Mixing
Show that the function $f$ is attain its minimum or maximum value and bounded on $mathbb R$
$begingroup$
Let the continuous function $f:mathbb R→mathbb R$ satisfying
$$lim_{xrightarrow -infty}{f(x)}=lim_{xrightarrow +infty}{f(x)}=0$$
Show that the function $f$ is:
(i) bounded;
(ii) attain its minimum or maximum value on $mathbb R$, that is, there exists a point $x_0in mathbb R$ such that $f(x_0)$ be the maximum or minimum value on $mathbb R$.
For the part (i), let
$$S={f(x) : x in mathbb R}$$
Since both $inf S$ and $sup S$ are finite, so the function $f$ is bounded.
For the part (ii), I would like to discuss the following four cases:
(a) when $x>M$, $f(x)<frac{sup S}{2} neq 0$ and $f(x)>frac{inf S}{2} neq 0$
(b) when $x<N$, $f(x)<frac{sup S}{2} neq 0$ and $f(x)>frac{inf S}{2} neq 0$
(c) Both $inf S= sup S=0$
(d) Either $inf S=0$ or $sup S=0$, then the function $f$ either attains its global maximum or global minimum.
How do I elaborate my idea for part (ii) (a) and (b) to show that any local extremum point on the closed interval $[N,M]$ is the global extremum point on $mathbb R$?
continuity proof-explanation
asked Jan 13 at 11:00
weilam06weilam06
9511
$endgroup$
add a comment |
$begingroup$
Let the continuous function $f:mathbb R→mathbb R$ satisfying
$$lim_{xrightarrow -infty}{f(x)}=lim_{xrightarrow +infty}{f(x)}=0$$
Show that the function $f$ is:
(i) bounded;
(ii) attain its minimum or maximum value on $mathbb R$, that is, there exists a point $x_0in mathbb R$ such that $f(x_0)$ be the maximum or minimum value on $mathbb R$.
For the part (i), let
$$S={f(x) : x in mathbb R}$$
Since both $inf S$ and $sup S$ are finite, so the function $f$ is bounded.
For the part (ii), I would like to discuss the following four cases:
(a) when $x>M$, $f(x)<frac{sup S}{2} neq 0$ and $f(x)>frac{inf S}{2} neq 0$
(b) when $x<N$, $f(x)<frac{sup S}{2} neq 0$ and $f(x)>frac{inf S}{2} neq 0$
(c) Both $inf S= sup S=0$
(d) Either $inf S=0$ or $sup S=0$, then the function $f$ either attains its global maximum or global minimum.
How do I elaborate my idea for part (ii) (a) and (b) to show that any local extremum point on the closed interval $[N,M]$ is the global extremum point on $mathbb R$?
continuity proof-explanation
asked Jan 13 at 11:00
weilam06weilam06
9511
$endgroup$
1
$begingroup$
In part $(i)$ I believe you need to explain why $inf S$ and $sup S$ are finite. In part $(ii)$ it's not clear what are $M,N$.
$endgroup$
– Yanko
Jan 13 at 11:03
$begingroup$
$M$ and $N$ are both real numbers. I tried to construct a closed interval and show that $f$ has attain its maximum or minimum value on $[M,N]$.
$endgroup$
– weilam06
Jan 13 at 11:05
$begingroup$
@weilam06 What bounds do you get from the two limits? Write out the definitions (or one particular case of them), and go from there. You don't need any cases.
$endgroup$
– user3482749
Jan 13 at 11:06
$begingroup$
That $inf S$ and $sup S$ are finite is basically what you have to show in part (i), so you can't just state it.
$endgroup$
– Henrik
Jan 13 at 11:07
add a comment |
$begingroup$
Let the continuous function $f:mathbb R→mathbb R$ satisfying
$$lim_{xrightarrow -infty}{f(x)}=lim_{xrightarrow +infty}{f(x)}=0$$
Show that the function $f$ is:
(i) bounded;
(ii) attain its minimum or maximum value on $mathbb R$, that is, there exists a point $x_0in mathbb R$ such that $f(x_0)$ be the maximum or minimum value on $mathbb R$.
For the part (i), let
$$S={f(x) : x in mathbb R}$$
Since both $inf S$ and $sup S$ are finite, so the function $f$ is bounded.
For the part (ii), I would like to discuss the following four cases:
(a) when $x>M$, $f(x)<frac{sup S}{2} neq 0$ and $f(x)>frac{inf S}{2} neq 0$
(b) when $x<N$, $f(x)<frac{sup S}{2} neq 0$ and $f(x)>frac{inf S}{2} neq 0$
(c) Both $inf S= sup S=0$
(d) Either $inf S=0$ or $sup S=0$, then the function $f$ either attains its global maximum or global minimum.
How do I elaborate my idea for part (ii) (a) and (b) to show that any local extremum point on the closed interval $[N,M]$ is the global extremum point on $mathbb R$?
continuity proof-explanation
asked Jan 13 at 11:00
weilam06weilam06
9511
$endgroup$
Let the continuous function $f:mathbb R→mathbb R$ satisfying
$$lim_{xrightarrow -infty}{f(x)}=lim_{xrightarrow +infty}{f(x)}=0$$
Show that the function $f$ is:
(i) bounded;
(ii) attain its minimum or maximum value on $mathbb R$, that is, there exists a point $x_0in mathbb R$ such that $f(x_0)$ be the maximum or minimum value on $mathbb R$.
For the part (i), let
$$S={f(x) : x in mathbb R}$$
Since both $inf S$ and $sup S$ are finite, so the function $f$ is bounded.
For the part (ii), I would like to discuss the following four cases:
(a) when $x>M$, $f(x)<frac{sup S}{2} neq 0$ and $f(x)>frac{inf S}{2} neq 0$
(b) when $x<N$, $f(x)<frac{sup S}{2} neq 0$ and $f(x)>frac{inf S}{2} neq 0$
(c) Both $inf S= sup S=0$
(d) Either $inf S=0$ or $sup S=0$, then the function $f$ either attains its global maximum or global minimum.
How do I elaborate my idea for part (ii) (a) and (b) to show that any local extremum point on the closed interval $[N,M]$ is the global extremum point on $mathbb R$?
continuity proof-explanation
continuity proof-explanation
asked Jan 13 at 11:00
weilam06weilam06
9511
asked Jan 13 at 11:00
weilam06weilam06
9511
asked Jan 13 at 11:00
weilam06weilam06
9511
asked Jan 13 at 11:00
weilam06weilam06
9511
asked Jan 13 at 11:00
weilam06weilam06
9511
9511
1
$begingroup$
In part $(i)$ I believe you need to explain why $inf S$ and $sup S$ are finite. In part $(ii)$ it's not clear what are $M,N$.
$endgroup$
– Yanko
Jan 13 at 11:03
$begingroup$
$M$ and $N$ are both real numbers. I tried to construct a closed interval and show that $f$ has attain its maximum or minimum value on $[M,N]$.
$endgroup$
– weilam06
Jan 13 at 11:05
$begingroup$
@weilam06 What bounds do you get from the two limits? Write out the definitions (or one particular case of them), and go from there. You don't need any cases.
$endgroup$
– user3482749
Jan 13 at 11:06
$begingroup$
That $inf S$ and $sup S$ are finite is basically what you have to show in part (i), so you can't just state it.
$endgroup$
– Henrik
Jan 13 at 11:07
add a comment |
1
$begingroup$
In part $(i)$ I believe you need to explain why $inf S$ and $sup S$ are finite. In part $(ii)$ it's not clear what are $M,N$.
$endgroup$
– Yanko
Jan 13 at 11:03
$begingroup$
$M$ and $N$ are both real numbers. I tried to construct a closed interval and show that $f$ has attain its maximum or minimum value on $[M,N]$.
$endgroup$
– weilam06
Jan 13 at 11:05
$begingroup$
@weilam06 What bounds do you get from the two limits? Write out the definitions (or one particular case of them), and go from there. You don't need any cases.
$endgroup$
– user3482749
Jan 13 at 11:06
$begingroup$
That $inf S$ and $sup S$ are finite is basically what you have to show in part (i), so you can't just state it.
$endgroup$
– Henrik
Jan 13 at 11:07
1
1
$begingroup$
In part $(i)$ I believe you need to explain why $inf S$ and $sup S$ are finite. In part $(ii)$ it's not clear what are $M,N$.
$endgroup$
– Yanko
Jan 13 at 11:03
$begingroup$
In part $(i)$ I believe you need to explain why $inf S$ and $sup S$ are finite. In part $(ii)$ it's not clear what are $M,N$.
$endgroup$
– Yanko
Jan 13 at 11:03
$begingroup$
$M$ and $N$ are both real numbers. I tried to construct a closed interval and show that $f$ has attain its maximum or minimum value on $[M,N]$.
$endgroup$
– weilam06
Jan 13 at 11:05
$begingroup$
$M$ and $N$ are both real numbers. I tried to construct a closed interval and show that $f$ has attain its maximum or minimum value on $[M,N]$.
$endgroup$
– weilam06
Jan 13 at 11:05
$begingroup$
@weilam06 What bounds do you get from the two limits? Write out the definitions (or one particular case of them), and go from there. You don't need any cases.
$endgroup$
– user3482749
Jan 13 at 11:06
$begingroup$
@weilam06 What bounds do you get from the two limits? Write out the definitions (or one particular case of them), and go from there. You don't need any cases.
$endgroup$
– user3482749
Jan 13 at 11:06
$begingroup$
That $inf S$ and $sup S$ are finite is basically what you have to show in part (i), so you can't just state it.
$endgroup$
– Henrik
Jan 13 at 11:07
$begingroup$
That $inf S$ and $sup S$ are finite is basically what you have to show in part (i), so you can't just state it.
$endgroup$
– Henrik
Jan 13 at 11:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In part (i), you should elaborate more on why $S$ is bounded. You shouldn't just claim it has a supremum and infimum; you'll need to appeal to continuity (specifically, the extreme value theorem).
In fact, let's show it, because I think it might give you a kickstart towards (ii). Since $lim_{x to pm infty} f(x) = 0$, we know that, choosing $varepsilon = 1$ arbitrarily, there is some $N, M$ such that
begin{align*}
x > N &implies |f(x)| < 1 \
x < M &implies |f(x)| < 1
end{align*}
Consider now the restriction of $f$ to $[M, N]$. Now, $f$ is a continuous function over a compact interval, and thus must be bounded, by the extreme value theorem. That is, there's some $K$ (which is any number larger than the absolute values of the maximum and minimum $f$ over the interval) such that
$$M le x le N implies |f(x)| < K.$$
Now, putting it together, we know that $|f(x)| < max {1, K }$ everywhere, thus showing $S$ is bounded (above and below).
Now, to do (ii), you're right about splitting into cases. First deal with the contant case. Otherwise $sup f > 0$ or $inf f < 0$. Deal with these cases separately. Try using the limit definition, this time with $varepsilon = frac{sup f}{2}$ for the first case, and use this to construct an appropriate $M$ and $N$. You're pretty close, you just need a little more rigour.
answered Jan 13 at 11:13
Theo BenditTheo Bendit
18.1k12152
$endgroup$
add a comment |
$begingroup$
My idea is to use the Bolzano-Weierstrass Theorem. Suppose $f$ has a non-zero supremum. By definition of supremum, there is a sequence $x_n$ such that $f(x_n) to M=sup S$. Note $x_n$ cannot be unbounded, otherwise there is a subsequence of it such that $y_n to infty$ and $f(y_n) to 0=M$.
Then $x_n$ is a bounded sequence, implying that there is a convergent subsequence. By continuity, the limit point is where the supremum is attained.
answered Jan 13 at 11:19
JerryJerry
422313
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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votes
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oldest
votes
$begingroup$
In part (i), you should elaborate more on why $S$ is bounded. You shouldn't just claim it has a supremum and infimum; you'll need to appeal to continuity (specifically, the extreme value theorem).
In fact, let's show it, because I think it might give you a kickstart towards (ii). Since $lim_{x to pm infty} f(x) = 0$, we know that, choosing $varepsilon = 1$ arbitrarily, there is some $N, M$ such that
begin{align*}
x > N &implies |f(x)| < 1 \
x < M &implies |f(x)| < 1
end{align*}
Consider now the restriction of $f$ to $[M, N]$. Now, $f$ is a continuous function over a compact interval, and thus must be bounded, by the extreme value theorem. That is, there's some $K$ (which is any number larger than the absolute values of the maximum and minimum $f$ over the interval) such that
$$M le x le N implies |f(x)| < K.$$
Now, putting it together, we know that $|f(x)| < max {1, K }$ everywhere, thus showing $S$ is bounded (above and below).
Now, to do (ii), you're right about splitting into cases. First deal with the contant case. Otherwise $sup f > 0$ or $inf f < 0$. Deal with these cases separately. Try using the limit definition, this time with $varepsilon = frac{sup f}{2}$ for the first case, and use this to construct an appropriate $M$ and $N$. You're pretty close, you just need a little more rigour.
answered Jan 13 at 11:13
Theo BenditTheo Bendit
18.1k12152
$endgroup$
add a comment |
$begingroup$
In part (i), you should elaborate more on why $S$ is bounded. You shouldn't just claim it has a supremum and infimum; you'll need to appeal to continuity (specifically, the extreme value theorem).
In fact, let's show it, because I think it might give you a kickstart towards (ii). Since $lim_{x to pm infty} f(x) = 0$, we know that, choosing $varepsilon = 1$ arbitrarily, there is some $N, M$ such that
begin{align*}
x > N &implies |f(x)| < 1 \
x < M &implies |f(x)| < 1
end{align*}
Consider now the restriction of $f$ to $[M, N]$. Now, $f$ is a continuous function over a compact interval, and thus must be bounded, by the extreme value theorem. That is, there's some $K$ (which is any number larger than the absolute values of the maximum and minimum $f$ over the interval) such that
$$M le x le N implies |f(x)| < K.$$
Now, putting it together, we know that $|f(x)| < max {1, K }$ everywhere, thus showing $S$ is bounded (above and below).
Now, to do (ii), you're right about splitting into cases. First deal with the contant case. Otherwise $sup f > 0$ or $inf f < 0$. Deal with these cases separately. Try using the limit definition, this time with $varepsilon = frac{sup f}{2}$ for the first case, and use this to construct an appropriate $M$ and $N$. You're pretty close, you just need a little more rigour.
answered Jan 13 at 11:13
Theo BenditTheo Bendit
18.1k12152
$endgroup$
add a comment |
$begingroup$
In part (i), you should elaborate more on why $S$ is bounded. You shouldn't just claim it has a supremum and infimum; you'll need to appeal to continuity (specifically, the extreme value theorem).
In fact, let's show it, because I think it might give you a kickstart towards (ii). Since $lim_{x to pm infty} f(x) = 0$, we know that, choosing $varepsilon = 1$ arbitrarily, there is some $N, M$ such that
begin{align*}
x > N &implies |f(x)| < 1 \
x < M &implies |f(x)| < 1
end{align*}
Consider now the restriction of $f$ to $[M, N]$. Now, $f$ is a continuous function over a compact interval, and thus must be bounded, by the extreme value theorem. That is, there's some $K$ (which is any number larger than the absolute values of the maximum and minimum $f$ over the interval) such that
$$M le x le N implies |f(x)| < K.$$
Now, putting it together, we know that $|f(x)| < max {1, K }$ everywhere, thus showing $S$ is bounded (above and below).
Now, to do (ii), you're right about splitting into cases. First deal with the contant case. Otherwise $sup f > 0$ or $inf f < 0$. Deal with these cases separately. Try using the limit definition, this time with $varepsilon = frac{sup f}{2}$ for the first case, and use this to construct an appropriate $M$ and $N$. You're pretty close, you just need a little more rigour.
answered Jan 13 at 11:13
Theo BenditTheo Bendit
18.1k12152
$endgroup$
In part (i), you should elaborate more on why $S$ is bounded. You shouldn't just claim it has a supremum and infimum; you'll need to appeal to continuity (specifically, the extreme value theorem).
In fact, let's show it, because I think it might give you a kickstart towards (ii). Since $lim_{x to pm infty} f(x) = 0$, we know that, choosing $varepsilon = 1$ arbitrarily, there is some $N, M$ such that
begin{align*}
x > N &implies |f(x)| < 1 \
x < M &implies |f(x)| < 1
end{align*}
Consider now the restriction of $f$ to $[M, N]$. Now, $f$ is a continuous function over a compact interval, and thus must be bounded, by the extreme value theorem. That is, there's some $K$ (which is any number larger than the absolute values of the maximum and minimum $f$ over the interval) such that
$$M le x le N implies |f(x)| < K.$$
Now, putting it together, we know that $|f(x)| < max {1, K }$ everywhere, thus showing $S$ is bounded (above and below).
Now, to do (ii), you're right about splitting into cases. First deal with the contant case. Otherwise $sup f > 0$ or $inf f < 0$. Deal with these cases separately. Try using the limit definition, this time with $varepsilon = frac{sup f}{2}$ for the first case, and use this to construct an appropriate $M$ and $N$. You're pretty close, you just need a little more rigour.
answered Jan 13 at 11:13
Theo BenditTheo Bendit
18.1k12152
answered Jan 13 at 11:13
Theo BenditTheo Bendit
18.1k12152
answered Jan 13 at 11:13
Theo BenditTheo Bendit
18.1k12152
answered Jan 13 at 11:13
Theo BenditTheo Bendit
18.1k12152
18.1k12152
add a comment |
add a comment |
$begingroup$
My idea is to use the Bolzano-Weierstrass Theorem. Suppose $f$ has a non-zero supremum. By definition of supremum, there is a sequence $x_n$ such that $f(x_n) to M=sup S$. Note $x_n$ cannot be unbounded, otherwise there is a subsequence of it such that $y_n to infty$ and $f(y_n) to 0=M$.
Then $x_n$ is a bounded sequence, implying that there is a convergent subsequence. By continuity, the limit point is where the supremum is attained.
answered Jan 13 at 11:19
JerryJerry
422313
$endgroup$
add a comment |
$begingroup$
My idea is to use the Bolzano-Weierstrass Theorem. Suppose $f$ has a non-zero supremum. By definition of supremum, there is a sequence $x_n$ such that $f(x_n) to M=sup S$. Note $x_n$ cannot be unbounded, otherwise there is a subsequence of it such that $y_n to infty$ and $f(y_n) to 0=M$.
Then $x_n$ is a bounded sequence, implying that there is a convergent subsequence. By continuity, the limit point is where the supremum is attained.
answered Jan 13 at 11:19
JerryJerry
422313
$endgroup$
add a comment |
$begingroup$
My idea is to use the Bolzano-Weierstrass Theorem. Suppose $f$ has a non-zero supremum. By definition of supremum, there is a sequence $x_n$ such that $f(x_n) to M=sup S$. Note $x_n$ cannot be unbounded, otherwise there is a subsequence of it such that $y_n to infty$ and $f(y_n) to 0=M$.
Then $x_n$ is a bounded sequence, implying that there is a convergent subsequence. By continuity, the limit point is where the supremum is attained.
answered Jan 13 at 11:19
JerryJerry
422313
$endgroup$
My idea is to use the Bolzano-Weierstrass Theorem. Suppose $f$ has a non-zero supremum. By definition of supremum, there is a sequence $x_n$ such that $f(x_n) to M=sup S$. Note $x_n$ cannot be unbounded, otherwise there is a subsequence of it such that $y_n to infty$ and $f(y_n) to 0=M$.
Then $x_n$ is a bounded sequence, implying that there is a convergent subsequence. By continuity, the limit point is where the supremum is attained.
answered Jan 13 at 11:19
JerryJerry
422313
answered Jan 13 at 11:19
JerryJerry
422313
answered Jan 13 at 11:19
JerryJerry
422313
answered Jan 13 at 11:19
JerryJerry
422313
422313
add a comment |
add a comment |
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1
$begingroup$
In part $(i)$ I believe you need to explain why $inf S$ and $sup S$ are finite. In part $(ii)$ it's not clear what are $M,N$.
$endgroup$
– Yanko
Jan 13 at 11:03
$begingroup$
$M$ and $N$ are both real numbers. I tried to construct a closed interval and show that $f$ has attain its maximum or minimum value on $[M,N]$.
$endgroup$
– weilam06
Jan 13 at 11:05
$begingroup$
@weilam06 What bounds do you get from the two limits? Write out the definitions (or one particular case of them), and go from there. You don't need any cases.
$endgroup$
– user3482749
Jan 13 at 11:06
$begingroup$
That $inf S$ and $sup S$ are finite is basically what you have to show in part (i), so you can't just state it.
$endgroup$
– Henrik
Jan 13 at 11:07