_2^2<||g-f^*||_2^2$$
which contradicts the best-approximation-property.
answered Jan 24 at 13:04
Peter MelechPeter Melech
2,687813
$endgroup$
$begingroup$
It's much more clear and general as a proof, for sure I will include it in my thesis with a footnote thanking you for the hint
$endgroup$
– RScrlli
Jan 24 at 15:27
$begingroup$
Thank You very much!
$endgroup$
– Peter Melech
Jan 25 at 7:49
add a comment |
$begingroup$
I would like to suggest a slightly different proof, because I don't feel well with the argument " by letting $tf:=f^*-f$".
Assume $langle f,g-f^*rangleneq 0$ for some $fin U$ (I call the subspace $U$), then for all $tinmathbb{R}$ one has:
$$||g-f^*||_2^2=||g-f^*+tf-tf||_2^2=||g-(f^*+tf)||_2^2+2langle g-(f^*+tf),tfrangle +||tf||_2^2$$
$$=||g-(f^*+tf)||_2^2+2tlangle g-f^*,frangle -2t^2langle f,frangle+||tf||_2 ^2$$
$$=|||g-(f^*+tf)||_2^2+2tlangle g-f^*,frangle-t^2||f||_2^2.$$
And now choose $t$ so that
$$2tlangle g-f^*,frangle-t^2||f||_2^2>0$$
which in case that $t$ has the same sign as $langle g-f^*,frangle$ is equivalent to
$$|t|<frac{2|langle g-f^*,frangle|}{||f||_2^2}.$$
(Note that $langle g-f^*,frangleneq 0$ is essential to that.)
Then:
$$|||g-(f^*+tf)||_2^2<||g-f^*||_2^2$$
which contradicts the best-approximation-property.
answered Jan 24 at 13:04
Peter MelechPeter Melech
2,687813
$endgroup$
I would like to suggest a slightly different proof, because I don't feel well with the argument " by letting $tf:=f^*-f$".
Assume $langle f,g-f^*rangleneq 0$ for some $fin U$ (I call the subspace $U$), then for all $tinmathbb{R}$ one has:
$$||g-f^*||_2^2=||g-f^*+tf-tf||_2^2=||g-(f^*+tf)||_2^2+2langle g-(f^*+tf),tfrangle +||tf||_2^2$$
$$=||g-(f^*+tf)||_2^2+2tlangle g-f^*,frangle -2t^2langle f,frangle+||tf||_2 ^2$$
$$=|||g-(f^*+tf)||_2^2+2tlangle g-f^*,frangle-t^2||f||_2^2.$$
And now choose $t$ so that
$$2tlangle g-f^*,frangle-t^2||f||_2^2>0$$
which in case that $t$ has the same sign as $langle g-f^*,frangle$ is equivalent to
$$|t|<frac{2|langle g-f^*,frangle|}{||f||_2^2}.$$
(Note that $langle g-f^*,frangleneq 0$ is essential to that.)
Then:
$$|||g-(f^*+tf)||_2^2<||g-f^*||_2^2$$
which contradicts the best-approximation-property.
answered Jan 24 at 13:04
Peter MelechPeter Melech
2,687813
edited Jan 24 at 13:12
answered Jan 24 at 13:04
Peter MelechPeter Melech
2,687813
answered Jan 24 at 13:04
Peter MelechPeter Melech
2,687813
answered Jan 24 at 13:04
Peter MelechPeter Melech
2,687813
2,687813
$begingroup$
It's much more clear and general as a proof, for sure I will include it in my thesis with a footnote thanking you for the hint
$endgroup$
– RScrlli
Jan 24 at 15:27
$begingroup$
Thank You very much!
$endgroup$
– Peter Melech
Jan 25 at 7:49
add a comment |
$begingroup$
It's much more clear and general as a proof, for sure I will include it in my thesis with a footnote thanking you for the hint
$endgroup$
– RScrlli
Jan 24 at 15:27
$begingroup$
Thank You very much!
$endgroup$
– Peter Melech
Jan 25 at 7:49
$begingroup$
It's much more clear and general as a proof, for sure I will include it in my thesis with a footnote thanking you for the hint
$endgroup$
– RScrlli
Jan 24 at 15:27
$begingroup$
It's much more clear and general as a proof, for sure I will include it in my thesis with a footnote thanking you for the hint
$endgroup$
– RScrlli
Jan 24 at 15:27
$begingroup$
Thank You very much!
$endgroup$
– Peter Melech
Jan 25 at 7:49
$begingroup$
Thank You very much!
$endgroup$
– Peter Melech
Jan 25 at 7:49
add a comment |
$begingroup$
If $t$ has same sign as $langle f, g-f^{*}rangle$ the inequality you want is $2|t||langle f, g-f^{*}rangle| >|t|^{2} |f|_2^{2}$. This is true whenever $|t| <frac {2|langle f, g-f^{*}rangle|} { |f|_2^{2}}$.
answered Jan 23 at 10:29
Kavi Rama MurthyKavi Rama Murthy
66.8k53067
$endgroup$
$begingroup$
by taking the absolute value everything makes sense, but the problem is that what I want to show is $2langle f^*-f,g-frangle+|f^*-f|_2^2>4langle tf,g-f^*rangle$.
$endgroup$
– RScrlli
Jan 23 at 10:40
$begingroup$
I've edited the question, showing the first part of the proof.
$endgroup$
– RScrlli
Jan 23 at 10:40
add a comment |
$begingroup$
If $t$ has same sign as $langle f, g-f^{*}rangle$ the inequality you want is $2|t||langle f, g-f^{*}rangle| >|t|^{2} |f|_2^{2}$. This is true whenever $|t| <frac {2|langle f, g-f^{*}rangle|} { |f|_2^{2}}$.
answered Jan 23 at 10:29
Kavi Rama MurthyKavi Rama Murthy
66.8k53067
$endgroup$
$begingroup$
by taking the absolute value everything makes sense, but the problem is that what I want to show is $2langle f^*-f,g-frangle+|f^*-f|_2^2>4langle tf,g-f^*rangle$.
$endgroup$
– RScrlli
Jan 23 at 10:40
$begingroup$
I've edited the question, showing the first part of the proof.
$endgroup$
– RScrlli
Jan 23 at 10:40
add a comment |
$begingroup$
If $t$ has same sign as $langle f, g-f^{*}rangle$ the inequality you want is $2|t||langle f, g-f^{*}rangle| >|t|^{2} |f|_2^{2}$. This is true whenever $|t| <frac {2|langle f, g-f^{*}rangle|} { |f|_2^{2}}$.
answered Jan 23 at 10:29
Kavi Rama MurthyKavi Rama Murthy
66.8k53067
$endgroup$
If $t$ has same sign as $langle f, g-f^{*}rangle$ the inequality you want is $2|t||langle f, g-f^{*}rangle| >|t|^{2} |f|_2^{2}$. This is true whenever $|t| <frac {2|langle f, g-f^{*}rangle|} { |f|_2^{2}}$.
answered Jan 23 at 10:29
Kavi Rama MurthyKavi Rama Murthy
66.8k53067
answered Jan 23 at 10:29
Kavi Rama MurthyKavi Rama Murthy
66.8k53067
answered Jan 23 at 10:29
Kavi Rama MurthyKavi Rama Murthy
66.8k53067
answered Jan 23 at 10:29
Kavi Rama MurthyKavi Rama Murthy
66.8k53067
66.8k53067
$begingroup$
by taking the absolute value everything makes sense, but the problem is that what I want to show is $2langle f^*-f,g-frangle+|f^*-f|_2^2>4langle tf,g-f^*rangle$.
$endgroup$
– RScrlli
Jan 23 at 10:40
$begingroup$
I've edited the question, showing the first part of the proof.
$endgroup$
– RScrlli
Jan 23 at 10:40
add a comment |
$begingroup$
by taking the absolute value everything makes sense, but the problem is that what I want to show is $2langle f^*-f,g-frangle+|f^*-f|_2^2>4langle tf,g-f^*rangle$.
$endgroup$
– RScrlli
Jan 23 at 10:40
$begingroup$
I've edited the question, showing the first part of the proof.
$endgroup$
– RScrlli
Jan 23 at 10:40
$begingroup$
by taking the absolute value everything makes sense, but the problem is that what I want to show is $2langle f^*-f,g-frangle+|f^*-f|_2^2>4langle tf,g-f^*rangle$.
$endgroup$
– RScrlli
Jan 23 at 10:40
$begingroup$
by taking the absolute value everything makes sense, but the problem is that what I want to show is $2langle f^*-f,g-frangle+|f^*-f|_2^2>4langle tf,g-f^*rangle$.
$endgroup$
– RScrlli
Jan 23 at 10:40
$begingroup$
I've edited the question, showing the first part of the proof.
$endgroup$
– RScrlli
Jan 23 at 10:40
$begingroup$
I've edited the question, showing the first part of the proof.
$endgroup$
– RScrlli
Jan 23 at 10:40
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%2f3084293%2fproving-that-a-given-function-f-is-the-best-least-square-approximation%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
$begingroup$
I suspect there are some typos, e.g. $langle f^*-f,g-f^*rangle$ in the first line after "we have that" and $||g-f||_2^2=||g-f^*||_2^2+||f^*-f||_2^2geq ||g-f^*||_2^2$
$endgroup$
– Peter Melech
Jan 23 at 16:33
$begingroup$
@PeterMelech I agree with you about the second typo, but the first one,( the inner product) looks okey to me, what do you think?
$endgroup$
– RScrlli
Jan 23 at 16:37
1
$begingroup$
I'm quite sure it has to be $f^*$ instead of $f$ in the right argument of the inner product, similarly: If $langle f,g-f^*rangleneq 0$ some lines under, when showing the other direction
$endgroup$
– Peter Melech
Jan 23 at 16:46
$begingroup$
Besides: wonder how" by letting $tf:=f^*-f$" is to be understood
$endgroup$
– Peter Melech
Jan 23 at 16:57
$begingroup$
@PeterMelech you are right, I've done a copy and paste from another document and I didn't realize that. Thanks!
$endgroup$
– RScrlli
Jan 23 at 16:57