Mixing
Prove or Disprove: Summation of two functions (at least one discontinuous) supports IVP, if both of them...
$begingroup$
Let, f and g be two functions on R, support IVP and at least one them is discontinuous. Then prove or disprove ( with example) whether f+g also supports IVP.
If f and g, both are continuous, then it is easy to say that, f+g supports intermediate-value-property. But here, we have to show, whether this result holds for discontinuous function or not.
real-analysis functional-analysis special-functions
asked Jan 17 at 14:00
Amlan Saha KunduAmlan Saha Kundu
83
$endgroup$
add a comment |
$begingroup$
Let, f and g be two functions on R, support IVP and at least one them is discontinuous. Then prove or disprove ( with example) whether f+g also supports IVP.
If f and g, both are continuous, then it is easy to say that, f+g supports intermediate-value-property. But here, we have to show, whether this result holds for discontinuous function or not.
real-analysis functional-analysis special-functions
asked Jan 17 at 14:00
Amlan Saha KunduAmlan Saha Kundu
83
$endgroup$
$begingroup$
I found a related answer: math.stackexchange.com/a/624120/58577 This shows that every function is the sum of two functions satisfying the IVP.
$endgroup$
– gerw
Jan 17 at 18:11
add a comment |
$begingroup$
Let, f and g be two functions on R, support IVP and at least one them is discontinuous. Then prove or disprove ( with example) whether f+g also supports IVP.
If f and g, both are continuous, then it is easy to say that, f+g supports intermediate-value-property. But here, we have to show, whether this result holds for discontinuous function or not.
real-analysis functional-analysis special-functions
asked Jan 17 at 14:00
Amlan Saha KunduAmlan Saha Kundu
83
$endgroup$
Let, f and g be two functions on R, support IVP and at least one them is discontinuous. Then prove or disprove ( with example) whether f+g also supports IVP.
If f and g, both are continuous, then it is easy to say that, f+g supports intermediate-value-property. But here, we have to show, whether this result holds for discontinuous function or not.
real-analysis functional-analysis special-functions
real-analysis functional-analysis special-functions
asked Jan 17 at 14:00
Amlan Saha KunduAmlan Saha Kundu
83
asked Jan 17 at 14:00
Amlan Saha KunduAmlan Saha Kundu
83
asked Jan 17 at 14:00
Amlan Saha KunduAmlan Saha Kundu
83
asked Jan 17 at 14:00
Amlan Saha KunduAmlan Saha Kundu
83
asked Jan 17 at 14:00
Amlan Saha KunduAmlan Saha Kundu
83
83
$begingroup$
I found a related answer: math.stackexchange.com/a/624120/58577 This shows that every function is the sum of two functions satisfying the IVP.
$endgroup$
– gerw
Jan 17 at 18:11
add a comment |
$begingroup$
I found a related answer: math.stackexchange.com/a/624120/58577 This shows that every function is the sum of two functions satisfying the IVP.
$endgroup$
– gerw
Jan 17 at 18:11
$begingroup$
I found a related answer: math.stackexchange.com/a/624120/58577 This shows that every function is the sum of two functions satisfying the IVP.
$endgroup$
– gerw
Jan 17 at 18:11
$begingroup$
I found a related answer: math.stackexchange.com/a/624120/58577 This shows that every function is the sum of two functions satisfying the IVP.
$endgroup$
– gerw
Jan 17 at 18:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the interval $I=[0,1]$.
$f(x)=x$ is continuous on $I$. The map
$$h(x)=begin{cases}
0 & mbox{for} &0 le x <1/2\
1 & mbox{for} &1/2 le x le 1
end{cases}$$
Doesn't support the IVP. However $g=h-f$ which is discontinuous does.
Unfortunately my answer is wrong...
answered Jan 17 at 14:16
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
$begingroup$
Thanks for your answer. But here, we have to prove whether, summation of two functions that support ivp (at least one discontinuous) supports ivp. Since h(x) is discontinuous but doesn't support IVP, this example is not valid.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:24
$begingroup$
Please read again. $f$ is continuous. $g$ is discontinuous and supports IVP. $f+g=h$ doesn't support IVP. This answers your question (negatively).
$endgroup$
– mathcounterexamples.net
Jan 17 at 14:26
$begingroup$
I have tried with f as x and g as - {x}, and come up with [x] which doesn't support IVP. Since, {x} doesn't support IVP, my example was not valid.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:26
$begingroup$
I am extremely sorry. Yes, it does. Thanks a lot.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:29
$begingroup$
I am confused. Why does $g$ satisfies IVP? In fact, $g(1/4) = -1/4$, $g(3/4) = 1/4$, but $g(s) < -1/4$ for $s in (1/4,1/2)$ and $g(s) > 1/4$ for $s in [1/2,3/4)$. In particular, there is no $s in (1/4,3/4)$ with $g(s) = 0$. Where is my mistake?
$endgroup$
– gerw
Jan 17 at 14:44
|
show 2 more comments
$begingroup$
For any $s in [-1,1]$, one can show that $f_s : mathbb R to mathbb R$ defined via
$$f_s(x):=
begin{cases} sin(1/x) & text{if } x ne 0 \ s & text{else}end{cases}$$
satisfies the IVP.
However, the difference of two such functions does clearly violate the IVP.
answered Jan 17 at 14:56
gerwgerw
19.6k11334
$endgroup$
$begingroup$
This indeed answers the question! I was looking to the case where one of the maps is continuous.
$endgroup$
– mathcounterexamples.net
Jan 17 at 15:09
$begingroup$
@mathcounterexamples.net: I have the feeling that this is not possible, but I am not sure.
$endgroup$
– gerw
Jan 17 at 15:22
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the interval $I=[0,1]$.
$f(x)=x$ is continuous on $I$. The map
$$h(x)=begin{cases}
0 & mbox{for} &0 le x <1/2\
1 & mbox{for} &1/2 le x le 1
end{cases}$$
Doesn't support the IVP. However $g=h-f$ which is discontinuous does.
Unfortunately my answer is wrong...
answered Jan 17 at 14:16
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
$begingroup$
Thanks for your answer. But here, we have to prove whether, summation of two functions that support ivp (at least one discontinuous) supports ivp. Since h(x) is discontinuous but doesn't support IVP, this example is not valid.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:24
$begingroup$
Please read again. $f$ is continuous. $g$ is discontinuous and supports IVP. $f+g=h$ doesn't support IVP. This answers your question (negatively).
$endgroup$
– mathcounterexamples.net
Jan 17 at 14:26
$begingroup$
I have tried with f as x and g as - {x}, and come up with [x] which doesn't support IVP. Since, {x} doesn't support IVP, my example was not valid.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:26
$begingroup$
I am extremely sorry. Yes, it does. Thanks a lot.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:29
$begingroup$
I am confused. Why does $g$ satisfies IVP? In fact, $g(1/4) = -1/4$, $g(3/4) = 1/4$, but $g(s) < -1/4$ for $s in (1/4,1/2)$ and $g(s) > 1/4$ for $s in [1/2,3/4)$. In particular, there is no $s in (1/4,3/4)$ with $g(s) = 0$. Where is my mistake?
$endgroup$
– gerw
Jan 17 at 14:44
|
show 2 more comments
$begingroup$
Consider the interval $I=[0,1]$.
$f(x)=x$ is continuous on $I$. The map
$$h(x)=begin{cases}
0 & mbox{for} &0 le x <1/2\
1 & mbox{for} &1/2 le x le 1
end{cases}$$
Doesn't support the IVP. However $g=h-f$ which is discontinuous does.
Unfortunately my answer is wrong...
answered Jan 17 at 14:16
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
$begingroup$
Thanks for your answer. But here, we have to prove whether, summation of two functions that support ivp (at least one discontinuous) supports ivp. Since h(x) is discontinuous but doesn't support IVP, this example is not valid.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:24
$begingroup$
Please read again. $f$ is continuous. $g$ is discontinuous and supports IVP. $f+g=h$ doesn't support IVP. This answers your question (negatively).
$endgroup$
– mathcounterexamples.net
Jan 17 at 14:26
$begingroup$
I have tried with f as x and g as - {x}, and come up with [x] which doesn't support IVP. Since, {x} doesn't support IVP, my example was not valid.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:26
$begingroup$
I am extremely sorry. Yes, it does. Thanks a lot.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:29
$begingroup$
I am confused. Why does $g$ satisfies IVP? In fact, $g(1/4) = -1/4$, $g(3/4) = 1/4$, but $g(s) < -1/4$ for $s in (1/4,1/2)$ and $g(s) > 1/4$ for $s in [1/2,3/4)$. In particular, there is no $s in (1/4,3/4)$ with $g(s) = 0$. Where is my mistake?
$endgroup$
– gerw
Jan 17 at 14:44
|
show 2 more comments
$begingroup$
Consider the interval $I=[0,1]$.
$f(x)=x$ is continuous on $I$. The map
$$h(x)=begin{cases}
0 & mbox{for} &0 le x <1/2\
1 & mbox{for} &1/2 le x le 1
end{cases}$$
Doesn't support the IVP. However $g=h-f$ which is discontinuous does.
Unfortunately my answer is wrong...
answered Jan 17 at 14:16
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
Consider the interval $I=[0,1]$.
$f(x)=x$ is continuous on $I$. The map
$$h(x)=begin{cases}
0 & mbox{for} &0 le x <1/2\
1 & mbox{for} &1/2 le x le 1
end{cases}$$
Doesn't support the IVP. However $g=h-f$ which is discontinuous does.
Unfortunately my answer is wrong...
answered Jan 17 at 14:16
mathcounterexamples.netmathcounterexamples.net
26.9k22157
edited Jan 17 at 16:56
answered Jan 17 at 14:16
mathcounterexamples.netmathcounterexamples.net
26.9k22157
answered Jan 17 at 14:16
mathcounterexamples.netmathcounterexamples.net
26.9k22157
answered Jan 17 at 14:16
mathcounterexamples.netmathcounterexamples.net
26.9k22157
26.9k22157
$begingroup$
Thanks for your answer. But here, we have to prove whether, summation of two functions that support ivp (at least one discontinuous) supports ivp. Since h(x) is discontinuous but doesn't support IVP, this example is not valid.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:24
$begingroup$
Please read again. $f$ is continuous. $g$ is discontinuous and supports IVP. $f+g=h$ doesn't support IVP. This answers your question (negatively).
$endgroup$
– mathcounterexamples.net
Jan 17 at 14:26
$begingroup$
I have tried with f as x and g as - {x}, and come up with [x] which doesn't support IVP. Since, {x} doesn't support IVP, my example was not valid.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:26
$begingroup$
I am extremely sorry. Yes, it does. Thanks a lot.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:29
$begingroup$
I am confused. Why does $g$ satisfies IVP? In fact, $g(1/4) = -1/4$, $g(3/4) = 1/4$, but $g(s) < -1/4$ for $s in (1/4,1/2)$ and $g(s) > 1/4$ for $s in [1/2,3/4)$. In particular, there is no $s in (1/4,3/4)$ with $g(s) = 0$. Where is my mistake?
$endgroup$
– gerw
Jan 17 at 14:44
|
show 2 more comments
$begingroup$
Thanks for your answer. But here, we have to prove whether, summation of two functions that support ivp (at least one discontinuous) supports ivp. Since h(x) is discontinuous but doesn't support IVP, this example is not valid.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:24
$begingroup$
Please read again. $f$ is continuous. $g$ is discontinuous and supports IVP. $f+g=h$ doesn't support IVP. This answers your question (negatively).
$endgroup$
– mathcounterexamples.net
Jan 17 at 14:26
$begingroup$
I have tried with f as x and g as - {x}, and come up with [x] which doesn't support IVP. Since, {x} doesn't support IVP, my example was not valid.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:26
$begingroup$
I am extremely sorry. Yes, it does. Thanks a lot.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:29
$begingroup$
I am confused. Why does $g$ satisfies IVP? In fact, $g(1/4) = -1/4$, $g(3/4) = 1/4$, but $g(s) < -1/4$ for $s in (1/4,1/2)$ and $g(s) > 1/4$ for $s in [1/2,3/4)$. In particular, there is no $s in (1/4,3/4)$ with $g(s) = 0$. Where is my mistake?
$endgroup$
– gerw
Jan 17 at 14:44
$begingroup$
Thanks for your answer. But here, we have to prove whether, summation of two functions that support ivp (at least one discontinuous) supports ivp. Since h(x) is discontinuous but doesn't support IVP, this example is not valid.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:24
$begingroup$
Thanks for your answer. But here, we have to prove whether, summation of two functions that support ivp (at least one discontinuous) supports ivp. Since h(x) is discontinuous but doesn't support IVP, this example is not valid.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:24
$begingroup$
Please read again. $f$ is continuous. $g$ is discontinuous and supports IVP. $f+g=h$ doesn't support IVP. This answers your question (negatively).
$endgroup$
– mathcounterexamples.net
Jan 17 at 14:26
$begingroup$
Please read again. $f$ is continuous. $g$ is discontinuous and supports IVP. $f+g=h$ doesn't support IVP. This answers your question (negatively).
$endgroup$
– mathcounterexamples.net
Jan 17 at 14:26
$begingroup$
I have tried with f as x and g as - {x}, and come up with [x] which doesn't support IVP. Since, {x} doesn't support IVP, my example was not valid.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:26
$begingroup$
I have tried with f as x and g as - {x}, and come up with [x] which doesn't support IVP. Since, {x} doesn't support IVP, my example was not valid.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:26
$begingroup$
I am extremely sorry. Yes, it does. Thanks a lot.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:29
$begingroup$
I am extremely sorry. Yes, it does. Thanks a lot.
$endgroup$
– Amlan Saha Kundu
Jan 17 at 14:29
$begingroup$
I am confused. Why does $g$ satisfies IVP? In fact, $g(1/4) = -1/4$, $g(3/4) = 1/4$, but $g(s) < -1/4$ for $s in (1/4,1/2)$ and $g(s) > 1/4$ for $s in [1/2,3/4)$. In particular, there is no $s in (1/4,3/4)$ with $g(s) = 0$. Where is my mistake?
$endgroup$
– gerw
Jan 17 at 14:44
$begingroup$
I am confused. Why does $g$ satisfies IVP? In fact, $g(1/4) = -1/4$, $g(3/4) = 1/4$, but $g(s) < -1/4$ for $s in (1/4,1/2)$ and $g(s) > 1/4$ for $s in [1/2,3/4)$. In particular, there is no $s in (1/4,3/4)$ with $g(s) = 0$. Where is my mistake?
$endgroup$
– gerw
Jan 17 at 14:44
|
show 2 more comments
$begingroup$
For any $s in [-1,1]$, one can show that $f_s : mathbb R to mathbb R$ defined via
$$f_s(x):=
begin{cases} sin(1/x) & text{if } x ne 0 \ s & text{else}end{cases}$$
satisfies the IVP.
However, the difference of two such functions does clearly violate the IVP.
answered Jan 17 at 14:56
gerwgerw
19.6k11334
$endgroup$
$begingroup$
This indeed answers the question! I was looking to the case where one of the maps is continuous.
$endgroup$
– mathcounterexamples.net
Jan 17 at 15:09
$begingroup$
@mathcounterexamples.net: I have the feeling that this is not possible, but I am not sure.
$endgroup$
– gerw
Jan 17 at 15:22
add a comment |
$begingroup$
For any $s in [-1,1]$, one can show that $f_s : mathbb R to mathbb R$ defined via
$$f_s(x):=
begin{cases} sin(1/x) & text{if } x ne 0 \ s & text{else}end{cases}$$
satisfies the IVP.
However, the difference of two such functions does clearly violate the IVP.
answered Jan 17 at 14:56
gerwgerw
19.6k11334
$endgroup$
$begingroup$
This indeed answers the question! I was looking to the case where one of the maps is continuous.
$endgroup$
– mathcounterexamples.net
Jan 17 at 15:09
$begingroup$
@mathcounterexamples.net: I have the feeling that this is not possible, but I am not sure.
$endgroup$
– gerw
Jan 17 at 15:22
add a comment |
$begingroup$
For any $s in [-1,1]$, one can show that $f_s : mathbb R to mathbb R$ defined via
$$f_s(x):=
begin{cases} sin(1/x) & text{if } x ne 0 \ s & text{else}end{cases}$$
satisfies the IVP.
However, the difference of two such functions does clearly violate the IVP.
answered Jan 17 at 14:56
gerwgerw
19.6k11334
$endgroup$
For any $s in [-1,1]$, one can show that $f_s : mathbb R to mathbb R$ defined via
$$f_s(x):=
begin{cases} sin(1/x) & text{if } x ne 0 \ s & text{else}end{cases}$$
satisfies the IVP.
However, the difference of two such functions does clearly violate the IVP.
answered Jan 17 at 14:56
gerwgerw
19.6k11334
answered Jan 17 at 14:56
gerwgerw
19.6k11334
answered Jan 17 at 14:56
gerwgerw
19.6k11334
answered Jan 17 at 14:56
gerwgerw
19.6k11334
19.6k11334
$begingroup$
This indeed answers the question! I was looking to the case where one of the maps is continuous.
$endgroup$
– mathcounterexamples.net
Jan 17 at 15:09
$begingroup$
@mathcounterexamples.net: I have the feeling that this is not possible, but I am not sure.
$endgroup$
– gerw
Jan 17 at 15:22
add a comment |
$begingroup$
This indeed answers the question! I was looking to the case where one of the maps is continuous.
$endgroup$
– mathcounterexamples.net
Jan 17 at 15:09
$begingroup$
@mathcounterexamples.net: I have the feeling that this is not possible, but I am not sure.
$endgroup$
– gerw
Jan 17 at 15:22
$begingroup$
This indeed answers the question! I was looking to the case where one of the maps is continuous.
$endgroup$
– mathcounterexamples.net
Jan 17 at 15:09
$begingroup$
This indeed answers the question! I was looking to the case where one of the maps is continuous.
$endgroup$
– mathcounterexamples.net
Jan 17 at 15:09
$begingroup$
@mathcounterexamples.net: I have the feeling that this is not possible, but I am not sure.
$endgroup$
– gerw
Jan 17 at 15:22
$begingroup$
@mathcounterexamples.net: I have the feeling that this is not possible, but I am not sure.
$endgroup$
– gerw
Jan 17 at 15:22
add a comment |
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$begingroup$
I found a related answer: math.stackexchange.com/a/624120/58577 This shows that every function is the sum of two functions satisfying the IVP.
$endgroup$
– gerw
Jan 17 at 18:11