Showing that a function has a minimum on a non compact interval
$begingroup$
The question states: Let $f: (-1,1) rightarrow R$ be cts and suppose $lim_{x to -1} f(x) = lim_{x to 1} f(x) = infty$. Show that $f$ has a minimum on $(-1,1)$
My attempt:
Given $lim_{x to -1} = infty$, we have $f(x) > M$ for all $M$ when $x > N_{1}$. Similarly, $lim_{x to 1} f(x) = infty$ implies that $f(x)>M$ for all $M$ when $x>N_{2}$.
Since $f$ is cts, it has a minimum on all compact intervals by the Maximum Theorem. Take the interval $[N_{1},N_{2}] = I$, then there exists an L s.t. $f(x) geq L$ for all $x in I$. If x $notin I$, take $M = L$. So we have $f(x) geq L$ for all $x$ which completes the proof.
Is my logic correct here?
real-analysis limits
asked Dec 12 '18 at 6:28
hkj447hkj447
334
$endgroup$
add a comment |
$begingroup$
The question states: Let $f: (-1,1) rightarrow R$ be cts and suppose $lim_{x to -1} f(x) = lim_{x to 1} f(x) = infty$. Show that $f$ has a minimum on $(-1,1)$
My attempt:
Given $lim_{x to -1} = infty$, we have $f(x) > M$ for all $M$ when $x > N_{1}$. Similarly, $lim_{x to 1} f(x) = infty$ implies that $f(x)>M$ for all $M$ when $x>N_{2}$.
Since $f$ is cts, it has a minimum on all compact intervals by the Maximum Theorem. Take the interval $[N_{1},N_{2}] = I$, then there exists an L s.t. $f(x) geq L$ for all $x in I$. If x $notin I$, take $M = L$. So we have $f(x) geq L$ for all $x$ which completes the proof.
Is my logic correct here?
real-analysis limits
asked Dec 12 '18 at 6:28
hkj447hkj447
334
$endgroup$
$begingroup$
You never say who $N_1$ and $N_2$ are; that paragraph is wrong, in part because you seem to be confusing limits at infinity with limits go to infinity. It is also wrong that “$f(x)>M$ for all $M$ when...” because that would require $f(x)$ to be larger than all real numbers, and no such number exists.
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:32
$begingroup$
Order matters. What you meant in that paragraph is: for each $M>0$ there exists $epsilon_{1,M}>0$ such that if $-1< x < -1+epsilon_{1,M}$, then $f(x)>M$. Similarly, for each $M>0$ there exists $epsilon_{2,M}>0$ such that if $1-epsilon_{2,M}<x<1$, then $f(x)>M$.”
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:34
$begingroup$
(Order in that you first specify $M$, then you specify the conditions on $x$, not the other way around). The basic idea can be made to work, but since you never specify what $N_1$ and $N_2$ are, and you don’t place them correctly inside of $(-1,1)$, what you write does not actually work. In addition, your $L$ is only asserted to be a lower bound, so you have not proven that it has a minimum, just that it is bounded below.
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:36
$begingroup$
Ah ok that makes sense, thank you.
$endgroup$
– hkj447
Dec 12 '18 at 7:02
add a comment |
$begingroup$
The question states: Let $f: (-1,1) rightarrow R$ be cts and suppose $lim_{x to -1} f(x) = lim_{x to 1} f(x) = infty$. Show that $f$ has a minimum on $(-1,1)$
My attempt:
Given $lim_{x to -1} = infty$, we have $f(x) > M$ for all $M$ when $x > N_{1}$. Similarly, $lim_{x to 1} f(x) = infty$ implies that $f(x)>M$ for all $M$ when $x>N_{2}$.
Since $f$ is cts, it has a minimum on all compact intervals by the Maximum Theorem. Take the interval $[N_{1},N_{2}] = I$, then there exists an L s.t. $f(x) geq L$ for all $x in I$. If x $notin I$, take $M = L$. So we have $f(x) geq L$ for all $x$ which completes the proof.
Is my logic correct here?
real-analysis limits
asked Dec 12 '18 at 6:28
hkj447hkj447
334
$endgroup$
The question states: Let $f: (-1,1) rightarrow R$ be cts and suppose $lim_{x to -1} f(x) = lim_{x to 1} f(x) = infty$. Show that $f$ has a minimum on $(-1,1)$
My attempt:
Given $lim_{x to -1} = infty$, we have $f(x) > M$ for all $M$ when $x > N_{1}$. Similarly, $lim_{x to 1} f(x) = infty$ implies that $f(x)>M$ for all $M$ when $x>N_{2}$.
Since $f$ is cts, it has a minimum on all compact intervals by the Maximum Theorem. Take the interval $[N_{1},N_{2}] = I$, then there exists an L s.t. $f(x) geq L$ for all $x in I$. If x $notin I$, take $M = L$. So we have $f(x) geq L$ for all $x$ which completes the proof.
Is my logic correct here?
real-analysis limits
real-analysis limits
asked Dec 12 '18 at 6:28
hkj447hkj447
334
asked Dec 12 '18 at 6:28
hkj447hkj447
334
asked Dec 12 '18 at 6:28
hkj447hkj447
334
asked Dec 12 '18 at 6:28
hkj447hkj447
334
asked Dec 12 '18 at 6:28
hkj447hkj447
334
334
$begingroup$
You never say who $N_1$ and $N_2$ are; that paragraph is wrong, in part because you seem to be confusing limits at infinity with limits go to infinity. It is also wrong that “$f(x)>M$ for all $M$ when...” because that would require $f(x)$ to be larger than all real numbers, and no such number exists.
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:32
$begingroup$
Order matters. What you meant in that paragraph is: for each $M>0$ there exists $epsilon_{1,M}>0$ such that if $-1< x < -1+epsilon_{1,M}$, then $f(x)>M$. Similarly, for each $M>0$ there exists $epsilon_{2,M}>0$ such that if $1-epsilon_{2,M}<x<1$, then $f(x)>M$.”
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:34
$begingroup$
(Order in that you first specify $M$, then you specify the conditions on $x$, not the other way around). The basic idea can be made to work, but since you never specify what $N_1$ and $N_2$ are, and you don’t place them correctly inside of $(-1,1)$, what you write does not actually work. In addition, your $L$ is only asserted to be a lower bound, so you have not proven that it has a minimum, just that it is bounded below.
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:36
$begingroup$
Ah ok that makes sense, thank you.
$endgroup$
– hkj447
Dec 12 '18 at 7:02
add a comment |
$begingroup$
You never say who $N_1$ and $N_2$ are; that paragraph is wrong, in part because you seem to be confusing limits at infinity with limits go to infinity. It is also wrong that “$f(x)>M$ for all $M$ when...” because that would require $f(x)$ to be larger than all real numbers, and no such number exists.
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:32
$begingroup$
Order matters. What you meant in that paragraph is: for each $M>0$ there exists $epsilon_{1,M}>0$ such that if $-1< x < -1+epsilon_{1,M}$, then $f(x)>M$. Similarly, for each $M>0$ there exists $epsilon_{2,M}>0$ such that if $1-epsilon_{2,M}<x<1$, then $f(x)>M$.”
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:34
$begingroup$
(Order in that you first specify $M$, then you specify the conditions on $x$, not the other way around). The basic idea can be made to work, but since you never specify what $N_1$ and $N_2$ are, and you don’t place them correctly inside of $(-1,1)$, what you write does not actually work. In addition, your $L$ is only asserted to be a lower bound, so you have not proven that it has a minimum, just that it is bounded below.
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:36
$begingroup$
Ah ok that makes sense, thank you.
$endgroup$
– hkj447
Dec 12 '18 at 7:02
$begingroup$
You never say who $N_1$ and $N_2$ are; that paragraph is wrong, in part because you seem to be confusing limits at infinity with limits go to infinity. It is also wrong that “$f(x)>M$ for all $M$ when...” because that would require $f(x)$ to be larger than all real numbers, and no such number exists.
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:32
$begingroup$
You never say who $N_1$ and $N_2$ are; that paragraph is wrong, in part because you seem to be confusing limits at infinity with limits go to infinity. It is also wrong that “$f(x)>M$ for all $M$ when...” because that would require $f(x)$ to be larger than all real numbers, and no such number exists.
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:32
$begingroup$
Order matters. What you meant in that paragraph is: for each $M>0$ there exists $epsilon_{1,M}>0$ such that if $-1< x < -1+epsilon_{1,M}$, then $f(x)>M$. Similarly, for each $M>0$ there exists $epsilon_{2,M}>0$ such that if $1-epsilon_{2,M}<x<1$, then $f(x)>M$.”
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:34
$begingroup$
Order matters. What you meant in that paragraph is: for each $M>0$ there exists $epsilon_{1,M}>0$ such that if $-1< x < -1+epsilon_{1,M}$, then $f(x)>M$. Similarly, for each $M>0$ there exists $epsilon_{2,M}>0$ such that if $1-epsilon_{2,M}<x<1$, then $f(x)>M$.”
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:34
$begingroup$
(Order in that you first specify $M$, then you specify the conditions on $x$, not the other way around). The basic idea can be made to work, but since you never specify what $N_1$ and $N_2$ are, and you don’t place them correctly inside of $(-1,1)$, what you write does not actually work. In addition, your $L$ is only asserted to be a lower bound, so you have not proven that it has a minimum, just that it is bounded below.
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:36
$begingroup$
(Order in that you first specify $M$, then you specify the conditions on $x$, not the other way around). The basic idea can be made to work, but since you never specify what $N_1$ and $N_2$ are, and you don’t place them correctly inside of $(-1,1)$, what you write does not actually work. In addition, your $L$ is only asserted to be a lower bound, so you have not proven that it has a minimum, just that it is bounded below.
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:36
$begingroup$
Ah ok that makes sense, thank you.
$endgroup$
– hkj447
Dec 12 '18 at 7:02
$begingroup$
Ah ok that makes sense, thank you.
$endgroup$
– hkj447
Dec 12 '18 at 7:02
add a comment |
1 Answer
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$begingroup$
Your argument does not make much sense. What do mean by $x >N_1$ and $x >N_2$. You are not taking limits as $ x to 1$and $x to -1$.
Here is a valid argument: let $alpha =inf {f(x): -1<x<1}$. Note that $alpha leq f(0) <infty$. The hypothesis tells you that if $delta in (0,1)$ is sufficiently small then $f(x) >alpha$ when $x in (-1,-1+delta)$ and also when $x in (1-delta, 1)$. Now I leave it to you to verify that the minimum of $f$ on the compact interval $[-1+delta, 1-delta]$ is also the minimum of $f$ on $(-1,1)$.
answered Dec 12 '18 at 6:36
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
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add a comment |
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1 Answer
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$begingroup$
Your argument does not make much sense. What do mean by $x >N_1$ and $x >N_2$. You are not taking limits as $ x to 1$and $x to -1$.
Here is a valid argument: let $alpha =inf {f(x): -1<x<1}$. Note that $alpha leq f(0) <infty$. The hypothesis tells you that if $delta in (0,1)$ is sufficiently small then $f(x) >alpha$ when $x in (-1,-1+delta)$ and also when $x in (1-delta, 1)$. Now I leave it to you to verify that the minimum of $f$ on the compact interval $[-1+delta, 1-delta]$ is also the minimum of $f$ on $(-1,1)$.
answered Dec 12 '18 at 6:36
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
$endgroup$
add a comment |
$begingroup$
Your argument does not make much sense. What do mean by $x >N_1$ and $x >N_2$. You are not taking limits as $ x to 1$and $x to -1$.
Here is a valid argument: let $alpha =inf {f(x): -1<x<1}$. Note that $alpha leq f(0) <infty$. The hypothesis tells you that if $delta in (0,1)$ is sufficiently small then $f(x) >alpha$ when $x in (-1,-1+delta)$ and also when $x in (1-delta, 1)$. Now I leave it to you to verify that the minimum of $f$ on the compact interval $[-1+delta, 1-delta]$ is also the minimum of $f$ on $(-1,1)$.
answered Dec 12 '18 at 6:36
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
$endgroup$
add a comment |
$begingroup$
Your argument does not make much sense. What do mean by $x >N_1$ and $x >N_2$. You are not taking limits as $ x to 1$and $x to -1$.
Here is a valid argument: let $alpha =inf {f(x): -1<x<1}$. Note that $alpha leq f(0) <infty$. The hypothesis tells you that if $delta in (0,1)$ is sufficiently small then $f(x) >alpha$ when $x in (-1,-1+delta)$ and also when $x in (1-delta, 1)$. Now I leave it to you to verify that the minimum of $f$ on the compact interval $[-1+delta, 1-delta]$ is also the minimum of $f$ on $(-1,1)$.
answered Dec 12 '18 at 6:36
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
$endgroup$
Your argument does not make much sense. What do mean by $x >N_1$ and $x >N_2$. You are not taking limits as $ x to 1$and $x to -1$.
Here is a valid argument: let $alpha =inf {f(x): -1<x<1}$. Note that $alpha leq f(0) <infty$. The hypothesis tells you that if $delta in (0,1)$ is sufficiently small then $f(x) >alpha$ when $x in (-1,-1+delta)$ and also when $x in (1-delta, 1)$. Now I leave it to you to verify that the minimum of $f$ on the compact interval $[-1+delta, 1-delta]$ is also the minimum of $f$ on $(-1,1)$.
answered Dec 12 '18 at 6:36
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
answered Dec 12 '18 at 6:36
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
answered Dec 12 '18 at 6:36
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
answered Dec 12 '18 at 6:36
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
57.9k42160
add a comment |
add a comment |
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$begingroup$
You never say who $N_1$ and $N_2$ are; that paragraph is wrong, in part because you seem to be confusing limits at infinity with limits go to infinity. It is also wrong that “$f(x)>M$ for all $M$ when...” because that would require $f(x)$ to be larger than all real numbers, and no such number exists.
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:32
$begingroup$
Order matters. What you meant in that paragraph is: for each $M>0$ there exists $epsilon_{1,M}>0$ such that if $-1< x < -1+epsilon_{1,M}$, then $f(x)>M$. Similarly, for each $M>0$ there exists $epsilon_{2,M}>0$ such that if $1-epsilon_{2,M}<x<1$, then $f(x)>M$.”
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:34
$begingroup$
(Order in that you first specify $M$, then you specify the conditions on $x$, not the other way around). The basic idea can be made to work, but since you never specify what $N_1$ and $N_2$ are, and you don’t place them correctly inside of $(-1,1)$, what you write does not actually work. In addition, your $L$ is only asserted to be a lower bound, so you have not proven that it has a minimum, just that it is bounded below.
$endgroup$
– Arturo Magidin
Dec 12 '18 at 6:36
$begingroup$
Ah ok that makes sense, thank you.
$endgroup$
– hkj447
Dec 12 '18 at 7:02