Proof regarding quasi-solutions
$begingroup$
I'm almost done proving a theorem regarding quasi-solutions, which I'm stating here:
$Let A:Xrightarrow Y be a compact linear operator and let rho>0. Then for each fin Y there exists a unique element varphi_0in X with ||varphi_0||leqrho satisfying$
$$||Avarphi_0-f||leq||Avarphi-f||$$
$for all varphiin X with ||varphi||leqrho.$
Here $X$ and $Y$ are Hilbert spaces. The part of the proof I'm struggling with is the following (uniqueness): We're considering the set $V:={Avarphi | ||varphi||leqrho}$ since for $varphi_0$ to be a quasi-solution with the constraint $rho$, then it has to be the best approximation to the set $V$.
I've already shown that $V$ is a convex set. The problem now is to show that there in the Hilbert space $Y$ exist at most one best approximation to $f$ with respect to the set $V$.
Here's my idea on how to do it (showing uniqueness of the quasi-solution):
If we consider two points $v_0,v_1in V$ in which we assume they both are the best approximation to some element $yin Y$, i.e., $||y-v_0||=||y-v_1||$. Now, we should consider the line segment between $v_0$ and $v_1$ (we know this is contained in $V$ since it is convex) and consider the distance from that to $y$, i.e., $||y-(tv_0+(1-t)v_1)||$ for $tin[0,1]$. My idea is now to show that one can find a unique $t$ for which the distance is the smallest. This will contradict the assumption that both $v_0$ and $v_1$ were the best approximations to $y$ since we've now found a new one.
Does this seem correct or am I missing something?
Thanks in advance!
real-analysis functional-analysis
asked Jan 8 at 9:37
JamesJames
12110
$endgroup$
add a comment |
$begingroup$
I'm almost done proving a theorem regarding quasi-solutions, which I'm stating here:
$Let A:Xrightarrow Y be a compact linear operator and let rho>0. Then for each fin Y there exists a unique element varphi_0in X with ||varphi_0||leqrho satisfying$
$$||Avarphi_0-f||leq||Avarphi-f||$$
$for all varphiin X with ||varphi||leqrho.$
Here $X$ and $Y$ are Hilbert spaces. The part of the proof I'm struggling with is the following (uniqueness): We're considering the set $V:={Avarphi | ||varphi||leqrho}$ since for $varphi_0$ to be a quasi-solution with the constraint $rho$, then it has to be the best approximation to the set $V$.
I've already shown that $V$ is a convex set. The problem now is to show that there in the Hilbert space $Y$ exist at most one best approximation to $f$ with respect to the set $V$.
Here's my idea on how to do it (showing uniqueness of the quasi-solution):
If we consider two points $v_0,v_1in V$ in which we assume they both are the best approximation to some element $yin Y$, i.e., $||y-v_0||=||y-v_1||$. Now, we should consider the line segment between $v_0$ and $v_1$ (we know this is contained in $V$ since it is convex) and consider the distance from that to $y$, i.e., $||y-(tv_0+(1-t)v_1)||$ for $tin[0,1]$. My idea is now to show that one can find a unique $t$ for which the distance is the smallest. This will contradict the assumption that both $v_0$ and $v_1$ were the best approximations to $y$ since we've now found a new one.
Does this seem correct or am I missing something?
Thanks in advance!
real-analysis functional-analysis
asked Jan 8 at 9:37
JamesJames
12110
$endgroup$
$begingroup$
you should state at the beginning, what spaces $X$ (and $Y$) are (probably Hilbert spaces). Then it is unclear whether your question is also about existence of $varphi_0$ or if you are only interested in uniqueness.
$endgroup$
– supinf
Jan 8 at 9:47
$begingroup$
Sorry, both $X$ and $Y$ are Hilbert spaces. I have shown existence of the quasi-solution so I'm only interested in uniqueness.
$endgroup$
– James
Jan 8 at 9:50
add a comment |
$begingroup$
I'm almost done proving a theorem regarding quasi-solutions, which I'm stating here:
$Let A:Xrightarrow Y be a compact linear operator and let rho>0. Then for each fin Y there exists a unique element varphi_0in X with ||varphi_0||leqrho satisfying$
$$||Avarphi_0-f||leq||Avarphi-f||$$
$for all varphiin X with ||varphi||leqrho.$
Here $X$ and $Y$ are Hilbert spaces. The part of the proof I'm struggling with is the following (uniqueness): We're considering the set $V:={Avarphi | ||varphi||leqrho}$ since for $varphi_0$ to be a quasi-solution with the constraint $rho$, then it has to be the best approximation to the set $V$.
I've already shown that $V$ is a convex set. The problem now is to show that there in the Hilbert space $Y$ exist at most one best approximation to $f$ with respect to the set $V$.
Here's my idea on how to do it (showing uniqueness of the quasi-solution):
If we consider two points $v_0,v_1in V$ in which we assume they both are the best approximation to some element $yin Y$, i.e., $||y-v_0||=||y-v_1||$. Now, we should consider the line segment between $v_0$ and $v_1$ (we know this is contained in $V$ since it is convex) and consider the distance from that to $y$, i.e., $||y-(tv_0+(1-t)v_1)||$ for $tin[0,1]$. My idea is now to show that one can find a unique $t$ for which the distance is the smallest. This will contradict the assumption that both $v_0$ and $v_1$ were the best approximations to $y$ since we've now found a new one.
Does this seem correct or am I missing something?
Thanks in advance!
real-analysis functional-analysis
asked Jan 8 at 9:37
JamesJames
12110
$endgroup$
I'm almost done proving a theorem regarding quasi-solutions, which I'm stating here:
$Let A:Xrightarrow Y be a compact linear operator and let rho>0. Then for each fin Y there exists a unique element varphi_0in X with ||varphi_0||leqrho satisfying$
$$||Avarphi_0-f||leq||Avarphi-f||$$
$for all varphiin X with ||varphi||leqrho.$
Here $X$ and $Y$ are Hilbert spaces. The part of the proof I'm struggling with is the following (uniqueness): We're considering the set $V:={Avarphi | ||varphi||leqrho}$ since for $varphi_0$ to be a quasi-solution with the constraint $rho$, then it has to be the best approximation to the set $V$.
I've already shown that $V$ is a convex set. The problem now is to show that there in the Hilbert space $Y$ exist at most one best approximation to $f$ with respect to the set $V$.
Here's my idea on how to do it (showing uniqueness of the quasi-solution):
If we consider two points $v_0,v_1in V$ in which we assume they both are the best approximation to some element $yin Y$, i.e., $||y-v_0||=||y-v_1||$. Now, we should consider the line segment between $v_0$ and $v_1$ (we know this is contained in $V$ since it is convex) and consider the distance from that to $y$, i.e., $||y-(tv_0+(1-t)v_1)||$ for $tin[0,1]$. My idea is now to show that one can find a unique $t$ for which the distance is the smallest. This will contradict the assumption that both $v_0$ and $v_1$ were the best approximations to $y$ since we've now found a new one.
Does this seem correct or am I missing something?
Thanks in advance!
real-analysis functional-analysis
real-analysis functional-analysis
asked Jan 8 at 9:37
JamesJames
12110
asked Jan 8 at 9:37
JamesJames
12110
edited Jan 8 at 9:51
James
asked Jan 8 at 9:37
JamesJames
12110
asked Jan 8 at 9:37
JamesJames
12110
asked Jan 8 at 9:37
JamesJames
12110
12110
$begingroup$
you should state at the beginning, what spaces $X$ (and $Y$) are (probably Hilbert spaces). Then it is unclear whether your question is also about existence of $varphi_0$ or if you are only interested in uniqueness.
$endgroup$
– supinf
Jan 8 at 9:47
$begingroup$
Sorry, both $X$ and $Y$ are Hilbert spaces. I have shown existence of the quasi-solution so I'm only interested in uniqueness.
$endgroup$
– James
Jan 8 at 9:50
add a comment |
$begingroup$
you should state at the beginning, what spaces $X$ (and $Y$) are (probably Hilbert spaces). Then it is unclear whether your question is also about existence of $varphi_0$ or if you are only interested in uniqueness.
$endgroup$
– supinf
Jan 8 at 9:47
$begingroup$
Sorry, both $X$ and $Y$ are Hilbert spaces. I have shown existence of the quasi-solution so I'm only interested in uniqueness.
$endgroup$
– James
Jan 8 at 9:50
$begingroup$
you should state at the beginning, what spaces $X$ (and $Y$) are (probably Hilbert spaces). Then it is unclear whether your question is also about existence of $varphi_0$ or if you are only interested in uniqueness.
$endgroup$
– supinf
Jan 8 at 9:47
$begingroup$
you should state at the beginning, what spaces $X$ (and $Y$) are (probably Hilbert spaces). Then it is unclear whether your question is also about existence of $varphi_0$ or if you are only interested in uniqueness.
$endgroup$
– supinf
Jan 8 at 9:47
$begingroup$
Sorry, both $X$ and $Y$ are Hilbert spaces. I have shown existence of the quasi-solution so I'm only interested in uniqueness.
$endgroup$
– James
Jan 8 at 9:50
$begingroup$
Sorry, both $X$ and $Y$ are Hilbert spaces. I have shown existence of the quasi-solution so I'm only interested in uniqueness.
$endgroup$
– James
Jan 8 at 9:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your idea to consider the line between $v_1,v_0$ is a good start.
However, it is often enough to only consider the central point $v_2:=tfrac12(v_0+v_1)$ to get a contradiction.
Here is a sketch on how to do this:
Using the parallelogram law, we have
$$
2|y-v_0|^2+2|y-v_1|^2 = |2y-v_0-v_1|^2+|v_0-v_1|^2
=4|y-v_2|^2+|v_0-v_1|^2.
$$
Now if you assume that $v_0neq v_1$ then it follows that $v_2$ is a better solution than $v_1$ or $v_0$.
answered Jan 8 at 10:06
supinfsupinf
6,7571129
$endgroup$
$begingroup$
Thank you for the help! I've tried it myself and I got the same as you with the parallelogram law. This is just to see if I understand why you can argue $v_2$ is a better choice than $v_0$ or $v_1$: We assumed that $||y-v_0||=||y-v_1||$ and $v_0neq v_1$, which means by the parallelogram law we have $$4||y-v_0||^2=4||y-v_2||^2+||v_0-v_1||$$ which means that $v_2$ is a better approximation, due to $||v_0-v_1||^2$ term, right?
$endgroup$
– James
Jan 8 at 10:38
$begingroup$
Or one could write the LHS of the parallelogram law with $v_1$, but the idea is that it shows that 4 times the distance between $y$ and $v_0$ or $v_1$ yields 4 times the distance of $y$ and $v_2$ plus an additional distance, namely the distance between $v_0$ and $v_1$, which means $v_2$ must approximate $y$ better.
$endgroup$
– James
Jan 8 at 10:43
1
$begingroup$
yes, that is correct
$endgroup$
– supinf
Jan 8 at 11:19
add a comment |
Your Answer
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%2f3065974%2fproof-regarding-quasi-solutions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your idea to consider the line between $v_1,v_0$ is a good start.
However, it is often enough to only consider the central point $v_2:=tfrac12(v_0+v_1)$ to get a contradiction.
Here is a sketch on how to do this:
Using the parallelogram law, we have
$$
2|y-v_0|^2+2|y-v_1|^2 = |2y-v_0-v_1|^2+|v_0-v_1|^2
=4|y-v_2|^2+|v_0-v_1|^2.
$$
Now if you assume that $v_0neq v_1$ then it follows that $v_2$ is a better solution than $v_1$ or $v_0$.
answered Jan 8 at 10:06
supinfsupinf
6,7571129
$endgroup$
$begingroup$
Thank you for the help! I've tried it myself and I got the same as you with the parallelogram law. This is just to see if I understand why you can argue $v_2$ is a better choice than $v_0$ or $v_1$: We assumed that $||y-v_0||=||y-v_1||$ and $v_0neq v_1$, which means by the parallelogram law we have $$4||y-v_0||^2=4||y-v_2||^2+||v_0-v_1||$$ which means that $v_2$ is a better approximation, due to $||v_0-v_1||^2$ term, right?
$endgroup$
– James
Jan 8 at 10:38
$begingroup$
Or one could write the LHS of the parallelogram law with $v_1$, but the idea is that it shows that 4 times the distance between $y$ and $v_0$ or $v_1$ yields 4 times the distance of $y$ and $v_2$ plus an additional distance, namely the distance between $v_0$ and $v_1$, which means $v_2$ must approximate $y$ better.
$endgroup$
– James
Jan 8 at 10:43
1
$begingroup$
yes, that is correct
$endgroup$
– supinf
Jan 8 at 11:19
add a comment |
$begingroup$
Your idea to consider the line between $v_1,v_0$ is a good start.
However, it is often enough to only consider the central point $v_2:=tfrac12(v_0+v_1)$ to get a contradiction.
Here is a sketch on how to do this:
Using the parallelogram law, we have
$$
2|y-v_0|^2+2|y-v_1|^2 = |2y-v_0-v_1|^2+|v_0-v_1|^2
=4|y-v_2|^2+|v_0-v_1|^2.
$$
Now if you assume that $v_0neq v_1$ then it follows that $v_2$ is a better solution than $v_1$ or $v_0$.
answered Jan 8 at 10:06
supinfsupinf
6,7571129
$endgroup$
$begingroup$
Thank you for the help! I've tried it myself and I got the same as you with the parallelogram law. This is just to see if I understand why you can argue $v_2$ is a better choice than $v_0$ or $v_1$: We assumed that $||y-v_0||=||y-v_1||$ and $v_0neq v_1$, which means by the parallelogram law we have $$4||y-v_0||^2=4||y-v_2||^2+||v_0-v_1||$$ which means that $v_2$ is a better approximation, due to $||v_0-v_1||^2$ term, right?
$endgroup$
– James
Jan 8 at 10:38
$begingroup$
Or one could write the LHS of the parallelogram law with $v_1$, but the idea is that it shows that 4 times the distance between $y$ and $v_0$ or $v_1$ yields 4 times the distance of $y$ and $v_2$ plus an additional distance, namely the distance between $v_0$ and $v_1$, which means $v_2$ must approximate $y$ better.
$endgroup$
– James
Jan 8 at 10:43
1
$begingroup$
yes, that is correct
$endgroup$
– supinf
Jan 8 at 11:19
add a comment |
$begingroup$
Your idea to consider the line between $v_1,v_0$ is a good start.
However, it is often enough to only consider the central point $v_2:=tfrac12(v_0+v_1)$ to get a contradiction.
Here is a sketch on how to do this:
Using the parallelogram law, we have
$$
2|y-v_0|^2+2|y-v_1|^2 = |2y-v_0-v_1|^2+|v_0-v_1|^2
=4|y-v_2|^2+|v_0-v_1|^2.
$$
Now if you assume that $v_0neq v_1$ then it follows that $v_2$ is a better solution than $v_1$ or $v_0$.
answered Jan 8 at 10:06
supinfsupinf
6,7571129
$endgroup$
Your idea to consider the line between $v_1,v_0$ is a good start.
However, it is often enough to only consider the central point $v_2:=tfrac12(v_0+v_1)$ to get a contradiction.
Here is a sketch on how to do this:
Using the parallelogram law, we have
$$
2|y-v_0|^2+2|y-v_1|^2 = |2y-v_0-v_1|^2+|v_0-v_1|^2
=4|y-v_2|^2+|v_0-v_1|^2.
$$
Now if you assume that $v_0neq v_1$ then it follows that $v_2$ is a better solution than $v_1$ or $v_0$.
answered Jan 8 at 10:06
supinfsupinf
6,7571129
answered Jan 8 at 10:06
supinfsupinf
6,7571129
answered Jan 8 at 10:06
supinfsupinf
6,7571129
answered Jan 8 at 10:06
supinfsupinf
6,7571129
6,7571129
$begingroup$
Thank you for the help! I've tried it myself and I got the same as you with the parallelogram law. This is just to see if I understand why you can argue $v_2$ is a better choice than $v_0$ or $v_1$: We assumed that $||y-v_0||=||y-v_1||$ and $v_0neq v_1$, which means by the parallelogram law we have $$4||y-v_0||^2=4||y-v_2||^2+||v_0-v_1||$$ which means that $v_2$ is a better approximation, due to $||v_0-v_1||^2$ term, right?
$endgroup$
– James
Jan 8 at 10:38
$begingroup$
Or one could write the LHS of the parallelogram law with $v_1$, but the idea is that it shows that 4 times the distance between $y$ and $v_0$ or $v_1$ yields 4 times the distance of $y$ and $v_2$ plus an additional distance, namely the distance between $v_0$ and $v_1$, which means $v_2$ must approximate $y$ better.
$endgroup$
– James
Jan 8 at 10:43
1
$begingroup$
yes, that is correct
$endgroup$
– supinf
Jan 8 at 11:19
add a comment |
$begingroup$
Thank you for the help! I've tried it myself and I got the same as you with the parallelogram law. This is just to see if I understand why you can argue $v_2$ is a better choice than $v_0$ or $v_1$: We assumed that $||y-v_0||=||y-v_1||$ and $v_0neq v_1$, which means by the parallelogram law we have $$4||y-v_0||^2=4||y-v_2||^2+||v_0-v_1||$$ which means that $v_2$ is a better approximation, due to $||v_0-v_1||^2$ term, right?
$endgroup$
– James
Jan 8 at 10:38
$begingroup$
Or one could write the LHS of the parallelogram law with $v_1$, but the idea is that it shows that 4 times the distance between $y$ and $v_0$ or $v_1$ yields 4 times the distance of $y$ and $v_2$ plus an additional distance, namely the distance between $v_0$ and $v_1$, which means $v_2$ must approximate $y$ better.
$endgroup$
– James
Jan 8 at 10:43
1
$begingroup$
yes, that is correct
$endgroup$
– supinf
Jan 8 at 11:19
$begingroup$
Thank you for the help! I've tried it myself and I got the same as you with the parallelogram law. This is just to see if I understand why you can argue $v_2$ is a better choice than $v_0$ or $v_1$: We assumed that $||y-v_0||=||y-v_1||$ and $v_0neq v_1$, which means by the parallelogram law we have $$4||y-v_0||^2=4||y-v_2||^2+||v_0-v_1||$$ which means that $v_2$ is a better approximation, due to $||v_0-v_1||^2$ term, right?
$endgroup$
– James
Jan 8 at 10:38
$begingroup$
Thank you for the help! I've tried it myself and I got the same as you with the parallelogram law. This is just to see if I understand why you can argue $v_2$ is a better choice than $v_0$ or $v_1$: We assumed that $||y-v_0||=||y-v_1||$ and $v_0neq v_1$, which means by the parallelogram law we have $$4||y-v_0||^2=4||y-v_2||^2+||v_0-v_1||$$ which means that $v_2$ is a better approximation, due to $||v_0-v_1||^2$ term, right?
$endgroup$
– James
Jan 8 at 10:38
$begingroup$
Or one could write the LHS of the parallelogram law with $v_1$, but the idea is that it shows that 4 times the distance between $y$ and $v_0$ or $v_1$ yields 4 times the distance of $y$ and $v_2$ plus an additional distance, namely the distance between $v_0$ and $v_1$, which means $v_2$ must approximate $y$ better.
$endgroup$
– James
Jan 8 at 10:43
$begingroup$
Or one could write the LHS of the parallelogram law with $v_1$, but the idea is that it shows that 4 times the distance between $y$ and $v_0$ or $v_1$ yields 4 times the distance of $y$ and $v_2$ plus an additional distance, namely the distance between $v_0$ and $v_1$, which means $v_2$ must approximate $y$ better.
$endgroup$
– James
Jan 8 at 10:43
1
1
$begingroup$
yes, that is correct
$endgroup$
– supinf
Jan 8 at 11:19
$begingroup$
yes, that is correct
$endgroup$
– supinf
Jan 8 at 11:19
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%2f3065974%2fproof-regarding-quasi-solutions%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
$begingroup$
you should state at the beginning, what spaces $X$ (and $Y$) are (probably Hilbert spaces). Then it is unclear whether your question is also about existence of $varphi_0$ or if you are only interested in uniqueness.
$endgroup$
– supinf
Jan 8 at 9:47
$begingroup$
Sorry, both $X$ and $Y$ are Hilbert spaces. I have shown existence of the quasi-solution so I'm only interested in uniqueness.
$endgroup$
– James
Jan 8 at 9:50