Mixing
Why do these two elements exist and satisfy this equation?
$begingroup$
From Convex optimization by Boyd and Vanderberghe:
In the below red box, why do there exist $y1,y2$ such that the equation follows? Doesn't it depend on what the domain of $f$ and what the function $f$ are?
proof-explanation convex-optimization
asked Jan 23 at 13:13
Oliver GOliver G
1,4331632
$endgroup$
add a comment |
$begingroup$
From Convex optimization by Boyd and Vanderberghe:
In the below red box, why do there exist $y1,y2$ such that the equation follows? Doesn't it depend on what the domain of $f$ and what the function $f$ are?
proof-explanation convex-optimization
asked Jan 23 at 13:13
Oliver GOliver G
1,4331632
$endgroup$
1
$begingroup$
Maybe it is easier to think about it as $|f(x_i,y_i)-g(x_i)| le epsilon$. It is basically saying, you can get to within epsilon of the infimum, for any epsilon. This follows from the definition of infimum.
$endgroup$
– Jaap Scherphuis
Jan 23 at 13:39
$begingroup$
@JaapScherphuis you should post that as an answer.
$endgroup$
– LinAlg
Jan 23 at 15:40
add a comment |
$begingroup$
From Convex optimization by Boyd and Vanderberghe:
In the below red box, why do there exist $y1,y2$ such that the equation follows? Doesn't it depend on what the domain of $f$ and what the function $f$ are?
proof-explanation convex-optimization
asked Jan 23 at 13:13
Oliver GOliver G
1,4331632
$endgroup$
From Convex optimization by Boyd and Vanderberghe:
In the below red box, why do there exist $y1,y2$ such that the equation follows? Doesn't it depend on what the domain of $f$ and what the function $f$ are?
proof-explanation convex-optimization
proof-explanation convex-optimization
asked Jan 23 at 13:13
Oliver GOliver G
1,4331632
asked Jan 23 at 13:13
Oliver GOliver G
1,4331632
edited Jan 23 at 13:26
Oliver G
asked Jan 23 at 13:13
Oliver GOliver G
1,4331632
asked Jan 23 at 13:13
Oliver GOliver G
1,4331632
asked Jan 23 at 13:13
Oliver GOliver G
1,4331632
1,4331632
1
$begingroup$
Maybe it is easier to think about it as $|f(x_i,y_i)-g(x_i)| le epsilon$. It is basically saying, you can get to within epsilon of the infimum, for any epsilon. This follows from the definition of infimum.
$endgroup$
– Jaap Scherphuis
Jan 23 at 13:39
$begingroup$
@JaapScherphuis you should post that as an answer.
$endgroup$
– LinAlg
Jan 23 at 15:40
add a comment |
1
$begingroup$
Maybe it is easier to think about it as $|f(x_i,y_i)-g(x_i)| le epsilon$. It is basically saying, you can get to within epsilon of the infimum, for any epsilon. This follows from the definition of infimum.
$endgroup$
– Jaap Scherphuis
Jan 23 at 13:39
$begingroup$
@JaapScherphuis you should post that as an answer.
$endgroup$
– LinAlg
Jan 23 at 15:40
1
1
$begingroup$
Maybe it is easier to think about it as $|f(x_i,y_i)-g(x_i)| le epsilon$. It is basically saying, you can get to within epsilon of the infimum, for any epsilon. This follows from the definition of infimum.
$endgroup$
– Jaap Scherphuis
Jan 23 at 13:39
$begingroup$
Maybe it is easier to think about it as $|f(x_i,y_i)-g(x_i)| le epsilon$. It is basically saying, you can get to within epsilon of the infimum, for any epsilon. This follows from the definition of infimum.
$endgroup$
– Jaap Scherphuis
Jan 23 at 13:39
$begingroup$
@JaapScherphuis you should post that as an answer.
$endgroup$
– LinAlg
Jan 23 at 15:40
$begingroup$
@JaapScherphuis you should post that as an answer.
$endgroup$
– LinAlg
Jan 23 at 15:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Maybe it is easier to think about the inequality as $$|f(x_i,y_i)−g(x_i)|≤epsilon$$
It is basically saying, you can get to within epsilon of the infimum, for any epsilon. This follows from the definition of infimum.
As for the domain, it seems to me that there is a bit of fudging in the calculation of the infimum. The infimum for a given $x$ coordinate is not necessarily calculated over the whole of $yin C$ but only over those values for which $f(x,y)$ makes sense, i.e. values $y$ such that $yin C$ and for which $(x,y)$ lies in the domain of $f$. This is kind of implied by how they define the domain of $g$ ("for some $y in C$").
With that proviso, the rest follows without any problems. An $y_1$ can be found such that it lies in $C$ and $(x_1,y_1)$ is in the domain of $f$, and for which $|f(x_1,y_1)−g(x_1)|le epsilon$. The same goes for $y_2$.
answered Jan 23 at 16:20
Jaap ScherphuisJaap Scherphuis
4,167717
$endgroup$
$begingroup$
I understand what it is saying but I'm asking why is this valid. Why do there exist $y_1,y_2$ such that the equation follows? Doesn't it depend on what the domain of $f$ and what the function $f$ are?
$endgroup$
– Oliver G
Jan 24 at 13:39
$begingroup$
@OliverG Strictly speaking I suppose the infimum is not necessarily calculated over the whole of $y in C$ but is only over those values for which $f(x,y)$ makes sense, i.e. values $y$ such that $y in C$ and $(x,y)$ in the domain of $f$. This is kind of implied by how they define the domain of $g$ ("for some $y in C$"). With that proviso, the rest follows, and the $y_1$ can be found such that it lies in $C$ and $(x_1,y_1)$ is in the domain of $f$, and $|f(x_1,y_1)−g(x_1)|≤epsilon$. Similarly for $y_2$.
$endgroup$
– Jaap Scherphuis
Jan 24 at 14:24
add a comment |
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$begingroup$
Maybe it is easier to think about the inequality as $$|f(x_i,y_i)−g(x_i)|≤epsilon$$
It is basically saying, you can get to within epsilon of the infimum, for any epsilon. This follows from the definition of infimum.
As for the domain, it seems to me that there is a bit of fudging in the calculation of the infimum. The infimum for a given $x$ coordinate is not necessarily calculated over the whole of $yin C$ but only over those values for which $f(x,y)$ makes sense, i.e. values $y$ such that $yin C$ and for which $(x,y)$ lies in the domain of $f$. This is kind of implied by how they define the domain of $g$ ("for some $y in C$").
With that proviso, the rest follows without any problems. An $y_1$ can be found such that it lies in $C$ and $(x_1,y_1)$ is in the domain of $f$, and for which $|f(x_1,y_1)−g(x_1)|le epsilon$. The same goes for $y_2$.
answered Jan 23 at 16:20
Jaap ScherphuisJaap Scherphuis
4,167717
$endgroup$
$begingroup$
I understand what it is saying but I'm asking why is this valid. Why do there exist $y_1,y_2$ such that the equation follows? Doesn't it depend on what the domain of $f$ and what the function $f$ are?
$endgroup$
– Oliver G
Jan 24 at 13:39
$begingroup$
@OliverG Strictly speaking I suppose the infimum is not necessarily calculated over the whole of $y in C$ but is only over those values for which $f(x,y)$ makes sense, i.e. values $y$ such that $y in C$ and $(x,y)$ in the domain of $f$. This is kind of implied by how they define the domain of $g$ ("for some $y in C$"). With that proviso, the rest follows, and the $y_1$ can be found such that it lies in $C$ and $(x_1,y_1)$ is in the domain of $f$, and $|f(x_1,y_1)−g(x_1)|≤epsilon$. Similarly for $y_2$.
$endgroup$
– Jaap Scherphuis
Jan 24 at 14:24
add a comment |
$begingroup$
Maybe it is easier to think about the inequality as $$|f(x_i,y_i)−g(x_i)|≤epsilon$$
It is basically saying, you can get to within epsilon of the infimum, for any epsilon. This follows from the definition of infimum.
As for the domain, it seems to me that there is a bit of fudging in the calculation of the infimum. The infimum for a given $x$ coordinate is not necessarily calculated over the whole of $yin C$ but only over those values for which $f(x,y)$ makes sense, i.e. values $y$ such that $yin C$ and for which $(x,y)$ lies in the domain of $f$. This is kind of implied by how they define the domain of $g$ ("for some $y in C$").
With that proviso, the rest follows without any problems. An $y_1$ can be found such that it lies in $C$ and $(x_1,y_1)$ is in the domain of $f$, and for which $|f(x_1,y_1)−g(x_1)|le epsilon$. The same goes for $y_2$.
answered Jan 23 at 16:20
Jaap ScherphuisJaap Scherphuis
4,167717
$endgroup$
$begingroup$
I understand what it is saying but I'm asking why is this valid. Why do there exist $y_1,y_2$ such that the equation follows? Doesn't it depend on what the domain of $f$ and what the function $f$ are?
$endgroup$
– Oliver G
Jan 24 at 13:39
$begingroup$
@OliverG Strictly speaking I suppose the infimum is not necessarily calculated over the whole of $y in C$ but is only over those values for which $f(x,y)$ makes sense, i.e. values $y$ such that $y in C$ and $(x,y)$ in the domain of $f$. This is kind of implied by how they define the domain of $g$ ("for some $y in C$"). With that proviso, the rest follows, and the $y_1$ can be found such that it lies in $C$ and $(x_1,y_1)$ is in the domain of $f$, and $|f(x_1,y_1)−g(x_1)|≤epsilon$. Similarly for $y_2$.
$endgroup$
– Jaap Scherphuis
Jan 24 at 14:24
add a comment |
$begingroup$
Maybe it is easier to think about the inequality as $$|f(x_i,y_i)−g(x_i)|≤epsilon$$
It is basically saying, you can get to within epsilon of the infimum, for any epsilon. This follows from the definition of infimum.
As for the domain, it seems to me that there is a bit of fudging in the calculation of the infimum. The infimum for a given $x$ coordinate is not necessarily calculated over the whole of $yin C$ but only over those values for which $f(x,y)$ makes sense, i.e. values $y$ such that $yin C$ and for which $(x,y)$ lies in the domain of $f$. This is kind of implied by how they define the domain of $g$ ("for some $y in C$").
With that proviso, the rest follows without any problems. An $y_1$ can be found such that it lies in $C$ and $(x_1,y_1)$ is in the domain of $f$, and for which $|f(x_1,y_1)−g(x_1)|le epsilon$. The same goes for $y_2$.
answered Jan 23 at 16:20
Jaap ScherphuisJaap Scherphuis
4,167717
$endgroup$
Maybe it is easier to think about the inequality as $$|f(x_i,y_i)−g(x_i)|≤epsilon$$
It is basically saying, you can get to within epsilon of the infimum, for any epsilon. This follows from the definition of infimum.
As for the domain, it seems to me that there is a bit of fudging in the calculation of the infimum. The infimum for a given $x$ coordinate is not necessarily calculated over the whole of $yin C$ but only over those values for which $f(x,y)$ makes sense, i.e. values $y$ such that $yin C$ and for which $(x,y)$ lies in the domain of $f$. This is kind of implied by how they define the domain of $g$ ("for some $y in C$").
With that proviso, the rest follows without any problems. An $y_1$ can be found such that it lies in $C$ and $(x_1,y_1)$ is in the domain of $f$, and for which $|f(x_1,y_1)−g(x_1)|le epsilon$. The same goes for $y_2$.
answered Jan 23 at 16:20
Jaap ScherphuisJaap Scherphuis
4,167717
edited Jan 24 at 16:26
answered Jan 23 at 16:20
Jaap ScherphuisJaap Scherphuis
4,167717
answered Jan 23 at 16:20
Jaap ScherphuisJaap Scherphuis
4,167717
answered Jan 23 at 16:20
Jaap ScherphuisJaap Scherphuis
4,167717
4,167717
$begingroup$
I understand what it is saying but I'm asking why is this valid. Why do there exist $y_1,y_2$ such that the equation follows? Doesn't it depend on what the domain of $f$ and what the function $f$ are?
$endgroup$
– Oliver G
Jan 24 at 13:39
$begingroup$
@OliverG Strictly speaking I suppose the infimum is not necessarily calculated over the whole of $y in C$ but is only over those values for which $f(x,y)$ makes sense, i.e. values $y$ such that $y in C$ and $(x,y)$ in the domain of $f$. This is kind of implied by how they define the domain of $g$ ("for some $y in C$"). With that proviso, the rest follows, and the $y_1$ can be found such that it lies in $C$ and $(x_1,y_1)$ is in the domain of $f$, and $|f(x_1,y_1)−g(x_1)|≤epsilon$. Similarly for $y_2$.
$endgroup$
– Jaap Scherphuis
Jan 24 at 14:24
add a comment |
$begingroup$
I understand what it is saying but I'm asking why is this valid. Why do there exist $y_1,y_2$ such that the equation follows? Doesn't it depend on what the domain of $f$ and what the function $f$ are?
$endgroup$
– Oliver G
Jan 24 at 13:39
$begingroup$
@OliverG Strictly speaking I suppose the infimum is not necessarily calculated over the whole of $y in C$ but is only over those values for which $f(x,y)$ makes sense, i.e. values $y$ such that $y in C$ and $(x,y)$ in the domain of $f$. This is kind of implied by how they define the domain of $g$ ("for some $y in C$"). With that proviso, the rest follows, and the $y_1$ can be found such that it lies in $C$ and $(x_1,y_1)$ is in the domain of $f$, and $|f(x_1,y_1)−g(x_1)|≤epsilon$. Similarly for $y_2$.
$endgroup$
– Jaap Scherphuis
Jan 24 at 14:24
$begingroup$
I understand what it is saying but I'm asking why is this valid. Why do there exist $y_1,y_2$ such that the equation follows? Doesn't it depend on what the domain of $f$ and what the function $f$ are?
$endgroup$
– Oliver G
Jan 24 at 13:39
$begingroup$
I understand what it is saying but I'm asking why is this valid. Why do there exist $y_1,y_2$ such that the equation follows? Doesn't it depend on what the domain of $f$ and what the function $f$ are?
$endgroup$
– Oliver G
Jan 24 at 13:39
$begingroup$
@OliverG Strictly speaking I suppose the infimum is not necessarily calculated over the whole of $y in C$ but is only over those values for which $f(x,y)$ makes sense, i.e. values $y$ such that $y in C$ and $(x,y)$ in the domain of $f$. This is kind of implied by how they define the domain of $g$ ("for some $y in C$"). With that proviso, the rest follows, and the $y_1$ can be found such that it lies in $C$ and $(x_1,y_1)$ is in the domain of $f$, and $|f(x_1,y_1)−g(x_1)|≤epsilon$. Similarly for $y_2$.
$endgroup$
– Jaap Scherphuis
Jan 24 at 14:24
$begingroup$
@OliverG Strictly speaking I suppose the infimum is not necessarily calculated over the whole of $y in C$ but is only over those values for which $f(x,y)$ makes sense, i.e. values $y$ such that $y in C$ and $(x,y)$ in the domain of $f$. This is kind of implied by how they define the domain of $g$ ("for some $y in C$"). With that proviso, the rest follows, and the $y_1$ can be found such that it lies in $C$ and $(x_1,y_1)$ is in the domain of $f$, and $|f(x_1,y_1)−g(x_1)|≤epsilon$. Similarly for $y_2$.
$endgroup$
– Jaap Scherphuis
Jan 24 at 14:24
add a comment |
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$begingroup$
Maybe it is easier to think about it as $|f(x_i,y_i)-g(x_i)| le epsilon$. It is basically saying, you can get to within epsilon of the infimum, for any epsilon. This follows from the definition of infimum.
$endgroup$
– Jaap Scherphuis
Jan 23 at 13:39
$begingroup$
@JaapScherphuis you should post that as an answer.
$endgroup$
– LinAlg
Jan 23 at 15:40