Mixing
If $f(x+1)=f(x)$ then?
$begingroup$
Let $f: mathbb{R} rightarrow mathbb{R}$ be a function such that $f(x+1) = f(x)$, $forall x in mathbb{R}$.
Then which of the following statement(s) is/are true?
$f$ is bounded.
$f$ is bounded if it is continuous.
$f$ is differentiable if it is continuous
$f$ is uniformly continuous if it is continuous
I took the example of
$$f(x) = begin{cases}
tan(pi x) & x neq displaystyle frac{n}{2}\
1 & x = displaystyle frac{n}{2}
end{cases}$$
$n in mathbb{Z}$. This function satisfies the given condition and $f(x)$ is unbounded, so we can exclude the first option.
Now since the example we took is not continuous at $x = frac{n}{2}$, I think if $f$ satisfies the given condition and is continuous then it is bounded. Is there any theorem which states: if $f$ is periodic and continuous then it is bounded? If yes, how to prove it?
I think we can even discard the third option by defining a triangle function
$$ f(x) =
begin{cases}
x & 0 leq x leq displaystyle frac{1}{2}\
f(x)=1-x & displaystyle frac{1}{2} < x leq 1
end{cases}$$
$$ f(x+k) = f(x), k in mathbb{Z} $$
now this function is continuous all over $mathbb{R}$ but not differentiable at $x = frac{n}{2}$.
The only left option is the 4th one. I know the basics of uniform continuity but not how to solve in this case. If $f$ is periodic and continuous, does this imply that $f$ is uniformly continuous? How to prove this if it is true?
real-analysis continuity uniform-continuity periodic-functions
edited Jan 14 at 13:06
Martin Sleziak
44.8k10118272
asked Jan 14 at 10:17
Nilesh KhatriNilesh Khatri
195
$endgroup$
|
show 1 more comment
$begingroup$
Let $f: mathbb{R} rightarrow mathbb{R}$ be a function such that $f(x+1) = f(x)$, $forall x in mathbb{R}$.
Then which of the following statement(s) is/are true?
$f$ is bounded.
$f$ is bounded if it is continuous.
$f$ is differentiable if it is continuous
$f$ is uniformly continuous if it is continuous
I took the example of
$$f(x) = begin{cases}
tan(pi x) & x neq displaystyle frac{n}{2}\
1 & x = displaystyle frac{n}{2}
end{cases}$$
$n in mathbb{Z}$. This function satisfies the given condition and $f(x)$ is unbounded, so we can exclude the first option.
Now since the example we took is not continuous at $x = frac{n}{2}$, I think if $f$ satisfies the given condition and is continuous then it is bounded. Is there any theorem which states: if $f$ is periodic and continuous then it is bounded? If yes, how to prove it?
I think we can even discard the third option by defining a triangle function
$$ f(x) =
begin{cases}
x & 0 leq x leq displaystyle frac{1}{2}\
f(x)=1-x & displaystyle frac{1}{2} < x leq 1
end{cases}$$
$$ f(x+k) = f(x), k in mathbb{Z} $$
now this function is continuous all over $mathbb{R}$ but not differentiable at $x = frac{n}{2}$.
The only left option is the 4th one. I know the basics of uniform continuity but not how to solve in this case. If $f$ is periodic and continuous, does this imply that $f$ is uniformly continuous? How to prove this if it is true?
real-analysis continuity uniform-continuity periodic-functions
edited Jan 14 at 13:06
Martin Sleziak
44.8k10118272
asked Jan 14 at 10:17
Nilesh KhatriNilesh Khatri
195
$endgroup$
1
$begingroup$
What have you tried so far?
$endgroup$
– Mindlack
Jan 14 at 10:20
1
$begingroup$
I took examples like f(x) = tan(pi*x)[that is periodic with period 1 and unbounded,so i think 1 option can be discarded]
$endgroup$
– Nilesh Khatri
Jan 14 at 10:24
1
$begingroup$
@NileshKhatri That's a good example to exclude option 1, except for a minor detail: It's not defined for $x = 0.5$. Easily fixable (you just define it to have your favourite value at that point), but it must be addressed. What about the others?
$endgroup$
– Arthur
Jan 14 at 10:26
1
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– Martin R
Jan 14 at 10:27
$begingroup$
For problem 2), do you know anything about a maximum/minimum of a continuous function on a finite, closed interval, like $[0,1]$? Can you use that and apply the periodicity premise to get a final answer for problem 2?
$endgroup$
– Ingix
Jan 14 at 11:17
|
show 1 more comment
$begingroup$
Let $f: mathbb{R} rightarrow mathbb{R}$ be a function such that $f(x+1) = f(x)$, $forall x in mathbb{R}$.
Then which of the following statement(s) is/are true?
$f$ is bounded.
$f$ is bounded if it is continuous.
$f$ is differentiable if it is continuous
$f$ is uniformly continuous if it is continuous
I took the example of
$$f(x) = begin{cases}
tan(pi x) & x neq displaystyle frac{n}{2}\
1 & x = displaystyle frac{n}{2}
end{cases}$$
$n in mathbb{Z}$. This function satisfies the given condition and $f(x)$ is unbounded, so we can exclude the first option.
Now since the example we took is not continuous at $x = frac{n}{2}$, I think if $f$ satisfies the given condition and is continuous then it is bounded. Is there any theorem which states: if $f$ is periodic and continuous then it is bounded? If yes, how to prove it?
I think we can even discard the third option by defining a triangle function
$$ f(x) =
begin{cases}
x & 0 leq x leq displaystyle frac{1}{2}\
f(x)=1-x & displaystyle frac{1}{2} < x leq 1
end{cases}$$
$$ f(x+k) = f(x), k in mathbb{Z} $$
now this function is continuous all over $mathbb{R}$ but not differentiable at $x = frac{n}{2}$.
The only left option is the 4th one. I know the basics of uniform continuity but not how to solve in this case. If $f$ is periodic and continuous, does this imply that $f$ is uniformly continuous? How to prove this if it is true?
real-analysis continuity uniform-continuity periodic-functions
edited Jan 14 at 13:06
Martin Sleziak
44.8k10118272
asked Jan 14 at 10:17
Nilesh KhatriNilesh Khatri
195
$endgroup$
Let $f: mathbb{R} rightarrow mathbb{R}$ be a function such that $f(x+1) = f(x)$, $forall x in mathbb{R}$.
Then which of the following statement(s) is/are true?
$f$ is bounded.
$f$ is bounded if it is continuous.
$f$ is differentiable if it is continuous
$f$ is uniformly continuous if it is continuous
I took the example of
$$f(x) = begin{cases}
tan(pi x) & x neq displaystyle frac{n}{2}\
1 & x = displaystyle frac{n}{2}
end{cases}$$
$n in mathbb{Z}$. This function satisfies the given condition and $f(x)$ is unbounded, so we can exclude the first option.
Now since the example we took is not continuous at $x = frac{n}{2}$, I think if $f$ satisfies the given condition and is continuous then it is bounded. Is there any theorem which states: if $f$ is periodic and continuous then it is bounded? If yes, how to prove it?
I think we can even discard the third option by defining a triangle function
$$ f(x) =
begin{cases}
x & 0 leq x leq displaystyle frac{1}{2}\
f(x)=1-x & displaystyle frac{1}{2} < x leq 1
end{cases}$$
$$ f(x+k) = f(x), k in mathbb{Z} $$
now this function is continuous all over $mathbb{R}$ but not differentiable at $x = frac{n}{2}$.
The only left option is the 4th one. I know the basics of uniform continuity but not how to solve in this case. If $f$ is periodic and continuous, does this imply that $f$ is uniformly continuous? How to prove this if it is true?
real-analysis continuity uniform-continuity periodic-functions
real-analysis continuity uniform-continuity periodic-functions
edited Jan 14 at 13:06
Martin Sleziak
44.8k10118272
asked Jan 14 at 10:17
Nilesh KhatriNilesh Khatri
195
edited Jan 14 at 13:06
Martin Sleziak
44.8k10118272
asked Jan 14 at 10:17
Nilesh KhatriNilesh Khatri
195
edited Jan 14 at 13:06
Martin Sleziak
44.8k10118272
edited Jan 14 at 13:06
Martin Sleziak
44.8k10118272
edited Jan 14 at 13:06
Martin Sleziak
44.8k10118272
44.8k10118272
asked Jan 14 at 10:17
Nilesh KhatriNilesh Khatri
195
asked Jan 14 at 10:17
Nilesh KhatriNilesh Khatri
195
asked Jan 14 at 10:17
Nilesh KhatriNilesh Khatri
195
195
1
$begingroup$
What have you tried so far?
$endgroup$
– Mindlack
Jan 14 at 10:20
1
$begingroup$
I took examples like f(x) = tan(pi*x)[that is periodic with period 1 and unbounded,so i think 1 option can be discarded]
$endgroup$
– Nilesh Khatri
Jan 14 at 10:24
1
$begingroup$
@NileshKhatri That's a good example to exclude option 1, except for a minor detail: It's not defined for $x = 0.5$. Easily fixable (you just define it to have your favourite value at that point), but it must be addressed. What about the others?
$endgroup$
– Arthur
Jan 14 at 10:26
1
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– Martin R
Jan 14 at 10:27
$begingroup$
For problem 2), do you know anything about a maximum/minimum of a continuous function on a finite, closed interval, like $[0,1]$? Can you use that and apply the periodicity premise to get a final answer for problem 2?
$endgroup$
– Ingix
Jan 14 at 11:17
|
show 1 more comment
1
$begingroup$
What have you tried so far?
$endgroup$
– Mindlack
Jan 14 at 10:20
1
$begingroup$
I took examples like f(x) = tan(pi*x)[that is periodic with period 1 and unbounded,so i think 1 option can be discarded]
$endgroup$
– Nilesh Khatri
Jan 14 at 10:24
1
$begingroup$
@NileshKhatri That's a good example to exclude option 1, except for a minor detail: It's not defined for $x = 0.5$. Easily fixable (you just define it to have your favourite value at that point), but it must be addressed. What about the others?
$endgroup$
– Arthur
Jan 14 at 10:26
1
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– Martin R
Jan 14 at 10:27
$begingroup$
For problem 2), do you know anything about a maximum/minimum of a continuous function on a finite, closed interval, like $[0,1]$? Can you use that and apply the periodicity premise to get a final answer for problem 2?
$endgroup$
– Ingix
Jan 14 at 11:17
1
1
$begingroup$
What have you tried so far?
$endgroup$
– Mindlack
Jan 14 at 10:20
$begingroup$
What have you tried so far?
$endgroup$
– Mindlack
Jan 14 at 10:20
1
1
$begingroup$
I took examples like f(x) = tan(pi*x)[that is periodic with period 1 and unbounded,so i think 1 option can be discarded]
$endgroup$
– Nilesh Khatri
Jan 14 at 10:24
$begingroup$
I took examples like f(x) = tan(pi*x)[that is periodic with period 1 and unbounded,so i think 1 option can be discarded]
$endgroup$
– Nilesh Khatri
Jan 14 at 10:24
1
1
$begingroup$
@NileshKhatri That's a good example to exclude option 1, except for a minor detail: It's not defined for $x = 0.5$. Easily fixable (you just define it to have your favourite value at that point), but it must be addressed. What about the others?
$endgroup$
– Arthur
Jan 14 at 10:26
$begingroup$
@NileshKhatri That's a good example to exclude option 1, except for a minor detail: It's not defined for $x = 0.5$. Easily fixable (you just define it to have your favourite value at that point), but it must be addressed. What about the others?
$endgroup$
– Arthur
Jan 14 at 10:26
1
1
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– Martin R
Jan 14 at 10:27
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– Martin R
Jan 14 at 10:27
$begingroup$
For problem 2), do you know anything about a maximum/minimum of a continuous function on a finite, closed interval, like $[0,1]$? Can you use that and apply the periodicity premise to get a final answer for problem 2?
$endgroup$
– Ingix
Jan 14 at 11:17
$begingroup$
For problem 2), do you know anything about a maximum/minimum of a continuous function on a finite, closed interval, like $[0,1]$? Can you use that and apply the periodicity premise to get a final answer for problem 2?
$endgroup$
– Ingix
Jan 14 at 11:17
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Point 2 and 4 essentially boil down to the following fact:
Consider
$$
f|_{[0,1]}
$$
that is $f$ restricted to the interval $[0,1]$. This interval is closed and bounded, hence compact.
You have that continuous functions on compact intervals are bounded and uniformly continuous. Can you conclude using the fact, that $f$ is periodic with period $1$?
answered Jan 14 at 13:09
F. ConradF. Conrad
1,253412
$endgroup$
$begingroup$
Does this theorem- continuous functions on compact intervals are bounded and uniformly continuous , is valid in complex plane too?
$endgroup$
– Nilesh Khatri
Jan 15 at 11:39
$begingroup$
Yes, continuous functions on compact sets are uniformly continuous, even in the complex plane. In fact, it even holds for any compact metric space; see math.stackexchange.com/questions/110573/… for a more detailed answer.
$endgroup$
– F. Conrad
Jan 15 at 12:24
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Point 2 and 4 essentially boil down to the following fact:
Consider
$$
f|_{[0,1]}
$$
that is $f$ restricted to the interval $[0,1]$. This interval is closed and bounded, hence compact.
You have that continuous functions on compact intervals are bounded and uniformly continuous. Can you conclude using the fact, that $f$ is periodic with period $1$?
answered Jan 14 at 13:09
F. ConradF. Conrad
1,253412
$endgroup$
$begingroup$
Does this theorem- continuous functions on compact intervals are bounded and uniformly continuous , is valid in complex plane too?
$endgroup$
– Nilesh Khatri
Jan 15 at 11:39
$begingroup$
Yes, continuous functions on compact sets are uniformly continuous, even in the complex plane. In fact, it even holds for any compact metric space; see math.stackexchange.com/questions/110573/… for a more detailed answer.
$endgroup$
– F. Conrad
Jan 15 at 12:24
add a comment |
$begingroup$
Point 2 and 4 essentially boil down to the following fact:
Consider
$$
f|_{[0,1]}
$$
that is $f$ restricted to the interval $[0,1]$. This interval is closed and bounded, hence compact.
You have that continuous functions on compact intervals are bounded and uniformly continuous. Can you conclude using the fact, that $f$ is periodic with period $1$?
answered Jan 14 at 13:09
F. ConradF. Conrad
1,253412
$endgroup$
$begingroup$
Does this theorem- continuous functions on compact intervals are bounded and uniformly continuous , is valid in complex plane too?
$endgroup$
– Nilesh Khatri
Jan 15 at 11:39
$begingroup$
Yes, continuous functions on compact sets are uniformly continuous, even in the complex plane. In fact, it even holds for any compact metric space; see math.stackexchange.com/questions/110573/… for a more detailed answer.
$endgroup$
– F. Conrad
Jan 15 at 12:24
add a comment |
$begingroup$
Point 2 and 4 essentially boil down to the following fact:
Consider
$$
f|_{[0,1]}
$$
that is $f$ restricted to the interval $[0,1]$. This interval is closed and bounded, hence compact.
You have that continuous functions on compact intervals are bounded and uniformly continuous. Can you conclude using the fact, that $f$ is periodic with period $1$?
answered Jan 14 at 13:09
F. ConradF. Conrad
1,253412
$endgroup$
Point 2 and 4 essentially boil down to the following fact:
Consider
$$
f|_{[0,1]}
$$
that is $f$ restricted to the interval $[0,1]$. This interval is closed and bounded, hence compact.
You have that continuous functions on compact intervals are bounded and uniformly continuous. Can you conclude using the fact, that $f$ is periodic with period $1$?
answered Jan 14 at 13:09
F. ConradF. Conrad
1,253412
answered Jan 14 at 13:09
F. ConradF. Conrad
1,253412
answered Jan 14 at 13:09
F. ConradF. Conrad
1,253412
answered Jan 14 at 13:09
F. ConradF. Conrad
1,253412
1,253412
$begingroup$
Does this theorem- continuous functions on compact intervals are bounded and uniformly continuous , is valid in complex plane too?
$endgroup$
– Nilesh Khatri
Jan 15 at 11:39
$begingroup$
Yes, continuous functions on compact sets are uniformly continuous, even in the complex plane. In fact, it even holds for any compact metric space; see math.stackexchange.com/questions/110573/… for a more detailed answer.
$endgroup$
– F. Conrad
Jan 15 at 12:24
add a comment |
$begingroup$
Does this theorem- continuous functions on compact intervals are bounded and uniformly continuous , is valid in complex plane too?
$endgroup$
– Nilesh Khatri
Jan 15 at 11:39
$begingroup$
Yes, continuous functions on compact sets are uniformly continuous, even in the complex plane. In fact, it even holds for any compact metric space; see math.stackexchange.com/questions/110573/… for a more detailed answer.
$endgroup$
– F. Conrad
Jan 15 at 12:24
$begingroup$
Does this theorem- continuous functions on compact intervals are bounded and uniformly continuous , is valid in complex plane too?
$endgroup$
– Nilesh Khatri
Jan 15 at 11:39
$begingroup$
Does this theorem- continuous functions on compact intervals are bounded and uniformly continuous , is valid in complex plane too?
$endgroup$
– Nilesh Khatri
Jan 15 at 11:39
$begingroup$
Yes, continuous functions on compact sets are uniformly continuous, even in the complex plane. In fact, it even holds for any compact metric space; see math.stackexchange.com/questions/110573/… for a more detailed answer.
$endgroup$
– F. Conrad
Jan 15 at 12:24
$begingroup$
Yes, continuous functions on compact sets are uniformly continuous, even in the complex plane. In fact, it even holds for any compact metric space; see math.stackexchange.com/questions/110573/… for a more detailed answer.
$endgroup$
– F. Conrad
Jan 15 at 12:24
add a comment |
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1
$begingroup$
What have you tried so far?
$endgroup$
– Mindlack
Jan 14 at 10:20
1
$begingroup$
I took examples like f(x) = tan(pi*x)[that is periodic with period 1 and unbounded,so i think 1 option can be discarded]
$endgroup$
– Nilesh Khatri
Jan 14 at 10:24
1
$begingroup$
@NileshKhatri That's a good example to exclude option 1, except for a minor detail: It's not defined for $x = 0.5$. Easily fixable (you just define it to have your favourite value at that point), but it must be addressed. What about the others?
$endgroup$
– Arthur
Jan 14 at 10:26
1
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– Martin R
Jan 14 at 10:27
$begingroup$
For problem 2), do you know anything about a maximum/minimum of a continuous function on a finite, closed interval, like $[0,1]$? Can you use that and apply the periodicity premise to get a final answer for problem 2?
$endgroup$
– Ingix
Jan 14 at 11:17