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Show that the support function of a bounded set is continuous.
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The support function of a set $A in mathbb{R}^n$ is defined as the following
$$
S_A(x)=sup_{y in A} x^Ty
$$
where $x in mathbb{R}^n$.
Show that the support function of a bounded set is continuous.
I tried the following:
Let $A$ be a bounded set in $mathbb{R}^n$ and $y in A$. So $|y| leq M$ where $M>0$. (If $M=0 rightarrow |y|=0 rightarrow y=0 rightarrow S_A(x)=0rightarrow S_A(x)$ is continuous $forall x$).
Let $|x-x_c|<delta=frac{epsilon}{M}, ,,,forall epsilon>0$ be a neighborhood of $x_c$ where $x_c in mathbb{R}^n$.
I need to show that
$$
|S_A(x)-S_A(x_c)|=|sup_{y in A} x^Ty-sup_{y in A} x_c^Ty|<epsilon
$$
How can I proceed?
real-analysis continuity metric-spaces uniform-continuity
asked Nov 8 at 19:36
Saeed
407110
add a comment |
up vote
1
down vote
favorite
The support function of a set $A in mathbb{R}^n$ is defined as the following
$$
S_A(x)=sup_{y in A} x^Ty
$$
where $x in mathbb{R}^n$.
Show that the support function of a bounded set is continuous.
I tried the following:
Let $A$ be a bounded set in $mathbb{R}^n$ and $y in A$. So $|y| leq M$ where $M>0$. (If $M=0 rightarrow |y|=0 rightarrow y=0 rightarrow S_A(x)=0rightarrow S_A(x)$ is continuous $forall x$).
Let $|x-x_c|<delta=frac{epsilon}{M}, ,,,forall epsilon>0$ be a neighborhood of $x_c$ where $x_c in mathbb{R}^n$.
I need to show that
$$
|S_A(x)-S_A(x_c)|=|sup_{y in A} x^Ty-sup_{y in A} x_c^Ty|<epsilon
$$
How can I proceed?
real-analysis continuity metric-spaces uniform-continuity
asked Nov 8 at 19:36
Saeed
407110
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The support function of a set $A in mathbb{R}^n$ is defined as the following
$$
S_A(x)=sup_{y in A} x^Ty
$$
where $x in mathbb{R}^n$.
Show that the support function of a bounded set is continuous.
I tried the following:
Let $A$ be a bounded set in $mathbb{R}^n$ and $y in A$. So $|y| leq M$ where $M>0$. (If $M=0 rightarrow |y|=0 rightarrow y=0 rightarrow S_A(x)=0rightarrow S_A(x)$ is continuous $forall x$).
Let $|x-x_c|<delta=frac{epsilon}{M}, ,,,forall epsilon>0$ be a neighborhood of $x_c$ where $x_c in mathbb{R}^n$.
I need to show that
$$
|S_A(x)-S_A(x_c)|=|sup_{y in A} x^Ty-sup_{y in A} x_c^Ty|<epsilon
$$
How can I proceed?
real-analysis continuity metric-spaces uniform-continuity
asked Nov 8 at 19:36
Saeed
407110
The support function of a set $A in mathbb{R}^n$ is defined as the following
$$
S_A(x)=sup_{y in A} x^Ty
$$
where $x in mathbb{R}^n$.
Show that the support function of a bounded set is continuous.
I tried the following:
Let $A$ be a bounded set in $mathbb{R}^n$ and $y in A$. So $|y| leq M$ where $M>0$. (If $M=0 rightarrow |y|=0 rightarrow y=0 rightarrow S_A(x)=0rightarrow S_A(x)$ is continuous $forall x$).
Let $|x-x_c|<delta=frac{epsilon}{M}, ,,,forall epsilon>0$ be a neighborhood of $x_c$ where $x_c in mathbb{R}^n$.
I need to show that
$$
|S_A(x)-S_A(x_c)|=|sup_{y in A} x^Ty-sup_{y in A} x_c^Ty|<epsilon
$$
How can I proceed?
real-analysis continuity metric-spaces uniform-continuity
real-analysis continuity metric-spaces uniform-continuity
asked Nov 8 at 19:36
Saeed
407110
asked Nov 8 at 19:36
Saeed
407110
asked Nov 8 at 19:36
Saeed
407110
asked Nov 8 at 19:36
Saeed
407110
asked Nov 8 at 19:36
Saeed
407110
407110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Using Cauchy-Schwarz inequality we can bound $(x^{T}-x_c^{T})y$
$(x^{T}-x_c^{T})yleq |x^{T}-x_c^{T}||y| $
Since $|x^{T}-x_c^{T}| < delta$, we have the following
So
$$(x^{T}-x_c^{T})y leq |x^{T}-x_c^{T}||y| < delta |y|$$
So $x_c^{T}y < delta |y|+x_c^{T}y$ for all $y$ in $A$.
Take sup over $y$
$$sup_{y in A} x_c^{T}y < sup_{y in A} (delta |y|+x_c^{T}y) leq sup_{y in A} delta |y|+ sup_{y in A}x_c^{T}y$$
So
$$S_A(x) < delta M + S_A(x_c)$$
Do the same for $(x_c^{T}-x^{T})y$ to get
$$S_A(x_c) < delta M + S_A(x)$$
Combine them to get the following
$|S_A(x)-S_A(x_c)|<delta M=epsilon$.
answered 23 hours ago
Sepide
1227
add a comment |
up vote
0
down vote
Here's a solution which reduces to another problem on this site: Supremum is continuous over equicontinuous family of functions
Step 1: define for each $ain A$ the function $f_a(x) = x cdot a$. Each of these is continuous.
Step 2: Show that $S_A$ is the supremum of the family ${f_a}$
Step 3: Show that because $A$ is bounded, the family is not only continuous but equicontinuous.
Then conclude using the link above!
answered Nov 8 at 21:20
user25959
1,016714
I want to proof it in a way that I explained. Could you help me to do that.
– Saeed
Nov 8 at 21:25
You will need to do an $epsilon/3$-style argument. Basically rewrite $|S_A(x)-S_A(x_c)|$ as $|S_A(x)-x cdot y +x cdot y -x_ccdot y +x_ccdot y -S_A(x_c)|$ for some $y$ which almost attains the supremum $S_A(x_c)$ and use the triangle inequality and equicontinuity.
– user25959
Nov 8 at 22:42
Let me I change your $y$ to $z$. Then how can I treat $|sup_{y in A}x^Ty-x^Tz|$?
– Saeed
Nov 8 at 23:57
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Using Cauchy-Schwarz inequality we can bound $(x^{T}-x_c^{T})y$
$(x^{T}-x_c^{T})yleq |x^{T}-x_c^{T}||y| $
Since $|x^{T}-x_c^{T}| < delta$, we have the following
So
$$(x^{T}-x_c^{T})y leq |x^{T}-x_c^{T}||y| < delta |y|$$
So $x_c^{T}y < delta |y|+x_c^{T}y$ for all $y$ in $A$.
Take sup over $y$
$$sup_{y in A} x_c^{T}y < sup_{y in A} (delta |y|+x_c^{T}y) leq sup_{y in A} delta |y|+ sup_{y in A}x_c^{T}y$$
So
$$S_A(x) < delta M + S_A(x_c)$$
Do the same for $(x_c^{T}-x^{T})y$ to get
$$S_A(x_c) < delta M + S_A(x)$$
Combine them to get the following
$|S_A(x)-S_A(x_c)|<delta M=epsilon$.
answered 23 hours ago
Sepide
1227
add a comment |
up vote
1
down vote
accepted
Using Cauchy-Schwarz inequality we can bound $(x^{T}-x_c^{T})y$
$(x^{T}-x_c^{T})yleq |x^{T}-x_c^{T}||y| $
Since $|x^{T}-x_c^{T}| < delta$, we have the following
So
$$(x^{T}-x_c^{T})y leq |x^{T}-x_c^{T}||y| < delta |y|$$
So $x_c^{T}y < delta |y|+x_c^{T}y$ for all $y$ in $A$.
Take sup over $y$
$$sup_{y in A} x_c^{T}y < sup_{y in A} (delta |y|+x_c^{T}y) leq sup_{y in A} delta |y|+ sup_{y in A}x_c^{T}y$$
So
$$S_A(x) < delta M + S_A(x_c)$$
Do the same for $(x_c^{T}-x^{T})y$ to get
$$S_A(x_c) < delta M + S_A(x)$$
Combine them to get the following
$|S_A(x)-S_A(x_c)|<delta M=epsilon$.
answered 23 hours ago
Sepide
1227
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Using Cauchy-Schwarz inequality we can bound $(x^{T}-x_c^{T})y$
$(x^{T}-x_c^{T})yleq |x^{T}-x_c^{T}||y| $
Since $|x^{T}-x_c^{T}| < delta$, we have the following
So
$$(x^{T}-x_c^{T})y leq |x^{T}-x_c^{T}||y| < delta |y|$$
So $x_c^{T}y < delta |y|+x_c^{T}y$ for all $y$ in $A$.
Take sup over $y$
$$sup_{y in A} x_c^{T}y < sup_{y in A} (delta |y|+x_c^{T}y) leq sup_{y in A} delta |y|+ sup_{y in A}x_c^{T}y$$
So
$$S_A(x) < delta M + S_A(x_c)$$
Do the same for $(x_c^{T}-x^{T})y$ to get
$$S_A(x_c) < delta M + S_A(x)$$
Combine them to get the following
$|S_A(x)-S_A(x_c)|<delta M=epsilon$.
answered 23 hours ago
Sepide
1227
Using Cauchy-Schwarz inequality we can bound $(x^{T}-x_c^{T})y$
$(x^{T}-x_c^{T})yleq |x^{T}-x_c^{T}||y| $
Since $|x^{T}-x_c^{T}| < delta$, we have the following
So
$$(x^{T}-x_c^{T})y leq |x^{T}-x_c^{T}||y| < delta |y|$$
So $x_c^{T}y < delta |y|+x_c^{T}y$ for all $y$ in $A$.
Take sup over $y$
$$sup_{y in A} x_c^{T}y < sup_{y in A} (delta |y|+x_c^{T}y) leq sup_{y in A} delta |y|+ sup_{y in A}x_c^{T}y$$
So
$$S_A(x) < delta M + S_A(x_c)$$
Do the same for $(x_c^{T}-x^{T})y$ to get
$$S_A(x_c) < delta M + S_A(x)$$
Combine them to get the following
$|S_A(x)-S_A(x_c)|<delta M=epsilon$.
answered 23 hours ago
Sepide
1227
answered 23 hours ago
Sepide
1227
answered 23 hours ago
Sepide
1227
answered 23 hours ago
Sepide
1227
1227
add a comment |
add a comment |
up vote
0
down vote
Here's a solution which reduces to another problem on this site: Supremum is continuous over equicontinuous family of functions
Step 1: define for each $ain A$ the function $f_a(x) = x cdot a$. Each of these is continuous.
Step 2: Show that $S_A$ is the supremum of the family ${f_a}$
Step 3: Show that because $A$ is bounded, the family is not only continuous but equicontinuous.
Then conclude using the link above!
answered Nov 8 at 21:20
user25959
1,016714
I want to proof it in a way that I explained. Could you help me to do that.
– Saeed
Nov 8 at 21:25
You will need to do an $epsilon/3$-style argument. Basically rewrite $|S_A(x)-S_A(x_c)|$ as $|S_A(x)-x cdot y +x cdot y -x_ccdot y +x_ccdot y -S_A(x_c)|$ for some $y$ which almost attains the supremum $S_A(x_c)$ and use the triangle inequality and equicontinuity.
– user25959
Nov 8 at 22:42
Let me I change your $y$ to $z$. Then how can I treat $|sup_{y in A}x^Ty-x^Tz|$?
– Saeed
Nov 8 at 23:57
add a comment |
up vote
0
down vote
Here's a solution which reduces to another problem on this site: Supremum is continuous over equicontinuous family of functions
Step 1: define for each $ain A$ the function $f_a(x) = x cdot a$. Each of these is continuous.
Step 2: Show that $S_A$ is the supremum of the family ${f_a}$
Step 3: Show that because $A$ is bounded, the family is not only continuous but equicontinuous.
Then conclude using the link above!
answered Nov 8 at 21:20
user25959
1,016714
I want to proof it in a way that I explained. Could you help me to do that.
– Saeed
Nov 8 at 21:25
You will need to do an $epsilon/3$-style argument. Basically rewrite $|S_A(x)-S_A(x_c)|$ as $|S_A(x)-x cdot y +x cdot y -x_ccdot y +x_ccdot y -S_A(x_c)|$ for some $y$ which almost attains the supremum $S_A(x_c)$ and use the triangle inequality and equicontinuity.
– user25959
Nov 8 at 22:42
Let me I change your $y$ to $z$. Then how can I treat $|sup_{y in A}x^Ty-x^Tz|$?
– Saeed
Nov 8 at 23:57
add a comment |
up vote
0
down vote
up vote
0
down vote
Here's a solution which reduces to another problem on this site: Supremum is continuous over equicontinuous family of functions
Step 1: define for each $ain A$ the function $f_a(x) = x cdot a$. Each of these is continuous.
Step 2: Show that $S_A$ is the supremum of the family ${f_a}$
Step 3: Show that because $A$ is bounded, the family is not only continuous but equicontinuous.
Then conclude using the link above!
answered Nov 8 at 21:20
user25959
1,016714
Here's a solution which reduces to another problem on this site: Supremum is continuous over equicontinuous family of functions
Step 1: define for each $ain A$ the function $f_a(x) = x cdot a$. Each of these is continuous.
Step 2: Show that $S_A$ is the supremum of the family ${f_a}$
Step 3: Show that because $A$ is bounded, the family is not only continuous but equicontinuous.
Then conclude using the link above!
answered Nov 8 at 21:20
user25959
1,016714
answered Nov 8 at 21:20
user25959
1,016714
answered Nov 8 at 21:20
user25959
1,016714
answered Nov 8 at 21:20
user25959
1,016714
1,016714
I want to proof it in a way that I explained. Could you help me to do that.
– Saeed
Nov 8 at 21:25
You will need to do an $epsilon/3$-style argument. Basically rewrite $|S_A(x)-S_A(x_c)|$ as $|S_A(x)-x cdot y +x cdot y -x_ccdot y +x_ccdot y -S_A(x_c)|$ for some $y$ which almost attains the supremum $S_A(x_c)$ and use the triangle inequality and equicontinuity.
– user25959
Nov 8 at 22:42
Let me I change your $y$ to $z$. Then how can I treat $|sup_{y in A}x^Ty-x^Tz|$?
– Saeed
Nov 8 at 23:57
add a comment |
I want to proof it in a way that I explained. Could you help me to do that.
– Saeed
Nov 8 at 21:25
You will need to do an $epsilon/3$-style argument. Basically rewrite $|S_A(x)-S_A(x_c)|$ as $|S_A(x)-x cdot y +x cdot y -x_ccdot y +x_ccdot y -S_A(x_c)|$ for some $y$ which almost attains the supremum $S_A(x_c)$ and use the triangle inequality and equicontinuity.
– user25959
Nov 8 at 22:42
Let me I change your $y$ to $z$. Then how can I treat $|sup_{y in A}x^Ty-x^Tz|$?
– Saeed
Nov 8 at 23:57
I want to proof it in a way that I explained. Could you help me to do that.
– Saeed
Nov 8 at 21:25
I want to proof it in a way that I explained. Could you help me to do that.
– Saeed
Nov 8 at 21:25
You will need to do an $epsilon/3$-style argument. Basically rewrite $|S_A(x)-S_A(x_c)|$ as $|S_A(x)-x cdot y +x cdot y -x_ccdot y +x_ccdot y -S_A(x_c)|$ for some $y$ which almost attains the supremum $S_A(x_c)$ and use the triangle inequality and equicontinuity.
– user25959
Nov 8 at 22:42
You will need to do an $epsilon/3$-style argument. Basically rewrite $|S_A(x)-S_A(x_c)|$ as $|S_A(x)-x cdot y +x cdot y -x_ccdot y +x_ccdot y -S_A(x_c)|$ for some $y$ which almost attains the supremum $S_A(x_c)$ and use the triangle inequality and equicontinuity.
– user25959
Nov 8 at 22:42
Let me I change your $y$ to $z$. Then how can I treat $|sup_{y in A}x^Ty-x^Tz|$?
– Saeed
Nov 8 at 23:57
Let me I change your $y$ to $z$. Then how can I treat $|sup_{y in A}x^Ty-x^Tz|$?
– Saeed
Nov 8 at 23:57
add a comment |
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