Proof lower level sets of continuous strictly convex function are closed
$begingroup$
I need to prove that the following set:
$$mathcal L:={xin mathbb{R}^n , : , f(x) leq f(x_0) }$$
which apparently is called lower level set, is compact. I've already proved boundedness and I just need to show closeness.
The function $f$ is twice continuously differentiable on all of $mathbb{R}^n$ and so it is continuous everywhere.
I am looking for a "simple" proof, i.e. one that uses the notion of continuity and the notion of closeness/openness, NOT the notion of semi-continuity, which is used in every single proof I've found.
My Attempt
$f$ is continuous on all of $mathbb{R}^n$ means that $forall xinmathbb{R}^n$:
$$forall epsilon >0, , exists delta > 0 , forall yin mathbb{R}^n: ||x - y|| < delta Longrightarrow ||f(x)-f(y)||< epsilon$$
On the other hand, $mathcal L$ is closed means that its complement wrt $mathbb{R}^n$ is open, that is:
$$forall xinbar{mathcal{L}}, exists epsilon > 0 , forall yin mathbb{R}^n , : , ||x - y||<epsilon Longrightarrow yin bar{mathcal{L}}$$
Now using continuity we know:
$$||x-y|| < delta Longrightarrow ||f(x)-f(y)||<epsilon Longrightarrow -epsilon < f(x) - f(y) < epsilon$$
And so $Longrightarrow f(y)-epsilon < f(x) < epsilon+f(y)$. How do I continue? Is this correct?
real-analysis calculus optimization
asked Dec 4 '18 at 18:57
Euler_SalterEuler_Salter
2,0671336
$endgroup$
add a comment |
$begingroup$
I need to prove that the following set:
$$mathcal L:={xin mathbb{R}^n , : , f(x) leq f(x_0) }$$
which apparently is called lower level set, is compact. I've already proved boundedness and I just need to show closeness.
The function $f$ is twice continuously differentiable on all of $mathbb{R}^n$ and so it is continuous everywhere.
I am looking for a "simple" proof, i.e. one that uses the notion of continuity and the notion of closeness/openness, NOT the notion of semi-continuity, which is used in every single proof I've found.
My Attempt
$f$ is continuous on all of $mathbb{R}^n$ means that $forall xinmathbb{R}^n$:
$$forall epsilon >0, , exists delta > 0 , forall yin mathbb{R}^n: ||x - y|| < delta Longrightarrow ||f(x)-f(y)||< epsilon$$
On the other hand, $mathcal L$ is closed means that its complement wrt $mathbb{R}^n$ is open, that is:
$$forall xinbar{mathcal{L}}, exists epsilon > 0 , forall yin mathbb{R}^n , : , ||x - y||<epsilon Longrightarrow yin bar{mathcal{L}}$$
Now using continuity we know:
$$||x-y|| < delta Longrightarrow ||f(x)-f(y)||<epsilon Longrightarrow -epsilon < f(x) - f(y) < epsilon$$
And so $Longrightarrow f(y)-epsilon < f(x) < epsilon+f(y)$. How do I continue? Is this correct?
real-analysis calculus optimization
asked Dec 4 '18 at 18:57
Euler_SalterEuler_Salter
2,0671336
$endgroup$
add a comment |
$begingroup$
I need to prove that the following set:
$$mathcal L:={xin mathbb{R}^n , : , f(x) leq f(x_0) }$$
which apparently is called lower level set, is compact. I've already proved boundedness and I just need to show closeness.
The function $f$ is twice continuously differentiable on all of $mathbb{R}^n$ and so it is continuous everywhere.
I am looking for a "simple" proof, i.e. one that uses the notion of continuity and the notion of closeness/openness, NOT the notion of semi-continuity, which is used in every single proof I've found.
My Attempt
$f$ is continuous on all of $mathbb{R}^n$ means that $forall xinmathbb{R}^n$:
$$forall epsilon >0, , exists delta > 0 , forall yin mathbb{R}^n: ||x - y|| < delta Longrightarrow ||f(x)-f(y)||< epsilon$$
On the other hand, $mathcal L$ is closed means that its complement wrt $mathbb{R}^n$ is open, that is:
$$forall xinbar{mathcal{L}}, exists epsilon > 0 , forall yin mathbb{R}^n , : , ||x - y||<epsilon Longrightarrow yin bar{mathcal{L}}$$
Now using continuity we know:
$$||x-y|| < delta Longrightarrow ||f(x)-f(y)||<epsilon Longrightarrow -epsilon < f(x) - f(y) < epsilon$$
And so $Longrightarrow f(y)-epsilon < f(x) < epsilon+f(y)$. How do I continue? Is this correct?
real-analysis calculus optimization
asked Dec 4 '18 at 18:57
Euler_SalterEuler_Salter
2,0671336
$endgroup$
I need to prove that the following set:
$$mathcal L:={xin mathbb{R}^n , : , f(x) leq f(x_0) }$$
which apparently is called lower level set, is compact. I've already proved boundedness and I just need to show closeness.
The function $f$ is twice continuously differentiable on all of $mathbb{R}^n$ and so it is continuous everywhere.
I am looking for a "simple" proof, i.e. one that uses the notion of continuity and the notion of closeness/openness, NOT the notion of semi-continuity, which is used in every single proof I've found.
My Attempt
$f$ is continuous on all of $mathbb{R}^n$ means that $forall xinmathbb{R}^n$:
$$forall epsilon >0, , exists delta > 0 , forall yin mathbb{R}^n: ||x - y|| < delta Longrightarrow ||f(x)-f(y)||< epsilon$$
On the other hand, $mathcal L$ is closed means that its complement wrt $mathbb{R}^n$ is open, that is:
$$forall xinbar{mathcal{L}}, exists epsilon > 0 , forall yin mathbb{R}^n , : , ||x - y||<epsilon Longrightarrow yin bar{mathcal{L}}$$
Now using continuity we know:
$$||x-y|| < delta Longrightarrow ||f(x)-f(y)||<epsilon Longrightarrow -epsilon < f(x) - f(y) < epsilon$$
And so $Longrightarrow f(y)-epsilon < f(x) < epsilon+f(y)$. How do I continue? Is this correct?
real-analysis calculus optimization
real-analysis calculus optimization
asked Dec 4 '18 at 18:57
Euler_SalterEuler_Salter
2,0671336
asked Dec 4 '18 at 18:57
Euler_SalterEuler_Salter
2,0671336
asked Dec 4 '18 at 18:57
Euler_SalterEuler_Salter
2,0671336
asked Dec 4 '18 at 18:57
Euler_SalterEuler_Salter
2,0671336
asked Dec 4 '18 at 18:57
Euler_SalterEuler_Salter
2,0671336
2,0671336
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If we define
$$
g(x)=f(x)-f(x_0)
$$
a continuous function, the set is
$$
g^{-1}((-infty,0])
$$
the preimage of a closed set under a continuous map.
If you are wedded to epsilon's and delta's, then pick $xin mathcal{L}^c$, then
$$
g(x)>0
$$
and for $epsilon=frac{g(x)}{2}$, we have for some $delta>0$ corresponding to this $epsilon$,
$$
g(y)>0
$$
for all $yin (x-delta,x+delta)$. So $(x-delta,x+delta)subset mathcal{L}^c$.
answered Dec 4 '18 at 19:03
qbertqbert
22.1k32561
$endgroup$
$begingroup$
Can you please expand a little bit more? It seems interesting but I’m struggling to understand
$endgroup$
– Euler_Salter
Dec 4 '18 at 19:06
$begingroup$
I added a solution using the epsilon-delta definition of continuity. In general in topology, a continuous function is one which has the property that preimages of open sets are open. This then implies that preimages of closed sets are closed.
$endgroup$
– qbert
Dec 4 '18 at 19:08
add a comment |
Your Answer
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If we define
$$
g(x)=f(x)-f(x_0)
$$
a continuous function, the set is
$$
g^{-1}((-infty,0])
$$
the preimage of a closed set under a continuous map.
If you are wedded to epsilon's and delta's, then pick $xin mathcal{L}^c$, then
$$
g(x)>0
$$
and for $epsilon=frac{g(x)}{2}$, we have for some $delta>0$ corresponding to this $epsilon$,
$$
g(y)>0
$$
for all $yin (x-delta,x+delta)$. So $(x-delta,x+delta)subset mathcal{L}^c$.
answered Dec 4 '18 at 19:03
qbertqbert
22.1k32561
$endgroup$
$begingroup$
Can you please expand a little bit more? It seems interesting but I’m struggling to understand
$endgroup$
– Euler_Salter
Dec 4 '18 at 19:06
$begingroup$
I added a solution using the epsilon-delta definition of continuity. In general in topology, a continuous function is one which has the property that preimages of open sets are open. This then implies that preimages of closed sets are closed.
$endgroup$
– qbert
Dec 4 '18 at 19:08
add a comment |
$begingroup$
If we define
$$
g(x)=f(x)-f(x_0)
$$
a continuous function, the set is
$$
g^{-1}((-infty,0])
$$
the preimage of a closed set under a continuous map.
If you are wedded to epsilon's and delta's, then pick $xin mathcal{L}^c$, then
$$
g(x)>0
$$
and for $epsilon=frac{g(x)}{2}$, we have for some $delta>0$ corresponding to this $epsilon$,
$$
g(y)>0
$$
for all $yin (x-delta,x+delta)$. So $(x-delta,x+delta)subset mathcal{L}^c$.
answered Dec 4 '18 at 19:03
qbertqbert
22.1k32561
$endgroup$
$begingroup$
Can you please expand a little bit more? It seems interesting but I’m struggling to understand
$endgroup$
– Euler_Salter
Dec 4 '18 at 19:06
$begingroup$
I added a solution using the epsilon-delta definition of continuity. In general in topology, a continuous function is one which has the property that preimages of open sets are open. This then implies that preimages of closed sets are closed.
$endgroup$
– qbert
Dec 4 '18 at 19:08
add a comment |
$begingroup$
If we define
$$
g(x)=f(x)-f(x_0)
$$
a continuous function, the set is
$$
g^{-1}((-infty,0])
$$
the preimage of a closed set under a continuous map.
If you are wedded to epsilon's and delta's, then pick $xin mathcal{L}^c$, then
$$
g(x)>0
$$
and for $epsilon=frac{g(x)}{2}$, we have for some $delta>0$ corresponding to this $epsilon$,
$$
g(y)>0
$$
for all $yin (x-delta,x+delta)$. So $(x-delta,x+delta)subset mathcal{L}^c$.
answered Dec 4 '18 at 19:03
qbertqbert
22.1k32561
$endgroup$
If we define
$$
g(x)=f(x)-f(x_0)
$$
a continuous function, the set is
$$
g^{-1}((-infty,0])
$$
the preimage of a closed set under a continuous map.
If you are wedded to epsilon's and delta's, then pick $xin mathcal{L}^c$, then
$$
g(x)>0
$$
and for $epsilon=frac{g(x)}{2}$, we have for some $delta>0$ corresponding to this $epsilon$,
$$
g(y)>0
$$
for all $yin (x-delta,x+delta)$. So $(x-delta,x+delta)subset mathcal{L}^c$.
answered Dec 4 '18 at 19:03
qbertqbert
22.1k32561
edited Dec 4 '18 at 19:07
answered Dec 4 '18 at 19:03
qbertqbert
22.1k32561
answered Dec 4 '18 at 19:03
qbertqbert
22.1k32561
answered Dec 4 '18 at 19:03
qbertqbert
22.1k32561
22.1k32561
$begingroup$
Can you please expand a little bit more? It seems interesting but I’m struggling to understand
$endgroup$
– Euler_Salter
Dec 4 '18 at 19:06
$begingroup$
I added a solution using the epsilon-delta definition of continuity. In general in topology, a continuous function is one which has the property that preimages of open sets are open. This then implies that preimages of closed sets are closed.
$endgroup$
– qbert
Dec 4 '18 at 19:08
add a comment |
$begingroup$
Can you please expand a little bit more? It seems interesting but I’m struggling to understand
$endgroup$
– Euler_Salter
Dec 4 '18 at 19:06
$begingroup$
I added a solution using the epsilon-delta definition of continuity. In general in topology, a continuous function is one which has the property that preimages of open sets are open. This then implies that preimages of closed sets are closed.
$endgroup$
– qbert
Dec 4 '18 at 19:08
$begingroup$
Can you please expand a little bit more? It seems interesting but I’m struggling to understand
$endgroup$
– Euler_Salter
Dec 4 '18 at 19:06
$begingroup$
Can you please expand a little bit more? It seems interesting but I’m struggling to understand
$endgroup$
– Euler_Salter
Dec 4 '18 at 19:06
$begingroup$
I added a solution using the epsilon-delta definition of continuity. In general in topology, a continuous function is one which has the property that preimages of open sets are open. This then implies that preimages of closed sets are closed.
$endgroup$
– qbert
Dec 4 '18 at 19:08
$begingroup$
I added a solution using the epsilon-delta definition of continuity. In general in topology, a continuous function is one which has the property that preimages of open sets are open. This then implies that preimages of closed sets are closed.
$endgroup$
– qbert
Dec 4 '18 at 19:08
add a comment |
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