Bounded functions and infimum/supremum
Question from my homework:
A function $f colon S to mathbb{R}$ is said to be decreasing on the domain $S subset mathbb{R}$ if for every $x, y in S$ with $y > x$, we have $f(y) < f(x)$.
Show that if $f colon mathbb{R} to mathbb{R}$ is decreasing and bounded on $mathbb{R}$, then the limits $lim_{x to infty} f(x)$ and $lim_{x to -infty} f(x)$ both exist.
Caution: While this claim is clearly related to the theorem that bounded monotone sequences always converge, it is not an immediate corollary of it.
You will have to use the definition of the limit directly.
(Original picture of the problem here.)
My proof:
Let $ S = {f(x) | x in mathbb{R} } $, since $f$ is bounded this set has both an infimum and supremum. (Is this true? I tried to prove it, but have no idea as it seems fairly obvious.) Now, I claim that $lim_{xto infty} f(x) = inf(S) = L$.
By definition, given $epsilon > 0$, there exists $K in mathbb{R}$ such that $x>K implies |f(x) - L| < epsilon$.
Now since $f(x) geq L$ by definition of the infimum, choose $x_1$ such that $f(x_1) < L + epsilon$ for $x_1 > K$. But since $f$ is a decreasing function we have $f(x) leq f(x_1)$, and thus
$$
L - epsilon < L leq f(x) leq f(x_1) < L + epsilon
$$
which proves our claim.
I use a similar argument for the supremum. Could anyone help me out on making this more rigorous as I feel as though I've skipped some steps.
analysis supremum-and-infimum
edited Aug 27 '18 at 11:17
Jendrik Stelzner
7,82121339
asked Dec 4 '13 at 20:22
DH.
44115
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
Question from my homework:
A function $f colon S to mathbb{R}$ is said to be decreasing on the domain $S subset mathbb{R}$ if for every $x, y in S$ with $y > x$, we have $f(y) < f(x)$.
Show that if $f colon mathbb{R} to mathbb{R}$ is decreasing and bounded on $mathbb{R}$, then the limits $lim_{x to infty} f(x)$ and $lim_{x to -infty} f(x)$ both exist.
Caution: While this claim is clearly related to the theorem that bounded monotone sequences always converge, it is not an immediate corollary of it.
You will have to use the definition of the limit directly.
(Original picture of the problem here.)
My proof:
Let $ S = {f(x) | x in mathbb{R} } $, since $f$ is bounded this set has both an infimum and supremum. (Is this true? I tried to prove it, but have no idea as it seems fairly obvious.) Now, I claim that $lim_{xto infty} f(x) = inf(S) = L$.
By definition, given $epsilon > 0$, there exists $K in mathbb{R}$ such that $x>K implies |f(x) - L| < epsilon$.
Now since $f(x) geq L$ by definition of the infimum, choose $x_1$ such that $f(x_1) < L + epsilon$ for $x_1 > K$. But since $f$ is a decreasing function we have $f(x) leq f(x_1)$, and thus
$$
L - epsilon < L leq f(x) leq f(x_1) < L + epsilon
$$
which proves our claim.
I use a similar argument for the supremum. Could anyone help me out on making this more rigorous as I feel as though I've skipped some steps.
analysis supremum-and-infimum
edited Aug 27 '18 at 11:17
Jendrik Stelzner
7,82121339
asked Dec 4 '13 at 20:22
DH.
44115
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Any bounded set in the real numbers has a sup and an inf. If they are in the set themselves is a different story.
– LASV
Dec 4 '13 at 20:36
About supremum and infimum you are ok. Now, you are wrong in your way to prove that the limit exists,because you don't prove that such a $K$ exists. All you have to do is to prove this existence.
– Haha
Dec 4 '13 at 20:42
I don't able to see your question
– user355797
Jul 26 '16 at 13:32
anyone edit so that I try to solve that question
– user355797
Jul 26 '16 at 13:33
add a comment |
Question from my homework:
A function $f colon S to mathbb{R}$ is said to be decreasing on the domain $S subset mathbb{R}$ if for every $x, y in S$ with $y > x$, we have $f(y) < f(x)$.
Show that if $f colon mathbb{R} to mathbb{R}$ is decreasing and bounded on $mathbb{R}$, then the limits $lim_{x to infty} f(x)$ and $lim_{x to -infty} f(x)$ both exist.
Caution: While this claim is clearly related to the theorem that bounded monotone sequences always converge, it is not an immediate corollary of it.
You will have to use the definition of the limit directly.
(Original picture of the problem here.)
My proof:
Let $ S = {f(x) | x in mathbb{R} } $, since $f$ is bounded this set has both an infimum and supremum. (Is this true? I tried to prove it, but have no idea as it seems fairly obvious.) Now, I claim that $lim_{xto infty} f(x) = inf(S) = L$.
By definition, given $epsilon > 0$, there exists $K in mathbb{R}$ such that $x>K implies |f(x) - L| < epsilon$.
Now since $f(x) geq L$ by definition of the infimum, choose $x_1$ such that $f(x_1) < L + epsilon$ for $x_1 > K$. But since $f$ is a decreasing function we have $f(x) leq f(x_1)$, and thus
$$
L - epsilon < L leq f(x) leq f(x_1) < L + epsilon
$$
which proves our claim.
I use a similar argument for the supremum. Could anyone help me out on making this more rigorous as I feel as though I've skipped some steps.
analysis supremum-and-infimum
edited Aug 27 '18 at 11:17
Jendrik Stelzner
7,82121339
asked Dec 4 '13 at 20:22
DH.
44115
Question from my homework:
A function $f colon S to mathbb{R}$ is said to be decreasing on the domain $S subset mathbb{R}$ if for every $x, y in S$ with $y > x$, we have $f(y) < f(x)$.
Show that if $f colon mathbb{R} to mathbb{R}$ is decreasing and bounded on $mathbb{R}$, then the limits $lim_{x to infty} f(x)$ and $lim_{x to -infty} f(x)$ both exist.
Caution: While this claim is clearly related to the theorem that bounded monotone sequences always converge, it is not an immediate corollary of it.
You will have to use the definition of the limit directly.
(Original picture of the problem here.)
My proof:
Let $ S = {f(x) | x in mathbb{R} } $, since $f$ is bounded this set has both an infimum and supremum. (Is this true? I tried to prove it, but have no idea as it seems fairly obvious.) Now, I claim that $lim_{xto infty} f(x) = inf(S) = L$.
By definition, given $epsilon > 0$, there exists $K in mathbb{R}$ such that $x>K implies |f(x) - L| < epsilon$.
Now since $f(x) geq L$ by definition of the infimum, choose $x_1$ such that $f(x_1) < L + epsilon$ for $x_1 > K$. But since $f$ is a decreasing function we have $f(x) leq f(x_1)$, and thus
$$
L - epsilon < L leq f(x) leq f(x_1) < L + epsilon
$$
which proves our claim.
I use a similar argument for the supremum. Could anyone help me out on making this more rigorous as I feel as though I've skipped some steps.
analysis supremum-and-infimum
analysis supremum-and-infimum
edited Aug 27 '18 at 11:17
Jendrik Stelzner
7,82121339
asked Dec 4 '13 at 20:22
DH.
44115
edited Aug 27 '18 at 11:17
Jendrik Stelzner
7,82121339
asked Dec 4 '13 at 20:22
DH.
44115
edited Aug 27 '18 at 11:17
Jendrik Stelzner
7,82121339
edited Aug 27 '18 at 11:17
Jendrik Stelzner
7,82121339
edited Aug 27 '18 at 11:17
Jendrik Stelzner
7,82121339
7,82121339
asked Dec 4 '13 at 20:22
DH.
44115
asked Dec 4 '13 at 20:22
DH.
44115
asked Dec 4 '13 at 20:22
DH.
44115
44115
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Any bounded set in the real numbers has a sup and an inf. If they are in the set themselves is a different story.
– LASV
Dec 4 '13 at 20:36
About supremum and infimum you are ok. Now, you are wrong in your way to prove that the limit exists,because you don't prove that such a $K$ exists. All you have to do is to prove this existence.
– Haha
Dec 4 '13 at 20:42
I don't able to see your question
– user355797
Jul 26 '16 at 13:32
anyone edit so that I try to solve that question
– user355797
Jul 26 '16 at 13:33
add a comment |
Any bounded set in the real numbers has a sup and an inf. If they are in the set themselves is a different story.
– LASV
Dec 4 '13 at 20:36
About supremum and infimum you are ok. Now, you are wrong in your way to prove that the limit exists,because you don't prove that such a $K$ exists. All you have to do is to prove this existence.
– Haha
Dec 4 '13 at 20:42
I don't able to see your question
– user355797
Jul 26 '16 at 13:32
anyone edit so that I try to solve that question
– user355797
Jul 26 '16 at 13:33
Any bounded set in the real numbers has a sup and an inf. If they are in the set themselves is a different story.
– LASV
Dec 4 '13 at 20:36
Any bounded set in the real numbers has a sup and an inf. If they are in the set themselves is a different story.
– LASV
Dec 4 '13 at 20:36
About supremum and infimum you are ok. Now, you are wrong in your way to prove that the limit exists,because you don't prove that such a $K$ exists. All you have to do is to prove this existence.
– Haha
Dec 4 '13 at 20:42
About supremum and infimum you are ok. Now, you are wrong in your way to prove that the limit exists,because you don't prove that such a $K$ exists. All you have to do is to prove this existence.
– Haha
Dec 4 '13 at 20:42
I don't able to see your question
– user355797
Jul 26 '16 at 13:32
I don't able to see your question
– user355797
Jul 26 '16 at 13:32
anyone edit so that I try to solve that question
– user355797
Jul 26 '16 at 13:33
anyone edit so that I try to solve that question
– user355797
Jul 26 '16 at 13:33
add a comment |
1 Answer
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I think you are almost there. I would proceed like this:
For $epsilon > 0$ be a given, there exists an $x^{*}$ such that $f(x^{*}) < L + epsilon $. So let $K = |x^{*}| + 1 > 0$, then for all $x$'s such that $x > K$ we have: $f(x) < f(K) < f(x^{*}) < L + epsilon$. But $L < f(x)$ since it is $text{inf}{f(x): x in mathbb{R}}$. So
$L - epsilon < L < f(x) < L + epsilon$. This means $|f(x) - L| < epsilon$ and we're done.
answered Dec 4 '13 at 20:51
DeepSea
70.9k54487
add a comment |
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I think you are almost there. I would proceed like this:
For $epsilon > 0$ be a given, there exists an $x^{*}$ such that $f(x^{*}) < L + epsilon $. So let $K = |x^{*}| + 1 > 0$, then for all $x$'s such that $x > K$ we have: $f(x) < f(K) < f(x^{*}) < L + epsilon$. But $L < f(x)$ since it is $text{inf}{f(x): x in mathbb{R}}$. So
$L - epsilon < L < f(x) < L + epsilon$. This means $|f(x) - L| < epsilon$ and we're done.
answered Dec 4 '13 at 20:51
DeepSea
70.9k54487
add a comment |
I think you are almost there. I would proceed like this:
For $epsilon > 0$ be a given, there exists an $x^{*}$ such that $f(x^{*}) < L + epsilon $. So let $K = |x^{*}| + 1 > 0$, then for all $x$'s such that $x > K$ we have: $f(x) < f(K) < f(x^{*}) < L + epsilon$. But $L < f(x)$ since it is $text{inf}{f(x): x in mathbb{R}}$. So
$L - epsilon < L < f(x) < L + epsilon$. This means $|f(x) - L| < epsilon$ and we're done.
answered Dec 4 '13 at 20:51
DeepSea
70.9k54487
add a comment |
I think you are almost there. I would proceed like this:
For $epsilon > 0$ be a given, there exists an $x^{*}$ such that $f(x^{*}) < L + epsilon $. So let $K = |x^{*}| + 1 > 0$, then for all $x$'s such that $x > K$ we have: $f(x) < f(K) < f(x^{*}) < L + epsilon$. But $L < f(x)$ since it is $text{inf}{f(x): x in mathbb{R}}$. So
$L - epsilon < L < f(x) < L + epsilon$. This means $|f(x) - L| < epsilon$ and we're done.
answered Dec 4 '13 at 20:51
DeepSea
70.9k54487
I think you are almost there. I would proceed like this:
For $epsilon > 0$ be a given, there exists an $x^{*}$ such that $f(x^{*}) < L + epsilon $. So let $K = |x^{*}| + 1 > 0$, then for all $x$'s such that $x > K$ we have: $f(x) < f(K) < f(x^{*}) < L + epsilon$. But $L < f(x)$ since it is $text{inf}{f(x): x in mathbb{R}}$. So
$L - epsilon < L < f(x) < L + epsilon$. This means $|f(x) - L| < epsilon$ and we're done.
answered Dec 4 '13 at 20:51
DeepSea
70.9k54487
edited Jul 26 '16 at 23:13
answered Dec 4 '13 at 20:51
DeepSea
70.9k54487
answered Dec 4 '13 at 20:51
DeepSea
70.9k54487
answered Dec 4 '13 at 20:51
DeepSea
70.9k54487
70.9k54487
add a comment |
add a comment |
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Any bounded set in the real numbers has a sup and an inf. If they are in the set themselves is a different story.
– LASV
Dec 4 '13 at 20:36
About supremum and infimum you are ok. Now, you are wrong in your way to prove that the limit exists,because you don't prove that such a $K$ exists. All you have to do is to prove this existence.
– Haha
Dec 4 '13 at 20:42
I don't able to see your question
– user355797
Jul 26 '16 at 13:32
anyone edit so that I try to solve that question
– user355797
Jul 26 '16 at 13:33