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Surjectivity of the integral operator
Clash Royale CLAN TAG#URR8PPP
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Let $[a,b]subseteqmathbbR$ be a closed interval and $mathcalR[a,b]$ denote the linear space of Riemann integrable functions. It is well known that the integral operator
$$R(f)(x):=(Rf)(a)+int_a^xf(t),dt$$
maps elements of $mathcalR[a,b]$ to the space of continuous functions on $[a,b]$ i.e. $R:mathcalR[a,b]to C[a,b]$.
My questions are: Is this map surjective? If not, is there a counterexample?
real-analysis continuity riemann-integration
edited 2 days ago
John Ma
37.5k93669
asked Aug 3 at 13:26
Arian
5,235817
add a comment |Â
up vote
1
down vote
favorite
Let $[a,b]subseteqmathbbR$ be a closed interval and $mathcalR[a,b]$ denote the linear space of Riemann integrable functions. It is well known that the integral operator
$$R(f)(x):=(Rf)(a)+int_a^xf(t),dt$$
maps elements of $mathcalR[a,b]$ to the space of continuous functions on $[a,b]$ i.e. $R:mathcalR[a,b]to C[a,b]$.
My questions are: Is this map surjective? If not, is there a counterexample?
real-analysis continuity riemann-integration
edited 2 days ago
John Ma
37.5k93669
asked Aug 3 at 13:26
Arian
5,235817
I presume that you want elements of $C[a,b]$ to vanish at $a$; otherwise, the map isn't injective for an even more trivial reason---any nonzero constant function is not in the image of $R$.
– Batominovski
Aug 3 at 14:26
@Batominovski thanks for pointing this out. Now each constant function has as preimage the identically zero function. And of course as you said if we want $R$ to be also injective we should invoke $(Rf)(a)=0$ for all $f$. Seems like the Weierstrass function as suggested by Quantic is a valid counterexample?
– Arian
Aug 3 at 15:45
Opps, I meant surjective, not injective. And yes, Quantic_Solver gave a good example, given that you shift the function to be zero at $a$.
– Batominovski
Aug 3 at 15:46
The definition of $R(f)$ has trouble: What does $R(f)(a)$ even mean?
– zhw.
2 days ago
@zhw. The OP wants to take the quotient of $C[a,b]$ by constant functions, but I agree that the presentation here looks confusing.
– Batominovski
2 days ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $[a,b]subseteqmathbbR$ be a closed interval and $mathcalR[a,b]$ denote the linear space of Riemann integrable functions. It is well known that the integral operator
$$R(f)(x):=(Rf)(a)+int_a^xf(t),dt$$
maps elements of $mathcalR[a,b]$ to the space of continuous functions on $[a,b]$ i.e. $R:mathcalR[a,b]to C[a,b]$.
My questions are: Is this map surjective? If not, is there a counterexample?
real-analysis continuity riemann-integration
edited 2 days ago
John Ma
37.5k93669
asked Aug 3 at 13:26
Arian
5,235817
Let $[a,b]subseteqmathbbR$ be a closed interval and $mathcalR[a,b]$ denote the linear space of Riemann integrable functions. It is well known that the integral operator
$$R(f)(x):=(Rf)(a)+int_a^xf(t),dt$$
maps elements of $mathcalR[a,b]$ to the space of continuous functions on $[a,b]$ i.e. $R:mathcalR[a,b]to C[a,b]$.
My questions are: Is this map surjective? If not, is there a counterexample?
real-analysis continuity riemann-integration
edited 2 days ago
John Ma
37.5k93669
asked Aug 3 at 13:26
Arian
5,235817
edited 2 days ago
John Ma
37.5k93669
edited 2 days ago
John Ma
37.5k93669
edited 2 days ago
John Ma
37.5k93669
37.5k93669
asked Aug 3 at 13:26
Arian
5,235817
asked Aug 3 at 13:26
Arian
5,235817
asked Aug 3 at 13:26
Arian
5,235817
5,235817
I presume that you want elements of $C[a,b]$ to vanish at $a$; otherwise, the map isn't injective for an even more trivial reason---any nonzero constant function is not in the image of $R$.
– Batominovski
Aug 3 at 14:26
@Batominovski thanks for pointing this out. Now each constant function has as preimage the identically zero function. And of course as you said if we want $R$ to be also injective we should invoke $(Rf)(a)=0$ for all $f$. Seems like the Weierstrass function as suggested by Quantic is a valid counterexample?
– Arian
Aug 3 at 15:45
Opps, I meant surjective, not injective. And yes, Quantic_Solver gave a good example, given that you shift the function to be zero at $a$.
– Batominovski
Aug 3 at 15:46
The definition of $R(f)$ has trouble: What does $R(f)(a)$ even mean?
– zhw.
2 days ago
@zhw. The OP wants to take the quotient of $C[a,b]$ by constant functions, but I agree that the presentation here looks confusing.
– Batominovski
2 days ago
add a comment |Â
I presume that you want elements of $C[a,b]$ to vanish at $a$; otherwise, the map isn't injective for an even more trivial reason---any nonzero constant function is not in the image of $R$.
– Batominovski
Aug 3 at 14:26
@Batominovski thanks for pointing this out. Now each constant function has as preimage the identically zero function. And of course as you said if we want $R$ to be also injective we should invoke $(Rf)(a)=0$ for all $f$. Seems like the Weierstrass function as suggested by Quantic is a valid counterexample?
– Arian
Aug 3 at 15:45
Opps, I meant surjective, not injective. And yes, Quantic_Solver gave a good example, given that you shift the function to be zero at $a$.
– Batominovski
Aug 3 at 15:46
The definition of $R(f)$ has trouble: What does $R(f)(a)$ even mean?
– zhw.
2 days ago
@zhw. The OP wants to take the quotient of $C[a,b]$ by constant functions, but I agree that the presentation here looks confusing.
– Batominovski
2 days ago
I presume that you want elements of $C[a,b]$ to vanish at $a$; otherwise, the map isn't injective for an even more trivial reason---any nonzero constant function is not in the image of $R$.
– Batominovski
Aug 3 at 14:26
I presume that you want elements of $C[a,b]$ to vanish at $a$; otherwise, the map isn't injective for an even more trivial reason---any nonzero constant function is not in the image of $R$.
– Batominovski
Aug 3 at 14:26
@Batominovski thanks for pointing this out. Now each constant function has as preimage the identically zero function. And of course as you said if we want $R$ to be also injective we should invoke $(Rf)(a)=0$ for all $f$. Seems like the Weierstrass function as suggested by Quantic is a valid counterexample?
– Arian
Aug 3 at 15:45
@Batominovski thanks for pointing this out. Now each constant function has as preimage the identically zero function. And of course as you said if we want $R$ to be also injective we should invoke $(Rf)(a)=0$ for all $f$. Seems like the Weierstrass function as suggested by Quantic is a valid counterexample?
– Arian
Aug 3 at 15:45
Opps, I meant surjective, not injective. And yes, Quantic_Solver gave a good example, given that you shift the function to be zero at $a$.
– Batominovski
Aug 3 at 15:46
Opps, I meant surjective, not injective. And yes, Quantic_Solver gave a good example, given that you shift the function to be zero at $a$.
– Batominovski
Aug 3 at 15:46
The definition of $R(f)$ has trouble: What does $R(f)(a)$ even mean?
– zhw.
2 days ago
The definition of $R(f)$ has trouble: What does $R(f)(a)$ even mean?
– zhw.
2 days ago
@zhw. The OP wants to take the quotient of $C[a,b]$ by constant functions, but I agree that the presentation here looks confusing.
– Batominovski
2 days ago
@zhw. The OP wants to take the quotient of $C[a,b]$ by constant functions, but I agree that the presentation here looks confusing.
– Batominovski
2 days ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
0
down vote
accepted
Assume that $mathcalCbig([a,b]big)$ is taken modulo constant functions. Any function in the image of $R$ is differentiable everywhere except on a set of Lebesgue measure $0$, but not every continuous function has this property. A properly shifted and rescaled version of Weierstrass function as suggested by Quintic_Solver is a continuous function which is differentiable nowhere, so it does not lie in $textim(R)$.
However, you can shift and rescale the Cantor function to fit your interval $[a,b]$ to get a continuous function on $[a,b]$ that is not in the image of $R$, but the Cantor function is differentiable everywhere outside the Cantor set, which has measure $0$. Therefore, the condition that a function is differentiable outside a set of Lebesgue measure $0$ is also insufficient to be in $textim(R)$. Indeed, $textim(R)$ consists of only (but not all) absolutely continuous functions on $[a,b]$ vanishing at $a$ (and the Cantor function is not absolutely continuous).
Being absolutely continuous is also insufficient. As zhw. illustrated, the function $g(x):=sqrtx-a$ is absolutely continuous, but any of its weak derivative is unbounded. Therefore, $g'$ is not Riemann-integrable. Every absolutely continuous function has a weak derivative. We know that every function in $textim(R)$ is absolutely continuous. Thus, each function in $textim(R)$ has a bounded weak derivative, whence Lipschitz continuous.
To get all the image of $R$, we note that a (real- or complex-valued) function $f$ on $[a,b]$ is Riemann-integrable if and only if it is bounded and continuous outside a set of Lebesgue measure $0$. Hence, $Fintextim(R)$ iff a weak derivative $F'$ of $F$ is bounded and continuous outside a set of Lebesgue measure $0$ ($F'$ exists because $F$ is absolutely continuous). This translates as follows: $textim(R)$ consists of all functions $F$ that is Lipschitz continuous and continuously differentiable everywhere outside a set of Lebesgue measure $0$.
answered Aug 3 at 14:19
Batominovski
22.6k22776
1
A step function is Riemann integrable, but its integral function won't be differentiable where the step function jumps.
– Daniel Schepler
Aug 3 at 20:49
add a comment |Â
up vote
1
down vote
It's easy to see every function in the range is Lipschitz, but there are plenty of functions in $C[a,b]$ that are not Lipschitz. For example, $g(x)=sqrt x-a.$ Hence the integral operator is not surjective.
answered 2 days ago
zhw.
65k42769
add a comment |Â
up vote
0
down vote
The map isn't surjective. A counterexample is the Weierstrass function which is continuous.
answered Aug 3 at 14:19
Quantic_Solver
214
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Assume that $mathcalCbig([a,b]big)$ is taken modulo constant functions. Any function in the image of $R$ is differentiable everywhere except on a set of Lebesgue measure $0$, but not every continuous function has this property. A properly shifted and rescaled version of Weierstrass function as suggested by Quintic_Solver is a continuous function which is differentiable nowhere, so it does not lie in $textim(R)$.
However, you can shift and rescale the Cantor function to fit your interval $[a,b]$ to get a continuous function on $[a,b]$ that is not in the image of $R$, but the Cantor function is differentiable everywhere outside the Cantor set, which has measure $0$. Therefore, the condition that a function is differentiable outside a set of Lebesgue measure $0$ is also insufficient to be in $textim(R)$. Indeed, $textim(R)$ consists of only (but not all) absolutely continuous functions on $[a,b]$ vanishing at $a$ (and the Cantor function is not absolutely continuous).
Being absolutely continuous is also insufficient. As zhw. illustrated, the function $g(x):=sqrtx-a$ is absolutely continuous, but any of its weak derivative is unbounded. Therefore, $g'$ is not Riemann-integrable. Every absolutely continuous function has a weak derivative. We know that every function in $textim(R)$ is absolutely continuous. Thus, each function in $textim(R)$ has a bounded weak derivative, whence Lipschitz continuous.
To get all the image of $R$, we note that a (real- or complex-valued) function $f$ on $[a,b]$ is Riemann-integrable if and only if it is bounded and continuous outside a set of Lebesgue measure $0$. Hence, $Fintextim(R)$ iff a weak derivative $F'$ of $F$ is bounded and continuous outside a set of Lebesgue measure $0$ ($F'$ exists because $F$ is absolutely continuous). This translates as follows: $textim(R)$ consists of all functions $F$ that is Lipschitz continuous and continuously differentiable everywhere outside a set of Lebesgue measure $0$.
answered Aug 3 at 14:19
Batominovski
22.6k22776
1
A step function is Riemann integrable, but its integral function won't be differentiable where the step function jumps.
– Daniel Schepler
Aug 3 at 20:49
add a comment |Â
up vote
0
down vote
accepted
Assume that $mathcalCbig([a,b]big)$ is taken modulo constant functions. Any function in the image of $R$ is differentiable everywhere except on a set of Lebesgue measure $0$, but not every continuous function has this property. A properly shifted and rescaled version of Weierstrass function as suggested by Quintic_Solver is a continuous function which is differentiable nowhere, so it does not lie in $textim(R)$.
However, you can shift and rescale the Cantor function to fit your interval $[a,b]$ to get a continuous function on $[a,b]$ that is not in the image of $R$, but the Cantor function is differentiable everywhere outside the Cantor set, which has measure $0$. Therefore, the condition that a function is differentiable outside a set of Lebesgue measure $0$ is also insufficient to be in $textim(R)$. Indeed, $textim(R)$ consists of only (but not all) absolutely continuous functions on $[a,b]$ vanishing at $a$ (and the Cantor function is not absolutely continuous).
Being absolutely continuous is also insufficient. As zhw. illustrated, the function $g(x):=sqrtx-a$ is absolutely continuous, but any of its weak derivative is unbounded. Therefore, $g'$ is not Riemann-integrable. Every absolutely continuous function has a weak derivative. We know that every function in $textim(R)$ is absolutely continuous. Thus, each function in $textim(R)$ has a bounded weak derivative, whence Lipschitz continuous.
To get all the image of $R$, we note that a (real- or complex-valued) function $f$ on $[a,b]$ is Riemann-integrable if and only if it is bounded and continuous outside a set of Lebesgue measure $0$. Hence, $Fintextim(R)$ iff a weak derivative $F'$ of $F$ is bounded and continuous outside a set of Lebesgue measure $0$ ($F'$ exists because $F$ is absolutely continuous). This translates as follows: $textim(R)$ consists of all functions $F$ that is Lipschitz continuous and continuously differentiable everywhere outside a set of Lebesgue measure $0$.
answered Aug 3 at 14:19
Batominovski
22.6k22776
1
A step function is Riemann integrable, but its integral function won't be differentiable where the step function jumps.
– Daniel Schepler
Aug 3 at 20:49
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Assume that $mathcalCbig([a,b]big)$ is taken modulo constant functions. Any function in the image of $R$ is differentiable everywhere except on a set of Lebesgue measure $0$, but not every continuous function has this property. A properly shifted and rescaled version of Weierstrass function as suggested by Quintic_Solver is a continuous function which is differentiable nowhere, so it does not lie in $textim(R)$.
However, you can shift and rescale the Cantor function to fit your interval $[a,b]$ to get a continuous function on $[a,b]$ that is not in the image of $R$, but the Cantor function is differentiable everywhere outside the Cantor set, which has measure $0$. Therefore, the condition that a function is differentiable outside a set of Lebesgue measure $0$ is also insufficient to be in $textim(R)$. Indeed, $textim(R)$ consists of only (but not all) absolutely continuous functions on $[a,b]$ vanishing at $a$ (and the Cantor function is not absolutely continuous).
Being absolutely continuous is also insufficient. As zhw. illustrated, the function $g(x):=sqrtx-a$ is absolutely continuous, but any of its weak derivative is unbounded. Therefore, $g'$ is not Riemann-integrable. Every absolutely continuous function has a weak derivative. We know that every function in $textim(R)$ is absolutely continuous. Thus, each function in $textim(R)$ has a bounded weak derivative, whence Lipschitz continuous.
To get all the image of $R$, we note that a (real- or complex-valued) function $f$ on $[a,b]$ is Riemann-integrable if and only if it is bounded and continuous outside a set of Lebesgue measure $0$. Hence, $Fintextim(R)$ iff a weak derivative $F'$ of $F$ is bounded and continuous outside a set of Lebesgue measure $0$ ($F'$ exists because $F$ is absolutely continuous). This translates as follows: $textim(R)$ consists of all functions $F$ that is Lipschitz continuous and continuously differentiable everywhere outside a set of Lebesgue measure $0$.
answered Aug 3 at 14:19
Batominovski
22.6k22776
Assume that $mathcalCbig([a,b]big)$ is taken modulo constant functions. Any function in the image of $R$ is differentiable everywhere except on a set of Lebesgue measure $0$, but not every continuous function has this property. A properly shifted and rescaled version of Weierstrass function as suggested by Quintic_Solver is a continuous function which is differentiable nowhere, so it does not lie in $textim(R)$.
However, you can shift and rescale the Cantor function to fit your interval $[a,b]$ to get a continuous function on $[a,b]$ that is not in the image of $R$, but the Cantor function is differentiable everywhere outside the Cantor set, which has measure $0$. Therefore, the condition that a function is differentiable outside a set of Lebesgue measure $0$ is also insufficient to be in $textim(R)$. Indeed, $textim(R)$ consists of only (but not all) absolutely continuous functions on $[a,b]$ vanishing at $a$ (and the Cantor function is not absolutely continuous).
Being absolutely continuous is also insufficient. As zhw. illustrated, the function $g(x):=sqrtx-a$ is absolutely continuous, but any of its weak derivative is unbounded. Therefore, $g'$ is not Riemann-integrable. Every absolutely continuous function has a weak derivative. We know that every function in $textim(R)$ is absolutely continuous. Thus, each function in $textim(R)$ has a bounded weak derivative, whence Lipschitz continuous.
To get all the image of $R$, we note that a (real- or complex-valued) function $f$ on $[a,b]$ is Riemann-integrable if and only if it is bounded and continuous outside a set of Lebesgue measure $0$. Hence, $Fintextim(R)$ iff a weak derivative $F'$ of $F$ is bounded and continuous outside a set of Lebesgue measure $0$ ($F'$ exists because $F$ is absolutely continuous). This translates as follows: $textim(R)$ consists of all functions $F$ that is Lipschitz continuous and continuously differentiable everywhere outside a set of Lebesgue measure $0$.
answered Aug 3 at 14:19
Batominovski
22.6k22776
edited 2 days ago
answered Aug 3 at 14:19
Batominovski
22.6k22776
answered Aug 3 at 14:19
Batominovski
22.6k22776
answered Aug 3 at 14:19
Batominovski
22.6k22776
22.6k22776
1
A step function is Riemann integrable, but its integral function won't be differentiable where the step function jumps.
– Daniel Schepler
Aug 3 at 20:49
add a comment |Â
1
A step function is Riemann integrable, but its integral function won't be differentiable where the step function jumps.
– Daniel Schepler
Aug 3 at 20:49
1
1
A step function is Riemann integrable, but its integral function won't be differentiable where the step function jumps.
– Daniel Schepler
Aug 3 at 20:49
A step function is Riemann integrable, but its integral function won't be differentiable where the step function jumps.
– Daniel Schepler
Aug 3 at 20:49
add a comment |Â
up vote
1
down vote
It's easy to see every function in the range is Lipschitz, but there are plenty of functions in $C[a,b]$ that are not Lipschitz. For example, $g(x)=sqrt x-a.$ Hence the integral operator is not surjective.
answered 2 days ago
zhw.
65k42769
add a comment |Â
up vote
1
down vote
It's easy to see every function in the range is Lipschitz, but there are plenty of functions in $C[a,b]$ that are not Lipschitz. For example, $g(x)=sqrt x-a.$ Hence the integral operator is not surjective.
answered 2 days ago
zhw.
65k42769
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It's easy to see every function in the range is Lipschitz, but there are plenty of functions in $C[a,b]$ that are not Lipschitz. For example, $g(x)=sqrt x-a.$ Hence the integral operator is not surjective.
answered 2 days ago
zhw.
65k42769
It's easy to see every function in the range is Lipschitz, but there are plenty of functions in $C[a,b]$ that are not Lipschitz. For example, $g(x)=sqrt x-a.$ Hence the integral operator is not surjective.
answered 2 days ago
zhw.
65k42769
answered 2 days ago
zhw.
65k42769
answered 2 days ago
zhw.
65k42769
answered 2 days ago
zhw.
65k42769
65k42769
add a comment |Â
add a comment |Â
up vote
0
down vote
The map isn't surjective. A counterexample is the Weierstrass function which is continuous.
answered Aug 3 at 14:19
Quantic_Solver
214
add a comment |Â
up vote
0
down vote
The map isn't surjective. A counterexample is the Weierstrass function which is continuous.
answered Aug 3 at 14:19
Quantic_Solver
214
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The map isn't surjective. A counterexample is the Weierstrass function which is continuous.
answered Aug 3 at 14:19
Quantic_Solver
214
The map isn't surjective. A counterexample is the Weierstrass function which is continuous.
answered Aug 3 at 14:19
Quantic_Solver
214
answered Aug 3 at 14:19
Quantic_Solver
214
answered Aug 3 at 14:19
Quantic_Solver
214
answered Aug 3 at 14:19
Quantic_Solver
214
214
add a comment |Â
add a comment |Â
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I presume that you want elements of $C[a,b]$ to vanish at $a$; otherwise, the map isn't injective for an even more trivial reason---any nonzero constant function is not in the image of $R$.
– Batominovski
Aug 3 at 14:26
@Batominovski thanks for pointing this out. Now each constant function has as preimage the identically zero function. And of course as you said if we want $R$ to be also injective we should invoke $(Rf)(a)=0$ for all $f$. Seems like the Weierstrass function as suggested by Quantic is a valid counterexample?
– Arian
Aug 3 at 15:45
Opps, I meant surjective, not injective. And yes, Quantic_Solver gave a good example, given that you shift the function to be zero at $a$.
– Batominovski
Aug 3 at 15:46
The definition of $R(f)$ has trouble: What does $R(f)(a)$ even mean?
– zhw.
2 days ago
@zhw. The OP wants to take the quotient of $C[a,b]$ by constant functions, but I agree that the presentation here looks confusing.
– Batominovski
2 days ago