prove $f$ is bounded given $f(t^2+u)=tf(t)+f(u)$
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I have been trying to solve the functional equation $f:Bbb R to
Bbb R$ $f(t^2+u)=tf(t)+f(u)$. So far i have managed to show that $f$ is additive i.e. $f(a+b)=f(a)+f(b)$ which means that the condition can be simplified to $f(t^2)=tf(t)$. To show that $f$ is linear i need to show that it is bounded above or below on some interval. I have been trying to do this for some time but have gotten nowhere. Could someone please give me a hint?
functional-equations
asked Jan 6 at 23:50
B.MartinB.Martin
4271210
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add a comment |
$begingroup$
I have been trying to solve the functional equation $f:Bbb R to
Bbb R$ $f(t^2+u)=tf(t)+f(u)$. So far i have managed to show that $f$ is additive i.e. $f(a+b)=f(a)+f(b)$ which means that the condition can be simplified to $f(t^2)=tf(t)$. To show that $f$ is linear i need to show that it is bounded above or below on some interval. I have been trying to do this for some time but have gotten nowhere. Could someone please give me a hint?
functional-equations
asked Jan 6 at 23:50
B.MartinB.Martin
4271210
$endgroup$
$begingroup$
If you managed to show that $f$ is additive, then $f$ is a Cauchy functional equation, hence, it has only one family of solutions, namely $f(x)=cx$. Check en.wikipedia.org/wiki/Cauchy%27s_functional_equation
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– EuxhenH
Jan 7 at 0:00
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@EuxhenH It is not true that every solution of Cauchy equation has this form. OP is not assuming continuity.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 0:03
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@KaviRamaMurthy True. Forgive my absent-mindedness.
$endgroup$
– EuxhenH
Jan 7 at 0:06
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@EuxhenH My issue is trying to prove one of the conditions that allows me to use the solution to that Cauchy equation
$endgroup$
– B.Martin
Jan 7 at 0:10
add a comment |
$begingroup$
I have been trying to solve the functional equation $f:Bbb R to
Bbb R$ $f(t^2+u)=tf(t)+f(u)$. So far i have managed to show that $f$ is additive i.e. $f(a+b)=f(a)+f(b)$ which means that the condition can be simplified to $f(t^2)=tf(t)$. To show that $f$ is linear i need to show that it is bounded above or below on some interval. I have been trying to do this for some time but have gotten nowhere. Could someone please give me a hint?
functional-equations
asked Jan 6 at 23:50
B.MartinB.Martin
4271210
$endgroup$
I have been trying to solve the functional equation $f:Bbb R to
Bbb R$ $f(t^2+u)=tf(t)+f(u)$. So far i have managed to show that $f$ is additive i.e. $f(a+b)=f(a)+f(b)$ which means that the condition can be simplified to $f(t^2)=tf(t)$. To show that $f$ is linear i need to show that it is bounded above or below on some interval. I have been trying to do this for some time but have gotten nowhere. Could someone please give me a hint?
functional-equations
functional-equations
asked Jan 6 at 23:50
B.MartinB.Martin
4271210
asked Jan 6 at 23:50
B.MartinB.Martin
4271210
asked Jan 6 at 23:50
B.MartinB.Martin
4271210
asked Jan 6 at 23:50
B.MartinB.Martin
4271210
asked Jan 6 at 23:50
B.MartinB.Martin
4271210
4271210
$begingroup$
If you managed to show that $f$ is additive, then $f$ is a Cauchy functional equation, hence, it has only one family of solutions, namely $f(x)=cx$. Check en.wikipedia.org/wiki/Cauchy%27s_functional_equation
$endgroup$
– EuxhenH
Jan 7 at 0:00
$begingroup$
@EuxhenH It is not true that every solution of Cauchy equation has this form. OP is not assuming continuity.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 0:03
$begingroup$
@KaviRamaMurthy True. Forgive my absent-mindedness.
$endgroup$
– EuxhenH
Jan 7 at 0:06
$begingroup$
@EuxhenH My issue is trying to prove one of the conditions that allows me to use the solution to that Cauchy equation
$endgroup$
– B.Martin
Jan 7 at 0:10
add a comment |
$begingroup$
If you managed to show that $f$ is additive, then $f$ is a Cauchy functional equation, hence, it has only one family of solutions, namely $f(x)=cx$. Check en.wikipedia.org/wiki/Cauchy%27s_functional_equation
$endgroup$
– EuxhenH
Jan 7 at 0:00
$begingroup$
@EuxhenH It is not true that every solution of Cauchy equation has this form. OP is not assuming continuity.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 0:03
$begingroup$
@KaviRamaMurthy True. Forgive my absent-mindedness.
$endgroup$
– EuxhenH
Jan 7 at 0:06
$begingroup$
@EuxhenH My issue is trying to prove one of the conditions that allows me to use the solution to that Cauchy equation
$endgroup$
– B.Martin
Jan 7 at 0:10
$begingroup$
If you managed to show that $f$ is additive, then $f$ is a Cauchy functional equation, hence, it has only one family of solutions, namely $f(x)=cx$. Check en.wikipedia.org/wiki/Cauchy%27s_functional_equation
$endgroup$
– EuxhenH
Jan 7 at 0:00
$begingroup$
If you managed to show that $f$ is additive, then $f$ is a Cauchy functional equation, hence, it has only one family of solutions, namely $f(x)=cx$. Check en.wikipedia.org/wiki/Cauchy%27s_functional_equation
$endgroup$
– EuxhenH
Jan 7 at 0:00
$begingroup$
@EuxhenH It is not true that every solution of Cauchy equation has this form. OP is not assuming continuity.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 0:03
$begingroup$
@EuxhenH It is not true that every solution of Cauchy equation has this form. OP is not assuming continuity.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 0:03
$begingroup$
@KaviRamaMurthy True. Forgive my absent-mindedness.
$endgroup$
– EuxhenH
Jan 7 at 0:06
$begingroup$
@KaviRamaMurthy True. Forgive my absent-mindedness.
$endgroup$
– EuxhenH
Jan 7 at 0:06
$begingroup$
@EuxhenH My issue is trying to prove one of the conditions that allows me to use the solution to that Cauchy equation
$endgroup$
– B.Martin
Jan 7 at 0:10
$begingroup$
@EuxhenH My issue is trying to prove one of the conditions that allows me to use the solution to that Cauchy equation
$endgroup$
– B.Martin
Jan 7 at 0:10
add a comment |
1 Answer
1
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$af(a)+bf(b)+ bf(a) + af(b) = (a+b)(f(a)+f(b)) = (a+b)f(a+b) = f((a+b)^2)$
$=f(a^2+b^2+2ab)= f(a^2)+f(b^2)+f(2ab) = af(a)+bf(b)+f(2ab)$.
Thus $bf(a)+af(b)=f(2ab)$.
For $a=1$ this yields $bf(1)+f(b)=f(2b)=f(b+b)=f(b)+f(b)$.
Thus $f(b)=bf(1)$, so $f$ is linear.
answered Jan 7 at 0:15
A. PongráczA. Pongrácz
6,0821929
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$begingroup$
Thank you for your answer!
$endgroup$
– B.Martin
Jan 7 at 0:25
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You are welcome!
$endgroup$
– A. Pongrácz
Jan 7 at 0:40
add a comment |
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$begingroup$
$af(a)+bf(b)+ bf(a) + af(b) = (a+b)(f(a)+f(b)) = (a+b)f(a+b) = f((a+b)^2)$
$=f(a^2+b^2+2ab)= f(a^2)+f(b^2)+f(2ab) = af(a)+bf(b)+f(2ab)$.
Thus $bf(a)+af(b)=f(2ab)$.
For $a=1$ this yields $bf(1)+f(b)=f(2b)=f(b+b)=f(b)+f(b)$.
Thus $f(b)=bf(1)$, so $f$ is linear.
answered Jan 7 at 0:15
A. PongráczA. Pongrácz
6,0821929
$endgroup$
$begingroup$
Thank you for your answer!
$endgroup$
– B.Martin
Jan 7 at 0:25
$begingroup$
You are welcome!
$endgroup$
– A. Pongrácz
Jan 7 at 0:40
add a comment |
$begingroup$
$af(a)+bf(b)+ bf(a) + af(b) = (a+b)(f(a)+f(b)) = (a+b)f(a+b) = f((a+b)^2)$
$=f(a^2+b^2+2ab)= f(a^2)+f(b^2)+f(2ab) = af(a)+bf(b)+f(2ab)$.
Thus $bf(a)+af(b)=f(2ab)$.
For $a=1$ this yields $bf(1)+f(b)=f(2b)=f(b+b)=f(b)+f(b)$.
Thus $f(b)=bf(1)$, so $f$ is linear.
answered Jan 7 at 0:15
A. PongráczA. Pongrácz
6,0821929
$endgroup$
$begingroup$
Thank you for your answer!
$endgroup$
– B.Martin
Jan 7 at 0:25
$begingroup$
You are welcome!
$endgroup$
– A. Pongrácz
Jan 7 at 0:40
add a comment |
$begingroup$
$af(a)+bf(b)+ bf(a) + af(b) = (a+b)(f(a)+f(b)) = (a+b)f(a+b) = f((a+b)^2)$
$=f(a^2+b^2+2ab)= f(a^2)+f(b^2)+f(2ab) = af(a)+bf(b)+f(2ab)$.
Thus $bf(a)+af(b)=f(2ab)$.
For $a=1$ this yields $bf(1)+f(b)=f(2b)=f(b+b)=f(b)+f(b)$.
Thus $f(b)=bf(1)$, so $f$ is linear.
answered Jan 7 at 0:15
A. PongráczA. Pongrácz
6,0821929
$endgroup$
$af(a)+bf(b)+ bf(a) + af(b) = (a+b)(f(a)+f(b)) = (a+b)f(a+b) = f((a+b)^2)$
$=f(a^2+b^2+2ab)= f(a^2)+f(b^2)+f(2ab) = af(a)+bf(b)+f(2ab)$.
Thus $bf(a)+af(b)=f(2ab)$.
For $a=1$ this yields $bf(1)+f(b)=f(2b)=f(b+b)=f(b)+f(b)$.
Thus $f(b)=bf(1)$, so $f$ is linear.
answered Jan 7 at 0:15
A. PongráczA. Pongrácz
6,0821929
answered Jan 7 at 0:15
A. PongráczA. Pongrácz
6,0821929
answered Jan 7 at 0:15
A. PongráczA. Pongrácz
6,0821929
answered Jan 7 at 0:15
A. PongráczA. Pongrácz
6,0821929
6,0821929
$begingroup$
Thank you for your answer!
$endgroup$
– B.Martin
Jan 7 at 0:25
$begingroup$
You are welcome!
$endgroup$
– A. Pongrácz
Jan 7 at 0:40
add a comment |
$begingroup$
Thank you for your answer!
$endgroup$
– B.Martin
Jan 7 at 0:25
$begingroup$
You are welcome!
$endgroup$
– A. Pongrácz
Jan 7 at 0:40
$begingroup$
Thank you for your answer!
$endgroup$
– B.Martin
Jan 7 at 0:25
$begingroup$
Thank you for your answer!
$endgroup$
– B.Martin
Jan 7 at 0:25
$begingroup$
You are welcome!
$endgroup$
– A. Pongrácz
Jan 7 at 0:40
$begingroup$
You are welcome!
$endgroup$
– A. Pongrácz
Jan 7 at 0:40
add a comment |
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$begingroup$
If you managed to show that $f$ is additive, then $f$ is a Cauchy functional equation, hence, it has only one family of solutions, namely $f(x)=cx$. Check en.wikipedia.org/wiki/Cauchy%27s_functional_equation
$endgroup$
– EuxhenH
Jan 7 at 0:00
$begingroup$
@EuxhenH It is not true that every solution of Cauchy equation has this form. OP is not assuming continuity.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 0:03
$begingroup$
@KaviRamaMurthy True. Forgive my absent-mindedness.
$endgroup$
– EuxhenH
Jan 7 at 0:06
$begingroup$
@EuxhenH My issue is trying to prove one of the conditions that allows me to use the solution to that Cauchy equation
$endgroup$
– B.Martin
Jan 7 at 0:10