$int_0^pi f(x+t)dt=int_0^left [ npi right ]f(x+t)dt=0$
Clash Royale CLAN TAG#URR8PPP
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Determine the continuous functions $f:mathbbRrightarrow mathbbR$ so that: $$int_0^pi f(x+t)dt=int_0^lfloor npi rfloor f(x+t)dt=0,forall xin mathbbR,nin mathbbN$$
All I've found is that $f = 0$ verifies.
Thank you!
calculus functions definite-integrals continuity
edited Apr 24 '17 at 18:53
copper.hat
123k557156
asked Apr 24 '17 at 18:06
George R.
1,417411
add a comment |Â
up vote
-1
down vote
favorite
Determine the continuous functions $f:mathbbRrightarrow mathbbR$ so that: $$int_0^pi f(x+t)dt=int_0^lfloor npi rfloor f(x+t)dt=0,forall xin mathbbR,nin mathbbN$$
All I've found is that $f = 0$ verifies.
Thank you!
calculus functions definite-integrals continuity
edited Apr 24 '17 at 18:53
copper.hat
123k557156
asked Apr 24 '17 at 18:06
George R.
1,417411
In the integral limit, is $[pi n]$ just a parenthesis, or does it denote the greatest integer (floor) function, or something else?
– Daniel Fischer♦
Apr 24 '17 at 18:24
It's the floor function.
– George R.
Apr 24 '17 at 18:32
Do you know that the group of periods of a continuous function is closed?
– Daniel Fischer♦
Apr 24 '17 at 18:38
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Determine the continuous functions $f:mathbbRrightarrow mathbbR$ so that: $$int_0^pi f(x+t)dt=int_0^lfloor npi rfloor f(x+t)dt=0,forall xin mathbbR,nin mathbbN$$
All I've found is that $f = 0$ verifies.
Thank you!
calculus functions definite-integrals continuity
edited Apr 24 '17 at 18:53
copper.hat
123k557156
asked Apr 24 '17 at 18:06
George R.
1,417411
Determine the continuous functions $f:mathbbRrightarrow mathbbR$ so that: $$int_0^pi f(x+t)dt=int_0^lfloor npi rfloor f(x+t)dt=0,forall xin mathbbR,nin mathbbN$$
All I've found is that $f = 0$ verifies.
Thank you!
calculus functions definite-integrals continuity
edited Apr 24 '17 at 18:53
copper.hat
123k557156
asked Apr 24 '17 at 18:06
George R.
1,417411
edited Apr 24 '17 at 18:53
copper.hat
123k557156
edited Apr 24 '17 at 18:53
copper.hat
123k557156
edited Apr 24 '17 at 18:53
copper.hat
123k557156
123k557156
asked Apr 24 '17 at 18:06
George R.
1,417411
asked Apr 24 '17 at 18:06
George R.
1,417411
asked Apr 24 '17 at 18:06
George R.
1,417411
1,417411
In the integral limit, is $[pi n]$ just a parenthesis, or does it denote the greatest integer (floor) function, or something else?
– Daniel Fischer♦
Apr 24 '17 at 18:24
It's the floor function.
– George R.
Apr 24 '17 at 18:32
Do you know that the group of periods of a continuous function is closed?
– Daniel Fischer♦
Apr 24 '17 at 18:38
add a comment |Â
In the integral limit, is $[pi n]$ just a parenthesis, or does it denote the greatest integer (floor) function, or something else?
– Daniel Fischer♦
Apr 24 '17 at 18:24
It's the floor function.
– George R.
Apr 24 '17 at 18:32
Do you know that the group of periods of a continuous function is closed?
– Daniel Fischer♦
Apr 24 '17 at 18:38
In the integral limit, is $[pi n]$ just a parenthesis, or does it denote the greatest integer (floor) function, or something else?
– Daniel Fischer♦
Apr 24 '17 at 18:24
In the integral limit, is $[pi n]$ just a parenthesis, or does it denote the greatest integer (floor) function, or something else?
– Daniel Fischer♦
Apr 24 '17 at 18:24
It's the floor function.
– George R.
Apr 24 '17 at 18:32
It's the floor function.
– George R.
Apr 24 '17 at 18:32
Do you know that the group of periods of a continuous function is closed?
– Daniel Fischer♦
Apr 24 '17 at 18:38
Do you know that the group of periods of a continuous function is closed?
– Daniel Fischer♦
Apr 24 '17 at 18:38
add a comment |Â
1 Answer
1
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votes
up vote
2
down vote
If $f colon mathbbR to mathbbR$ is continuous, and $L > 0$ such that
$$F(x) := int_0^L f(x+t),dt = int_x^x+L f(t),dttag$ast$$$
is constant, then $L$ is a period of $f$. That is easily seen by differentiating $(ast)$,
$$0 = F'(x) = f(x+L) - f(x)$$
for all $xin mathbbR$.
Your condition thus implies that $pi$ and $3$ (choosing $n = 1$) are periods of $f$.
Hence $S = kcdot 3 + mcdot pi : k,m in mathbbZ$ is contained in the set of periods of $f$. Since for a continuous function $g colon mathbbR to mathbbR$ the set
$$P(g) = bigl a in mathbbR : (forall xin mathbbR)bigl(g(x+a) = g(x)bigr)bigr$$
of all periods of $g$ is closed, and $S subset P(f)$ is dense in $mathbbR$, it follows that $P(f) = mathbbR$, i.e. $f$ is constant. With $int_0^pi f(t),dt = 0$ it then follows that $fequiv 0$.
answered Apr 24 '17 at 20:45
Daniel Fischer♦
172k16156278
How do we know that $P(g)$ is closed?
– CuriousKid7
Aug 23 at 22:40
Depends on your favourite way to show that a set is closed. The complement is open: If $a notin P(g)$, then there is an $x in mathbbR$ with $f(x) neq f(x+a)$. By continuity of $b mapsto f(x+b)$ there is $varepsilon > 0$ with $f(x) neq f(x+b)$ for all $b$ with $lvert b-arvert < varepsilon$, so $(a-varepsilon, a+varepsilon) cap P(g) = varnothing$. It is the intersection of a family of closed sets: For $x in mathbbR$ let $T(x) = a in mathbbR : f(x+a) = f(x) = f^-1(f(x)) - x$. Then $P(g) = bigcap_x T(x)$.
– Daniel Fischer♦
Aug 25 at 12:32
What do you mean by $f^-1(f(x)) - x$?
– CuriousKid7
Aug 26 at 2:13
1
@CuriousKid7 Sorry, I too often forget that the $A pm x$ notation for translates isn't universally known. $$f^-1(f(x)) - x = y - x : y in f^-1(f(x)) = y - x : f(y) = f(x)$$ It's the set $ a : f(x+a) = f(x)$.
– Daniel Fischer♦
Aug 26 at 12:52
Ok, then it's the preimage of the closed set $f^-1(f(x))$ under addition by $x$, which is continuous. Sounds good.
– CuriousKid7
Aug 26 at 17:28
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If $f colon mathbbR to mathbbR$ is continuous, and $L > 0$ such that
$$F(x) := int_0^L f(x+t),dt = int_x^x+L f(t),dttag$ast$$$
is constant, then $L$ is a period of $f$. That is easily seen by differentiating $(ast)$,
$$0 = F'(x) = f(x+L) - f(x)$$
for all $xin mathbbR$.
Your condition thus implies that $pi$ and $3$ (choosing $n = 1$) are periods of $f$.
Hence $S = kcdot 3 + mcdot pi : k,m in mathbbZ$ is contained in the set of periods of $f$. Since for a continuous function $g colon mathbbR to mathbbR$ the set
$$P(g) = bigl a in mathbbR : (forall xin mathbbR)bigl(g(x+a) = g(x)bigr)bigr$$
of all periods of $g$ is closed, and $S subset P(f)$ is dense in $mathbbR$, it follows that $P(f) = mathbbR$, i.e. $f$ is constant. With $int_0^pi f(t),dt = 0$ it then follows that $fequiv 0$.
answered Apr 24 '17 at 20:45
Daniel Fischer♦
172k16156278
How do we know that $P(g)$ is closed?
– CuriousKid7
Aug 23 at 22:40
Depends on your favourite way to show that a set is closed. The complement is open: If $a notin P(g)$, then there is an $x in mathbbR$ with $f(x) neq f(x+a)$. By continuity of $b mapsto f(x+b)$ there is $varepsilon > 0$ with $f(x) neq f(x+b)$ for all $b$ with $lvert b-arvert < varepsilon$, so $(a-varepsilon, a+varepsilon) cap P(g) = varnothing$. It is the intersection of a family of closed sets: For $x in mathbbR$ let $T(x) = a in mathbbR : f(x+a) = f(x) = f^-1(f(x)) - x$. Then $P(g) = bigcap_x T(x)$.
– Daniel Fischer♦
Aug 25 at 12:32
What do you mean by $f^-1(f(x)) - x$?
– CuriousKid7
Aug 26 at 2:13
1
@CuriousKid7 Sorry, I too often forget that the $A pm x$ notation for translates isn't universally known. $$f^-1(f(x)) - x = y - x : y in f^-1(f(x)) = y - x : f(y) = f(x)$$ It's the set $ a : f(x+a) = f(x)$.
– Daniel Fischer♦
Aug 26 at 12:52
Ok, then it's the preimage of the closed set $f^-1(f(x))$ under addition by $x$, which is continuous. Sounds good.
– CuriousKid7
Aug 26 at 17:28
 |Â
show 2 more comments
up vote
2
down vote
If $f colon mathbbR to mathbbR$ is continuous, and $L > 0$ such that
$$F(x) := int_0^L f(x+t),dt = int_x^x+L f(t),dttag$ast$$$
is constant, then $L$ is a period of $f$. That is easily seen by differentiating $(ast)$,
$$0 = F'(x) = f(x+L) - f(x)$$
for all $xin mathbbR$.
Your condition thus implies that $pi$ and $3$ (choosing $n = 1$) are periods of $f$.
Hence $S = kcdot 3 + mcdot pi : k,m in mathbbZ$ is contained in the set of periods of $f$. Since for a continuous function $g colon mathbbR to mathbbR$ the set
$$P(g) = bigl a in mathbbR : (forall xin mathbbR)bigl(g(x+a) = g(x)bigr)bigr$$
of all periods of $g$ is closed, and $S subset P(f)$ is dense in $mathbbR$, it follows that $P(f) = mathbbR$, i.e. $f$ is constant. With $int_0^pi f(t),dt = 0$ it then follows that $fequiv 0$.
answered Apr 24 '17 at 20:45
Daniel Fischer♦
172k16156278
How do we know that $P(g)$ is closed?
– CuriousKid7
Aug 23 at 22:40
Depends on your favourite way to show that a set is closed. The complement is open: If $a notin P(g)$, then there is an $x in mathbbR$ with $f(x) neq f(x+a)$. By continuity of $b mapsto f(x+b)$ there is $varepsilon > 0$ with $f(x) neq f(x+b)$ for all $b$ with $lvert b-arvert < varepsilon$, so $(a-varepsilon, a+varepsilon) cap P(g) = varnothing$. It is the intersection of a family of closed sets: For $x in mathbbR$ let $T(x) = a in mathbbR : f(x+a) = f(x) = f^-1(f(x)) - x$. Then $P(g) = bigcap_x T(x)$.
– Daniel Fischer♦
Aug 25 at 12:32
What do you mean by $f^-1(f(x)) - x$?
– CuriousKid7
Aug 26 at 2:13
1
@CuriousKid7 Sorry, I too often forget that the $A pm x$ notation for translates isn't universally known. $$f^-1(f(x)) - x = y - x : y in f^-1(f(x)) = y - x : f(y) = f(x)$$ It's the set $ a : f(x+a) = f(x)$.
– Daniel Fischer♦
Aug 26 at 12:52
Ok, then it's the preimage of the closed set $f^-1(f(x))$ under addition by $x$, which is continuous. Sounds good.
– CuriousKid7
Aug 26 at 17:28
 |Â
show 2 more comments
up vote
2
down vote
up vote
2
down vote
If $f colon mathbbR to mathbbR$ is continuous, and $L > 0$ such that
$$F(x) := int_0^L f(x+t),dt = int_x^x+L f(t),dttag$ast$$$
is constant, then $L$ is a period of $f$. That is easily seen by differentiating $(ast)$,
$$0 = F'(x) = f(x+L) - f(x)$$
for all $xin mathbbR$.
Your condition thus implies that $pi$ and $3$ (choosing $n = 1$) are periods of $f$.
Hence $S = kcdot 3 + mcdot pi : k,m in mathbbZ$ is contained in the set of periods of $f$. Since for a continuous function $g colon mathbbR to mathbbR$ the set
$$P(g) = bigl a in mathbbR : (forall xin mathbbR)bigl(g(x+a) = g(x)bigr)bigr$$
of all periods of $g$ is closed, and $S subset P(f)$ is dense in $mathbbR$, it follows that $P(f) = mathbbR$, i.e. $f$ is constant. With $int_0^pi f(t),dt = 0$ it then follows that $fequiv 0$.
answered Apr 24 '17 at 20:45
Daniel Fischer♦
172k16156278
If $f colon mathbbR to mathbbR$ is continuous, and $L > 0$ such that
$$F(x) := int_0^L f(x+t),dt = int_x^x+L f(t),dttag$ast$$$
is constant, then $L$ is a period of $f$. That is easily seen by differentiating $(ast)$,
$$0 = F'(x) = f(x+L) - f(x)$$
for all $xin mathbbR$.
Your condition thus implies that $pi$ and $3$ (choosing $n = 1$) are periods of $f$.
Hence $S = kcdot 3 + mcdot pi : k,m in mathbbZ$ is contained in the set of periods of $f$. Since for a continuous function $g colon mathbbR to mathbbR$ the set
$$P(g) = bigl a in mathbbR : (forall xin mathbbR)bigl(g(x+a) = g(x)bigr)bigr$$
of all periods of $g$ is closed, and $S subset P(f)$ is dense in $mathbbR$, it follows that $P(f) = mathbbR$, i.e. $f$ is constant. With $int_0^pi f(t),dt = 0$ it then follows that $fequiv 0$.
answered Apr 24 '17 at 20:45
Daniel Fischer♦
172k16156278
edited Aug 26 at 17:45
answered Apr 24 '17 at 20:45
Daniel Fischer♦
172k16156278
answered Apr 24 '17 at 20:45
Daniel Fischer♦
172k16156278
answered Apr 24 '17 at 20:45
Daniel Fischer♦
172k16156278
172k16156278
How do we know that $P(g)$ is closed?
– CuriousKid7
Aug 23 at 22:40
Depends on your favourite way to show that a set is closed. The complement is open: If $a notin P(g)$, then there is an $x in mathbbR$ with $f(x) neq f(x+a)$. By continuity of $b mapsto f(x+b)$ there is $varepsilon > 0$ with $f(x) neq f(x+b)$ for all $b$ with $lvert b-arvert < varepsilon$, so $(a-varepsilon, a+varepsilon) cap P(g) = varnothing$. It is the intersection of a family of closed sets: For $x in mathbbR$ let $T(x) = a in mathbbR : f(x+a) = f(x) = f^-1(f(x)) - x$. Then $P(g) = bigcap_x T(x)$.
– Daniel Fischer♦
Aug 25 at 12:32
What do you mean by $f^-1(f(x)) - x$?
– CuriousKid7
Aug 26 at 2:13
1
@CuriousKid7 Sorry, I too often forget that the $A pm x$ notation for translates isn't universally known. $$f^-1(f(x)) - x = y - x : y in f^-1(f(x)) = y - x : f(y) = f(x)$$ It's the set $ a : f(x+a) = f(x)$.
– Daniel Fischer♦
Aug 26 at 12:52
Ok, then it's the preimage of the closed set $f^-1(f(x))$ under addition by $x$, which is continuous. Sounds good.
– CuriousKid7
Aug 26 at 17:28
 |Â
show 2 more comments
How do we know that $P(g)$ is closed?
– CuriousKid7
Aug 23 at 22:40
Depends on your favourite way to show that a set is closed. The complement is open: If $a notin P(g)$, then there is an $x in mathbbR$ with $f(x) neq f(x+a)$. By continuity of $b mapsto f(x+b)$ there is $varepsilon > 0$ with $f(x) neq f(x+b)$ for all $b$ with $lvert b-arvert < varepsilon$, so $(a-varepsilon, a+varepsilon) cap P(g) = varnothing$. It is the intersection of a family of closed sets: For $x in mathbbR$ let $T(x) = a in mathbbR : f(x+a) = f(x) = f^-1(f(x)) - x$. Then $P(g) = bigcap_x T(x)$.
– Daniel Fischer♦
Aug 25 at 12:32
What do you mean by $f^-1(f(x)) - x$?
– CuriousKid7
Aug 26 at 2:13
1
@CuriousKid7 Sorry, I too often forget that the $A pm x$ notation for translates isn't universally known. $$f^-1(f(x)) - x = y - x : y in f^-1(f(x)) = y - x : f(y) = f(x)$$ It's the set $ a : f(x+a) = f(x)$.
– Daniel Fischer♦
Aug 26 at 12:52
Ok, then it's the preimage of the closed set $f^-1(f(x))$ under addition by $x$, which is continuous. Sounds good.
– CuriousKid7
Aug 26 at 17:28
How do we know that $P(g)$ is closed?
– CuriousKid7
Aug 23 at 22:40
How do we know that $P(g)$ is closed?
– CuriousKid7
Aug 23 at 22:40
Depends on your favourite way to show that a set is closed. The complement is open: If $a notin P(g)$, then there is an $x in mathbbR$ with $f(x) neq f(x+a)$. By continuity of $b mapsto f(x+b)$ there is $varepsilon > 0$ with $f(x) neq f(x+b)$ for all $b$ with $lvert b-arvert < varepsilon$, so $(a-varepsilon, a+varepsilon) cap P(g) = varnothing$. It is the intersection of a family of closed sets: For $x in mathbbR$ let $T(x) = a in mathbbR : f(x+a) = f(x) = f^-1(f(x)) - x$. Then $P(g) = bigcap_x T(x)$.
– Daniel Fischer♦
Aug 25 at 12:32
Depends on your favourite way to show that a set is closed. The complement is open: If $a notin P(g)$, then there is an $x in mathbbR$ with $f(x) neq f(x+a)$. By continuity of $b mapsto f(x+b)$ there is $varepsilon > 0$ with $f(x) neq f(x+b)$ for all $b$ with $lvert b-arvert < varepsilon$, so $(a-varepsilon, a+varepsilon) cap P(g) = varnothing$. It is the intersection of a family of closed sets: For $x in mathbbR$ let $T(x) = a in mathbbR : f(x+a) = f(x) = f^-1(f(x)) - x$. Then $P(g) = bigcap_x T(x)$.
– Daniel Fischer♦
Aug 25 at 12:32
What do you mean by $f^-1(f(x)) - x$?
– CuriousKid7
Aug 26 at 2:13
What do you mean by $f^-1(f(x)) - x$?
– CuriousKid7
Aug 26 at 2:13
1
1
@CuriousKid7 Sorry, I too often forget that the $A pm x$ notation for translates isn't universally known. $$f^-1(f(x)) - x = y - x : y in f^-1(f(x)) = y - x : f(y) = f(x)$$ It's the set $ a : f(x+a) = f(x)$.
– Daniel Fischer♦
Aug 26 at 12:52
@CuriousKid7 Sorry, I too often forget that the $A pm x$ notation for translates isn't universally known. $$f^-1(f(x)) - x = y - x : y in f^-1(f(x)) = y - x : f(y) = f(x)$$ It's the set $ a : f(x+a) = f(x)$.
– Daniel Fischer♦
Aug 26 at 12:52
Ok, then it's the preimage of the closed set $f^-1(f(x))$ under addition by $x$, which is continuous. Sounds good.
– CuriousKid7
Aug 26 at 17:28
Ok, then it's the preimage of the closed set $f^-1(f(x))$ under addition by $x$, which is continuous. Sounds good.
– CuriousKid7
Aug 26 at 17:28
 |Â
show 2 more comments
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In the integral limit, is $[pi n]$ just a parenthesis, or does it denote the greatest integer (floor) function, or something else?
– Daniel Fischer♦
Apr 24 '17 at 18:24
It's the floor function.
– George R.
Apr 24 '17 at 18:32
Do you know that the group of periods of a continuous function is closed?
– Daniel Fischer♦
Apr 24 '17 at 18:38