Effective Domain Convex Function
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0
$begingroup$
Let $f:mathbb{R}^nrightarrow mathbb{R}cup {pm infty}$ and let $S(f):={x ∈R^n | f(x) < +∞}$ (effective domain)
How can I prove that $S(f)$ is convex doesn't imply that $f$ is convex?
Attempt:
I'm thinking of showing that $S(f)$ can vary depending whether $f(x)$ is a proper or improper function?
optimization convex-optimization
edited Jan 13 at 14:05
Yanko
7,8801830
asked Jan 13 at 13:39
mgrmgr
246
$endgroup$
|
show 6 more comments
0
$begingroup$
Let $f:mathbb{R}^nrightarrow mathbb{R}cup {pm infty}$ and let $S(f):={x ∈R^n | f(x) < +∞}$ (effective domain)
How can I prove that $S(f)$ is convex doesn't imply that $f$ is convex?
Attempt:
I'm thinking of showing that $S(f)$ can vary depending whether $f(x)$ is a proper or improper function?
optimization convex-optimization
edited Jan 13 at 14:05
Yanko
7,8801830
asked Jan 13 at 13:39
mgrmgr
246
$endgroup$
$begingroup$
Can you give an example where $S(f)not = mathbb{R}^n$?
$endgroup$
– Yanko
Jan 13 at 13:41
$begingroup$
I'm assuming f(x) is always a real number or $pm∞$ so, no?
$endgroup$
– mgr
Jan 13 at 13:45
$begingroup$
You should mention that $f:mathbb{R}^nrightarrow mathbb{R}cup {pm infty}$. Also please explain how $S(f)$ is a subset of $P(f)$, they have different dimensions...
$endgroup$
– Yanko
Jan 13 at 13:47
$begingroup$
You're right. My bad, edited. Ignore my assumptions; only assumed that $S(f)$ is a subset of $P(f)$ on $R^n$
$endgroup$
– mgr
Jan 13 at 13:52
2
$begingroup$
The solution is not that hard now. If $f$ never get the value infinity then $S(f)=mathbb{R}^n$ (which is convex). So just find a function $f:mathbb{R}^nrightarrowmathbb{R}$ which is not convex.
$endgroup$
– Yanko
Jan 13 at 14:08
|
show 6 more comments
0
0
0
$begingroup$
Let $f:mathbb{R}^nrightarrow mathbb{R}cup {pm infty}$ and let $S(f):={x ∈R^n | f(x) < +∞}$ (effective domain)
How can I prove that $S(f)$ is convex doesn't imply that $f$ is convex?
Attempt:
I'm thinking of showing that $S(f)$ can vary depending whether $f(x)$ is a proper or improper function?
optimization convex-optimization
edited Jan 13 at 14:05
Yanko
7,8801830
asked Jan 13 at 13:39
mgrmgr
246
$endgroup$
Let $f:mathbb{R}^nrightarrow mathbb{R}cup {pm infty}$ and let $S(f):={x ∈R^n | f(x) < +∞}$ (effective domain)
How can I prove that $S(f)$ is convex doesn't imply that $f$ is convex?
Attempt:
I'm thinking of showing that $S(f)$ can vary depending whether $f(x)$ is a proper or improper function?
optimization convex-optimization
optimization convex-optimization
edited Jan 13 at 14:05
Yanko
7,8801830
asked Jan 13 at 13:39
mgrmgr
246
edited Jan 13 at 14:05
Yanko
7,8801830
asked Jan 13 at 13:39
mgrmgr
246
edited Jan 13 at 14:05
Yanko
7,8801830
edited Jan 13 at 14:05
Yanko
7,8801830
edited Jan 13 at 14:05
Yanko
7,8801830
7,8801830
asked Jan 13 at 13:39
mgrmgr
246
asked Jan 13 at 13:39
mgrmgr
246
asked Jan 13 at 13:39
mgrmgr
246
246
$begingroup$
Can you give an example where $S(f)not = mathbb{R}^n$?
$endgroup$
– Yanko
Jan 13 at 13:41
$begingroup$
I'm assuming f(x) is always a real number or $pm∞$ so, no?
$endgroup$
– mgr
Jan 13 at 13:45
$begingroup$
You should mention that $f:mathbb{R}^nrightarrow mathbb{R}cup {pm infty}$. Also please explain how $S(f)$ is a subset of $P(f)$, they have different dimensions...
$endgroup$
– Yanko
Jan 13 at 13:47
$begingroup$
You're right. My bad, edited. Ignore my assumptions; only assumed that $S(f)$ is a subset of $P(f)$ on $R^n$
$endgroup$
– mgr
Jan 13 at 13:52
2
$begingroup$
The solution is not that hard now. If $f$ never get the value infinity then $S(f)=mathbb{R}^n$ (which is convex). So just find a function $f:mathbb{R}^nrightarrowmathbb{R}$ which is not convex.
$endgroup$
– Yanko
Jan 13 at 14:08
|
show 6 more comments
$begingroup$
Can you give an example where $S(f)not = mathbb{R}^n$?
$endgroup$
– Yanko
Jan 13 at 13:41
$begingroup$
I'm assuming f(x) is always a real number or $pm∞$ so, no?
$endgroup$
– mgr
Jan 13 at 13:45
$begingroup$
You should mention that $f:mathbb{R}^nrightarrow mathbb{R}cup {pm infty}$. Also please explain how $S(f)$ is a subset of $P(f)$, they have different dimensions...
$endgroup$
– Yanko
Jan 13 at 13:47
$begingroup$
You're right. My bad, edited. Ignore my assumptions; only assumed that $S(f)$ is a subset of $P(f)$ on $R^n$
$endgroup$
– mgr
Jan 13 at 13:52
2
$begingroup$
The solution is not that hard now. If $f$ never get the value infinity then $S(f)=mathbb{R}^n$ (which is convex). So just find a function $f:mathbb{R}^nrightarrowmathbb{R}$ which is not convex.
$endgroup$
– Yanko
Jan 13 at 14:08
$begingroup$
Can you give an example where $S(f)not = mathbb{R}^n$?
$endgroup$
– Yanko
Jan 13 at 13:41
$begingroup$
Can you give an example where $S(f)not = mathbb{R}^n$?
$endgroup$
– Yanko
Jan 13 at 13:41
$begingroup$
I'm assuming f(x) is always a real number or $pm∞$ so, no?
$endgroup$
– mgr
Jan 13 at 13:45
$begingroup$
I'm assuming f(x) is always a real number or $pm∞$ so, no?
$endgroup$
– mgr
Jan 13 at 13:45
$begingroup$
You should mention that $f:mathbb{R}^nrightarrow mathbb{R}cup {pm infty}$. Also please explain how $S(f)$ is a subset of $P(f)$, they have different dimensions...
$endgroup$
– Yanko
Jan 13 at 13:47
$begingroup$
You should mention that $f:mathbb{R}^nrightarrow mathbb{R}cup {pm infty}$. Also please explain how $S(f)$ is a subset of $P(f)$, they have different dimensions...
$endgroup$
– Yanko
Jan 13 at 13:47
$begingroup$
You're right. My bad, edited. Ignore my assumptions; only assumed that $S(f)$ is a subset of $P(f)$ on $R^n$
$endgroup$
– mgr
Jan 13 at 13:52
$begingroup$
You're right. My bad, edited. Ignore my assumptions; only assumed that $S(f)$ is a subset of $P(f)$ on $R^n$
$endgroup$
– mgr
Jan 13 at 13:52
2
2
$begingroup$
The solution is not that hard now. If $f$ never get the value infinity then $S(f)=mathbb{R}^n$ (which is convex). So just find a function $f:mathbb{R}^nrightarrowmathbb{R}$ which is not convex.
$endgroup$
– Yanko
Jan 13 at 14:08
$begingroup$
The solution is not that hard now. If $f$ never get the value infinity then $S(f)=mathbb{R}^n$ (which is convex). So just find a function $f:mathbb{R}^nrightarrowmathbb{R}$ which is not convex.
$endgroup$
– Yanko
Jan 13 at 14:08
|
show 6 more comments
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$begingroup$
Can you give an example where $S(f)not = mathbb{R}^n$?
$endgroup$
– Yanko
Jan 13 at 13:41
$begingroup$
I'm assuming f(x) is always a real number or $pm∞$ so, no?
$endgroup$
– mgr
Jan 13 at 13:45
$begingroup$
You should mention that $f:mathbb{R}^nrightarrow mathbb{R}cup {pm infty}$. Also please explain how $S(f)$ is a subset of $P(f)$, they have different dimensions...
$endgroup$
– Yanko
Jan 13 at 13:47
$begingroup$
You're right. My bad, edited. Ignore my assumptions; only assumed that $S(f)$ is a subset of $P(f)$ on $R^n$
$endgroup$
– mgr
Jan 13 at 13:52
2
$begingroup$
The solution is not that hard now. If $f$ never get the value infinity then $S(f)=mathbb{R}^n$ (which is convex). So just find a function $f:mathbb{R}^nrightarrowmathbb{R}$ which is not convex.
$endgroup$
– Yanko
Jan 13 at 14:08