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
1267
$endgroup$
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
1267
$endgroup$
$begingroup$
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
1267
$endgroup$
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
1267
asked Jan 8 at 9:37
JamesJames
1267
edited Jan 8 at 9:51
James
asked Jan 8 at 9:37
JamesJames
1267
asked Jan 8 at 9:37
JamesJames
1267
asked Jan 8 at 9:37
JamesJames
1267
1267
$begingroup$
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
6,1391028
$endgroup$
$begingroup$
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 |
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1 Answer
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$begingroup$
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
6,1391028
$endgroup$
$begingroup$
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
6,1391028
$endgroup$
$begingroup$
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
6,1391028
$endgroup$
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
6,1391028
answered Jan 8 at 10:06
supinfsupinf
6,1391028
answered Jan 8 at 10:06
supinfsupinf
6,1391028
answered Jan 8 at 10:06
supinfsupinf
6,1391028
6,1391028
$begingroup$
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 |
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$begingroup$
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