A function defined on a closed set such that $f geq M$ reach a minimumIf $x$ is not in $A$, a closed set in a Metric space then $d(x,A)>0$Let $X$ be a compact metric space. If $f:Xrightarrow mathbbR$ is lower semi-continuous, then $f$ is bounded from below and attains its infimum.A metric space such that all closed balls are compact is complete.Prove that $f$ reach an absolute minimumExistence of minimum on unbounded set and functionCoercive continuous function on a closed subset has a global minimum proofdoes a convex polynomial always reach its minimum value?Existence of minimum distance from a point to a sequentially compact set?If X is compact and Y is closed. Show that there exists $x_0 in X$, and $y_0 in Y$ such that $|x_0-y_0| leq |x-y|$ for all $x in X$, $y in Y$Show that $F = lambda x$ is closed where $K$ is a compact set.
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A function defined on a closed set such that $f geq M$ reach a minimum
If $x$ is not in $A$, a closed set in a Metric space then $d(x,A)>0$Let $X$ be a compact metric space. If $f:Xrightarrow mathbbR$ is lower semi-continuous, then $f$ is bounded from below and attains its infimum.A metric space such that all closed balls are compact is complete.Prove that $f$ reach an absolute minimumExistence of minimum on unbounded set and functionCoercive continuous function on a closed subset has a global minimum proofdoes a convex polynomial always reach its minimum value?Existence of minimum distance from a point to a sequentially compact set?If X is compact and Y is closed. Show that there exists $x_0 in X$, and $y_0 in Y$ such that $|x_0-y_0| leq |x-y|$ for all $x in X$, $y in Y$Show that $F = lambda x$ is closed where $K$ is a compact set.
$begingroup$
I am wondering if the following is true :
Let $f : C to mathbbR^p$ be a continuous function where $C$ is a closed set of $mathbbR^n$. Moreover there is $M$ such that for all $x in C, |f(x)| geq M$. Then does $f$ reach a minimum ?
I know that this with the assumption $C$ is a compact set. Yet here $C$ is only closed. So if we denote $ mu = inf $ then I need to find $alpha in C$ such that : $|f(alpha) | = mu$.
So there is a sequence $(|f(x_n)|)_n$ which converges to $mu$. So there is a sequence $(x_n)$ such that $| f(x_n) |$ converges to $mu$. Yet the problem here is that the sequence $(x_n)$ might not converges...
Thank you !
real-analysis sequences-and-series general-topology
asked Mar 10 at 19:57
bonjoufhajlbonjoufhajl
324
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add a comment |
$begingroup$
I am wondering if the following is true :
Let $f : C to mathbbR^p$ be a continuous function where $C$ is a closed set of $mathbbR^n$. Moreover there is $M$ such that for all $x in C, |f(x)| geq M$. Then does $f$ reach a minimum ?
I know that this with the assumption $C$ is a compact set. Yet here $C$ is only closed. So if we denote $ mu = inf $ then I need to find $alpha in C$ such that : $|f(alpha) | = mu$.
So there is a sequence $(|f(x_n)|)_n$ which converges to $mu$. So there is a sequence $(x_n)$ such that $| f(x_n) |$ converges to $mu$. Yet the problem here is that the sequence $(x_n)$ might not converges...
Thank you !
real-analysis sequences-and-series general-topology
asked Mar 10 at 19:57
bonjoufhajlbonjoufhajl
324
New contributor
bonjoufhajl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
$f$ can't reach a minimum nor maximum (at least for $p>1$), I think you mean if $||f||$ reaches it's minimum
$endgroup$
– Jakobian
Mar 10 at 20:00
add a comment |
$begingroup$
I am wondering if the following is true :
Let $f : C to mathbbR^p$ be a continuous function where $C$ is a closed set of $mathbbR^n$. Moreover there is $M$ such that for all $x in C, |f(x)| geq M$. Then does $f$ reach a minimum ?
I know that this with the assumption $C$ is a compact set. Yet here $C$ is only closed. So if we denote $ mu = inf $ then I need to find $alpha in C$ such that : $|f(alpha) | = mu$.
So there is a sequence $(|f(x_n)|)_n$ which converges to $mu$. So there is a sequence $(x_n)$ such that $| f(x_n) |$ converges to $mu$. Yet the problem here is that the sequence $(x_n)$ might not converges...
Thank you !
real-analysis sequences-and-series general-topology
asked Mar 10 at 19:57
bonjoufhajlbonjoufhajl
324
New contributor
bonjoufhajl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am wondering if the following is true :
Let $f : C to mathbbR^p$ be a continuous function where $C$ is a closed set of $mathbbR^n$. Moreover there is $M$ such that for all $x in C, |f(x)| geq M$. Then does $f$ reach a minimum ?
I know that this with the assumption $C$ is a compact set. Yet here $C$ is only closed. So if we denote $ mu = inf $ then I need to find $alpha in C$ such that : $|f(alpha) | = mu$.
So there is a sequence $(|f(x_n)|)_n$ which converges to $mu$. So there is a sequence $(x_n)$ such that $| f(x_n) |$ converges to $mu$. Yet the problem here is that the sequence $(x_n)$ might not converges...
Thank you !
real-analysis sequences-and-series general-topology
real-analysis sequences-and-series general-topology
asked Mar 10 at 19:57
bonjoufhajlbonjoufhajl
324
New contributor
bonjoufhajl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 19:57
bonjoufhajlbonjoufhajl
324
New contributor
bonjoufhajl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 19:57
bonjoufhajlbonjoufhajl
324
New contributor
bonjoufhajl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Mar 10 at 19:57
bonjoufhajlbonjoufhajl
324
asked Mar 10 at 19:57
bonjoufhajlbonjoufhajl
324
324
New contributor
bonjoufhajl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
bonjoufhajl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.
$begingroup$
$f$ can't reach a minimum nor maximum (at least for $p>1$), I think you mean if $||f||$ reaches it's minimum
$endgroup$
– Jakobian
Mar 10 at 20:00
add a comment |
$begingroup$
$f$ can't reach a minimum nor maximum (at least for $p>1$), I think you mean if $||f||$ reaches it's minimum
$endgroup$
– Jakobian
Mar 10 at 20:00
$begingroup$
$f$ can't reach a minimum nor maximum (at least for $p>1$), I think you mean if $||f||$ reaches it's minimum
$endgroup$
– Jakobian
Mar 10 at 20:00
$begingroup$
$f$ can't reach a minimum nor maximum (at least for $p>1$), I think you mean if $||f||$ reaches it's minimum
$endgroup$
– Jakobian
Mar 10 at 20:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The answer is no.
For example, the function $f : mathbbR rightarrow mathbbR$ defined by $f(x) = exp(x)$ is defined on a closed subset of $mathbbR$ ($mathbbR$ itself), is continuous, is always greater than $0$, but it does not reach a minimum.
Moreover, notice that the asumption $||f(x)|| geq M$ is useless because it is always true with $M=0$.
answered Mar 10 at 20:00
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
$begingroup$
Nice catch about $|f(x) | geq M$ being redundant! I suppose a better question would be to set $f:C to mathbbR$ and to let $f$ be bounded (for which of course our answers work the same).
$endgroup$
– Kezer
Mar 10 at 20:05
add a comment |
$begingroup$
No. Choose $mathbbR to mathbbR, x mapsto e^x$. It's bounded by $0$ but does not reach a minimum.
answered Mar 10 at 19:59
KezerKezer
1,458621
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
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active
oldest
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$begingroup$
The answer is no.
For example, the function $f : mathbbR rightarrow mathbbR$ defined by $f(x) = exp(x)$ is defined on a closed subset of $mathbbR$ ($mathbbR$ itself), is continuous, is always greater than $0$, but it does not reach a minimum.
Moreover, notice that the asumption $||f(x)|| geq M$ is useless because it is always true with $M=0$.
answered Mar 10 at 20:00
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
$begingroup$
Nice catch about $|f(x) | geq M$ being redundant! I suppose a better question would be to set $f:C to mathbbR$ and to let $f$ be bounded (for which of course our answers work the same).
$endgroup$
– Kezer
Mar 10 at 20:05
add a comment |
$begingroup$
The answer is no.
For example, the function $f : mathbbR rightarrow mathbbR$ defined by $f(x) = exp(x)$ is defined on a closed subset of $mathbbR$ ($mathbbR$ itself), is continuous, is always greater than $0$, but it does not reach a minimum.
Moreover, notice that the asumption $||f(x)|| geq M$ is useless because it is always true with $M=0$.
answered Mar 10 at 20:00
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
$begingroup$
Nice catch about $|f(x) | geq M$ being redundant! I suppose a better question would be to set $f:C to mathbbR$ and to let $f$ be bounded (for which of course our answers work the same).
$endgroup$
– Kezer
Mar 10 at 20:05
add a comment |
$begingroup$
The answer is no.
For example, the function $f : mathbbR rightarrow mathbbR$ defined by $f(x) = exp(x)$ is defined on a closed subset of $mathbbR$ ($mathbbR$ itself), is continuous, is always greater than $0$, but it does not reach a minimum.
Moreover, notice that the asumption $||f(x)|| geq M$ is useless because it is always true with $M=0$.
answered Mar 10 at 20:00
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
The answer is no.
For example, the function $f : mathbbR rightarrow mathbbR$ defined by $f(x) = exp(x)$ is defined on a closed subset of $mathbbR$ ($mathbbR$ itself), is continuous, is always greater than $0$, but it does not reach a minimum.
Moreover, notice that the asumption $||f(x)|| geq M$ is useless because it is always true with $M=0$.
answered Mar 10 at 20:00
TheSilverDoeTheSilverDoe
3,837112
answered Mar 10 at 20:00
TheSilverDoeTheSilverDoe
3,837112
answered Mar 10 at 20:00
TheSilverDoeTheSilverDoe
3,837112
answered Mar 10 at 20:00
TheSilverDoeTheSilverDoe
3,837112
3,837112
$begingroup$
Nice catch about $|f(x) | geq M$ being redundant! I suppose a better question would be to set $f:C to mathbbR$ and to let $f$ be bounded (for which of course our answers work the same).
$endgroup$
– Kezer
Mar 10 at 20:05
add a comment |
$begingroup$
Nice catch about $|f(x) | geq M$ being redundant! I suppose a better question would be to set $f:C to mathbbR$ and to let $f$ be bounded (for which of course our answers work the same).
$endgroup$
– Kezer
Mar 10 at 20:05
$begingroup$
Nice catch about $|f(x) | geq M$ being redundant! I suppose a better question would be to set $f:C to mathbbR$ and to let $f$ be bounded (for which of course our answers work the same).
$endgroup$
– Kezer
Mar 10 at 20:05
$begingroup$
Nice catch about $|f(x) | geq M$ being redundant! I suppose a better question would be to set $f:C to mathbbR$ and to let $f$ be bounded (for which of course our answers work the same).
$endgroup$
– Kezer
Mar 10 at 20:05
add a comment |
$begingroup$
No. Choose $mathbbR to mathbbR, x mapsto e^x$. It's bounded by $0$ but does not reach a minimum.
answered Mar 10 at 19:59
KezerKezer
1,458621
$endgroup$
add a comment |
$begingroup$
No. Choose $mathbbR to mathbbR, x mapsto e^x$. It's bounded by $0$ but does not reach a minimum.
answered Mar 10 at 19:59
KezerKezer
1,458621
$endgroup$
add a comment |
$begingroup$
No. Choose $mathbbR to mathbbR, x mapsto e^x$. It's bounded by $0$ but does not reach a minimum.
answered Mar 10 at 19:59
KezerKezer
1,458621
$endgroup$
No. Choose $mathbbR to mathbbR, x mapsto e^x$. It's bounded by $0$ but does not reach a minimum.
answered Mar 10 at 19:59
KezerKezer
1,458621
answered Mar 10 at 19:59
KezerKezer
1,458621
answered Mar 10 at 19:59
KezerKezer
1,458621
answered Mar 10 at 19:59
KezerKezer
1,458621
1,458621
add a comment |
add a comment |
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$begingroup$
$f$ can't reach a minimum nor maximum (at least for $p>1$), I think you mean if $||f||$ reaches it's minimum
$endgroup$
– Jakobian
Mar 10 at 20:00