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Proving that a function is constant from functional equation [duplicate]

$ int_0^x f(t)dt=int_0^axf(t)dt+ int_0^bxf(t)dt$ implies $f$ constantProve that $exists a<b$ s.t. $f(a)=f(b)=0$ when $int_0^1f(x)dx=int_0^1xf(x)dx=0$Proving constant function given the second derivative.Proving the existence of a point with a certain property for a continuous functionShow that the cumulative distribution function is uniformly continuous.Best minimum constant for a functional inequalityFunctional equation $f(x-F(x))=f(x)$Domain of integral of complex function from 0 to 1Problem: proving that function is constant$frac12^psum_k=1^2^pfbiggl(frack2^pbiggl)=int_0^1f(x)dx$ implies $f$ is constant$int_0^1f(x)g(x) dx=int_0^1f(x)dx int_0^1g(x)dx $ implies $f$ constant

1

$begingroup$

This question already has an answer here:

• 
  $ int_0^x f(t)dt=int_0^axf(t)dt+ int_0^bxf(t)dt$ implies $f$ constant
  
  3 answers
  
  

$a,b in (0,1)$ and $f:[0,1] to mathbb R$ is continuous functions s.t. $ int_0^x f(x)dx=int_0^axf(x)dx+ int_0^bxf(x)dx$ . Knowing that $a+b=1$, we have to prove that $f$ is constant.

Using the derivative,we get:
$f(x)=af(ax)+bf(bx)$

I managed to do it for the case $a=b=1/2$, but I don't know how to make it with $a,b$ arbitrary and $a,b in (0,1)$ $a+b=1$

calculus integration functions

share|cite|improve this question

edited Mar 13 at 11:23

Gaboru

asked Mar 13 at 10:18

GaboruGaboru

4428

$endgroup$

marked as duplicate by John Omielan, John Hughes, Lord Shark the Unknown, Lee David Chung Lin, Cesareo Mar 14 at 9:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note the OP asked the same question again about $2$ hours later at $ int_0^x f(t)dt=int_0^axf(t)dt+ int_0^bxf(t)dt$ implies $f$ constant.
  $endgroup$
  – John Omielan
  Mar 13 at 21:04
  

  
  
  
  

add a comment | 

1

$begingroup$

This question already has an answer here:

• 
  $ int_0^x f(t)dt=int_0^axf(t)dt+ int_0^bxf(t)dt$ implies $f$ constant
  
  3 answers
  
  

$a,b in (0,1)$ and $f:[0,1] to mathbb R$ is continuous functions s.t. $ int_0^x f(x)dx=int_0^axf(x)dx+ int_0^bxf(x)dx$ . Knowing that $a+b=1$, we have to prove that $f$ is constant.

Using the derivative,we get:
$f(x)=af(ax)+bf(bx)$

I managed to do it for the case $a=b=1/2$, but I don't know how to make it with $a,b$ arbitrary and $a,b in (0,1)$ $a+b=1$

calculus integration functions

share|cite|improve this question

edited Mar 13 at 11:23

Gaboru

asked Mar 13 at 10:18

GaboruGaboru

4428

$endgroup$

marked as duplicate by John Omielan, John Hughes, Lord Shark the Unknown, Lee David Chung Lin, Cesareo Mar 14 at 9:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note the OP asked the same question again about $2$ hours later at $ int_0^x f(t)dt=int_0^axf(t)dt+ int_0^bxf(t)dt$ implies $f$ constant.
  $endgroup$
  – John Omielan
  Mar 13 at 21:04
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  $ int_0^x f(t)dt=int_0^axf(t)dt+ int_0^bxf(t)dt$ implies $f$ constant
  
  3 answers
  
  

$a,b in (0,1)$ and $f:[0,1] to mathbb R$ is continuous functions s.t. $ int_0^x f(x)dx=int_0^axf(x)dx+ int_0^bxf(x)dx$ . Knowing that $a+b=1$, we have to prove that $f$ is constant.

Using the derivative,we get:
$f(x)=af(ax)+bf(bx)$

I managed to do it for the case $a=b=1/2$, but I don't know how to make it with $a,b$ arbitrary and $a,b in (0,1)$ $a+b=1$

calculus integration functions

share|cite|improve this question

edited Mar 13 at 11:23

Gaboru

asked Mar 13 at 10:18

GaboruGaboru

4428

$endgroup$

share|cite|improve this question

edited Mar 13 at 11:23

Gaboru

asked Mar 13 at 10:18

GaboruGaboru

4428

share|cite|improve this question

edited Mar 13 at 11:23

Gaboru

asked Mar 13 at 10:18

GaboruGaboru

4428

asked Mar 13 at 10:18

GaboruGaboru

4428

asked Mar 13 at 10:18

GaboruGaboru

4428

2 Answers
2

active

oldest

votes

2

$begingroup$

I think it should read:

$int_0^1 f(t)dt=int_0^axf(t)dt+ int_0^bxf(t)dt.$

Then $ f(x)=af(ax)+bf(bx)$ is not correct. Using derivatives, you get $0=af(ax)+bf(bx)$, since

$fracddxint_0^1 f(t)dt=0.$

share|cite|improve this answer

answered Mar 13 at 10:32

FredFred

48.5k11849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. I corrected.
  $endgroup$
  – Gaboru
  Mar 13 at 11:24
  

  
  
  
  

add a comment | 

0

$begingroup$

$int_0^1 f(t)dt = lim_a->0int_0^axf(t)dt +
lim_a->0int_0^bxf(t)dt = int_0^xf(t)dt$.

Taking derivatives of both sides, 0 = f(x).

share|cite|improve this answer

answered Mar 13 at 10:48

William ElliotWilliam Elliot

8,7222820

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. I corrected.
  $endgroup$
  – Gaboru
  Mar 13 at 11:24
  

  
  
  
  

add a comment | 

2

$begingroup$

I think it should read:

$int_0^1 f(t)dt=int_0^axf(t)dt+ int_0^bxf(t)dt.$

Then $ f(x)=af(ax)+bf(bx)$ is not correct. Using derivatives, you get $0=af(ax)+bf(bx)$, since

$fracddxint_0^1 f(t)dt=0.$

share|cite|improve this answer

answered Mar 13 at 10:32

FredFred

48.5k11849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. I corrected.
  $endgroup$
  – Gaboru
  Mar 13 at 11:24
  

  
  
  
  

add a comment | 

2

$begingroup$

I think it should read:

$int_0^1 f(t)dt=int_0^axf(t)dt+ int_0^bxf(t)dt.$

Then $ f(x)=af(ax)+bf(bx)$ is not correct. Using derivatives, you get $0=af(ax)+bf(bx)$, since

$fracddxint_0^1 f(t)dt=0.$

share|cite|improve this answer

answered Mar 13 at 10:32

FredFred

48.5k11849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. I corrected.
  $endgroup$
  – Gaboru
  Mar 13 at 11:24
  

  
  
  
  

add a comment | 

$begingroup$

I think it should read:

$int_0^1 f(t)dt=int_0^axf(t)dt+ int_0^bxf(t)dt.$

Then $ f(x)=af(ax)+bf(bx)$ is not correct. Using derivatives, you get $0=af(ax)+bf(bx)$, since

$fracddxint_0^1 f(t)dt=0.$

share|cite|improve this answer

answered Mar 13 at 10:32

FredFred

48.5k11849

$endgroup$

share|cite|improve this answer

answered Mar 13 at 10:32

FredFred

48.5k11849

answered Mar 13 at 10:32

FredFred

48.5k11849

answered Mar 13 at 10:32

FredFred

48.5k11849

0

$begingroup$

$int_0^1 f(t)dt = lim_a->0int_0^axf(t)dt +
lim_a->0int_0^bxf(t)dt = int_0^xf(t)dt$.

Taking derivatives of both sides, 0 = f(x).

share|cite|improve this answer

answered Mar 13 at 10:48

William ElliotWilliam Elliot

8,7222820

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. I corrected.
  $endgroup$
  – Gaboru
  Mar 13 at 11:24
  

  
  
  
  

add a comment | 

0

$begingroup$

$int_0^1 f(t)dt = lim_a->0int_0^axf(t)dt +
lim_a->0int_0^bxf(t)dt = int_0^xf(t)dt$.

Taking derivatives of both sides, 0 = f(x).

share|cite|improve this answer

answered Mar 13 at 10:48

William ElliotWilliam Elliot

8,7222820

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. I corrected.
  $endgroup$
  – Gaboru
  Mar 13 at 11:24
  

  
  
  
  

add a comment | 

$begingroup$

$int_0^1 f(t)dt = lim_a->0int_0^axf(t)dt +
lim_a->0int_0^bxf(t)dt = int_0^xf(t)dt$.

Taking derivatives of both sides, 0 = f(x).

share|cite|improve this answer

answered Mar 13 at 10:48

William ElliotWilliam Elliot

8,7222820

$endgroup$

share|cite|improve this answer

answered Mar 13 at 10:48

William ElliotWilliam Elliot

8,7222820

answered Mar 13 at 10:48

William ElliotWilliam Elliot

8,7222820

answered Mar 13 at 10:48

William ElliotWilliam Elliot

8,7222820