Mixing
Prove that $int_0^1 f^2(x)dx > frac{1}{6}$ , knowing that $f(f(x)) = x^2$.
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How can I show that $int_0^1 f^2(x)dx > frac{1}{6}$, where $f(x)$ is continuous for $xin[0,infty)$, and $f(f(x)) = x^2$?
I figured out that $f(x^2)=f(f(f(x)))=f^2(x)$ then $int_0^1f^2(x)dx = int_0^1f(x^2)dx$, but I don't know what to do next.
real-analysis integration
asked Jan 26 at 0:49
BanBan
403
$endgroup$
add a comment |
$begingroup$
How can I show that $int_0^1 f^2(x)dx > frac{1}{6}$, where $f(x)$ is continuous for $xin[0,infty)$, and $f(f(x)) = x^2$?
I figured out that $f(x^2)=f(f(f(x)))=f^2(x)$ then $int_0^1f^2(x)dx = int_0^1f(x^2)dx$, but I don't know what to do next.
real-analysis integration
asked Jan 26 at 0:49
BanBan
403
$endgroup$
add a comment |
$begingroup$
How can I show that $int_0^1 f^2(x)dx > frac{1}{6}$, where $f(x)$ is continuous for $xin[0,infty)$, and $f(f(x)) = x^2$?
I figured out that $f(x^2)=f(f(f(x)))=f^2(x)$ then $int_0^1f^2(x)dx = int_0^1f(x^2)dx$, but I don't know what to do next.
real-analysis integration
asked Jan 26 at 0:49
BanBan
403
$endgroup$
How can I show that $int_0^1 f^2(x)dx > frac{1}{6}$, where $f(x)$ is continuous for $xin[0,infty)$, and $f(f(x)) = x^2$?
I figured out that $f(x^2)=f(f(f(x)))=f^2(x)$ then $int_0^1f^2(x)dx = int_0^1f(x^2)dx$, but I don't know what to do next.
real-analysis integration
real-analysis integration
asked Jan 26 at 0:49
BanBan
403
asked Jan 26 at 0:49
BanBan
403
asked Jan 26 at 0:49
BanBan
403
asked Jan 26 at 0:49
BanBan
403
asked Jan 26 at 0:49
BanBan
403
403
add a comment |
add a comment |
1 Answer
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$begingroup$
It can be proven that if $f:[0,infty)mapsto [0,infty)$ is continuous and $f(f(x))=x^2$, then $f(x)$ is bounded (not strictly) between $x$ and $x^2$ for all $xin (0,1)$. I will not prove this rigorously here, and leave it as an exercise (it can be done using the continuity requirement and the generalized intermediate value theorem for two curves).
This shows that $f^2(x)$ is between $x^2$ and $x^4$, and that $int_0^1 f^2(x)dxge int_0^1 x^4=1/5gt 1/6$.
MORE DETAILS FOR THE PROOF:
Since $f(f(x))=x^2$ is injective, $f(x)$ must be injective, and thus either strictly increasing or strictly decreasing; I’ll let you rule out the latter case.
Suppose it is increasing. If $f(x)gt x$ for some $xin (0,1)$, then $f(f(x))gt f(x)$ and $x^2gt f(x)gt x$ which is impossible. If $f(x)lt x^2$ for some $xin (0,1)$, then $f(x)lt f(f(x))$ and $xlt f(x)lt x^2$ which is impossible. Thus $x^2le f(x)le x$.
answered Jan 26 at 1:40
FrpzzdFrpzzd
23k841110
$endgroup$
$begingroup$
$int_0^1 x^4 dx=1/5.$
$endgroup$
– user
Jan 26 at 1:44
$begingroup$
@user Ha. I honestly don’t know if that was a typo, or if I was just being stupid. :P
$endgroup$
– Frpzzd
Jan 26 at 1:45
1
$begingroup$
I would lay bets on a typo. It would be interesting to know if there are other continuous functions besides $f(x)=x^sqrt2$ which satisfy the functional equation.
$endgroup$
– user
Jan 26 at 2:05
1
$begingroup$
It never crossed my mind to use injectivity to prove that the function is strictly increasing, I understant it now, thanks. I wonder if the lower bound can be raised even higher, maybe somewhere around 1/4.(Seeing that for $f(x)=x^sqrt{2}$ the value of the integral is around 0.26 which is greater than 0.25).
$endgroup$
– Ban
Jan 26 at 3:13
add a comment |
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$begingroup$
It can be proven that if $f:[0,infty)mapsto [0,infty)$ is continuous and $f(f(x))=x^2$, then $f(x)$ is bounded (not strictly) between $x$ and $x^2$ for all $xin (0,1)$. I will not prove this rigorously here, and leave it as an exercise (it can be done using the continuity requirement and the generalized intermediate value theorem for two curves).
This shows that $f^2(x)$ is between $x^2$ and $x^4$, and that $int_0^1 f^2(x)dxge int_0^1 x^4=1/5gt 1/6$.
MORE DETAILS FOR THE PROOF:
Since $f(f(x))=x^2$ is injective, $f(x)$ must be injective, and thus either strictly increasing or strictly decreasing; I’ll let you rule out the latter case.
Suppose it is increasing. If $f(x)gt x$ for some $xin (0,1)$, then $f(f(x))gt f(x)$ and $x^2gt f(x)gt x$ which is impossible. If $f(x)lt x^2$ for some $xin (0,1)$, then $f(x)lt f(f(x))$ and $xlt f(x)lt x^2$ which is impossible. Thus $x^2le f(x)le x$.
answered Jan 26 at 1:40
FrpzzdFrpzzd
23k841110
$endgroup$
$begingroup$
$int_0^1 x^4 dx=1/5.$
$endgroup$
– user
Jan 26 at 1:44
$begingroup$
@user Ha. I honestly don’t know if that was a typo, or if I was just being stupid. :P
$endgroup$
– Frpzzd
Jan 26 at 1:45
1
$begingroup$
I would lay bets on a typo. It would be interesting to know if there are other continuous functions besides $f(x)=x^sqrt2$ which satisfy the functional equation.
$endgroup$
– user
Jan 26 at 2:05
1
$begingroup$
It never crossed my mind to use injectivity to prove that the function is strictly increasing, I understant it now, thanks. I wonder if the lower bound can be raised even higher, maybe somewhere around 1/4.(Seeing that for $f(x)=x^sqrt{2}$ the value of the integral is around 0.26 which is greater than 0.25).
$endgroup$
– Ban
Jan 26 at 3:13
add a comment |
$begingroup$
It can be proven that if $f:[0,infty)mapsto [0,infty)$ is continuous and $f(f(x))=x^2$, then $f(x)$ is bounded (not strictly) between $x$ and $x^2$ for all $xin (0,1)$. I will not prove this rigorously here, and leave it as an exercise (it can be done using the continuity requirement and the generalized intermediate value theorem for two curves).
This shows that $f^2(x)$ is between $x^2$ and $x^4$, and that $int_0^1 f^2(x)dxge int_0^1 x^4=1/5gt 1/6$.
MORE DETAILS FOR THE PROOF:
Since $f(f(x))=x^2$ is injective, $f(x)$ must be injective, and thus either strictly increasing or strictly decreasing; I’ll let you rule out the latter case.
Suppose it is increasing. If $f(x)gt x$ for some $xin (0,1)$, then $f(f(x))gt f(x)$ and $x^2gt f(x)gt x$ which is impossible. If $f(x)lt x^2$ for some $xin (0,1)$, then $f(x)lt f(f(x))$ and $xlt f(x)lt x^2$ which is impossible. Thus $x^2le f(x)le x$.
answered Jan 26 at 1:40
FrpzzdFrpzzd
23k841110
$endgroup$
$begingroup$
$int_0^1 x^4 dx=1/5.$
$endgroup$
– user
Jan 26 at 1:44
$begingroup$
@user Ha. I honestly don’t know if that was a typo, or if I was just being stupid. :P
$endgroup$
– Frpzzd
Jan 26 at 1:45
1
$begingroup$
I would lay bets on a typo. It would be interesting to know if there are other continuous functions besides $f(x)=x^sqrt2$ which satisfy the functional equation.
$endgroup$
– user
Jan 26 at 2:05
1
$begingroup$
It never crossed my mind to use injectivity to prove that the function is strictly increasing, I understant it now, thanks. I wonder if the lower bound can be raised even higher, maybe somewhere around 1/4.(Seeing that for $f(x)=x^sqrt{2}$ the value of the integral is around 0.26 which is greater than 0.25).
$endgroup$
– Ban
Jan 26 at 3:13
add a comment |
$begingroup$
It can be proven that if $f:[0,infty)mapsto [0,infty)$ is continuous and $f(f(x))=x^2$, then $f(x)$ is bounded (not strictly) between $x$ and $x^2$ for all $xin (0,1)$. I will not prove this rigorously here, and leave it as an exercise (it can be done using the continuity requirement and the generalized intermediate value theorem for two curves).
This shows that $f^2(x)$ is between $x^2$ and $x^4$, and that $int_0^1 f^2(x)dxge int_0^1 x^4=1/5gt 1/6$.
MORE DETAILS FOR THE PROOF:
Since $f(f(x))=x^2$ is injective, $f(x)$ must be injective, and thus either strictly increasing or strictly decreasing; I’ll let you rule out the latter case.
Suppose it is increasing. If $f(x)gt x$ for some $xin (0,1)$, then $f(f(x))gt f(x)$ and $x^2gt f(x)gt x$ which is impossible. If $f(x)lt x^2$ for some $xin (0,1)$, then $f(x)lt f(f(x))$ and $xlt f(x)lt x^2$ which is impossible. Thus $x^2le f(x)le x$.
answered Jan 26 at 1:40
FrpzzdFrpzzd
23k841110
$endgroup$
It can be proven that if $f:[0,infty)mapsto [0,infty)$ is continuous and $f(f(x))=x^2$, then $f(x)$ is bounded (not strictly) between $x$ and $x^2$ for all $xin (0,1)$. I will not prove this rigorously here, and leave it as an exercise (it can be done using the continuity requirement and the generalized intermediate value theorem for two curves).
This shows that $f^2(x)$ is between $x^2$ and $x^4$, and that $int_0^1 f^2(x)dxge int_0^1 x^4=1/5gt 1/6$.
MORE DETAILS FOR THE PROOF:
Since $f(f(x))=x^2$ is injective, $f(x)$ must be injective, and thus either strictly increasing or strictly decreasing; I’ll let you rule out the latter case.
Suppose it is increasing. If $f(x)gt x$ for some $xin (0,1)$, then $f(f(x))gt f(x)$ and $x^2gt f(x)gt x$ which is impossible. If $f(x)lt x^2$ for some $xin (0,1)$, then $f(x)lt f(f(x))$ and $xlt f(x)lt x^2$ which is impossible. Thus $x^2le f(x)le x$.
answered Jan 26 at 1:40
FrpzzdFrpzzd
23k841110
edited Jan 26 at 1:53
answered Jan 26 at 1:40
FrpzzdFrpzzd
23k841110
answered Jan 26 at 1:40
FrpzzdFrpzzd
23k841110
answered Jan 26 at 1:40
FrpzzdFrpzzd
23k841110
23k841110
$begingroup$
$int_0^1 x^4 dx=1/5.$
$endgroup$
– user
Jan 26 at 1:44
$begingroup$
@user Ha. I honestly don’t know if that was a typo, or if I was just being stupid. :P
$endgroup$
– Frpzzd
Jan 26 at 1:45
1
$begingroup$
I would lay bets on a typo. It would be interesting to know if there are other continuous functions besides $f(x)=x^sqrt2$ which satisfy the functional equation.
$endgroup$
– user
Jan 26 at 2:05
1
$begingroup$
It never crossed my mind to use injectivity to prove that the function is strictly increasing, I understant it now, thanks. I wonder if the lower bound can be raised even higher, maybe somewhere around 1/4.(Seeing that for $f(x)=x^sqrt{2}$ the value of the integral is around 0.26 which is greater than 0.25).
$endgroup$
– Ban
Jan 26 at 3:13
add a comment |
$begingroup$
$int_0^1 x^4 dx=1/5.$
$endgroup$
– user
Jan 26 at 1:44
$begingroup$
@user Ha. I honestly don’t know if that was a typo, or if I was just being stupid. :P
$endgroup$
– Frpzzd
Jan 26 at 1:45
1
$begingroup$
I would lay bets on a typo. It would be interesting to know if there are other continuous functions besides $f(x)=x^sqrt2$ which satisfy the functional equation.
$endgroup$
– user
Jan 26 at 2:05
1
$begingroup$
It never crossed my mind to use injectivity to prove that the function is strictly increasing, I understant it now, thanks. I wonder if the lower bound can be raised even higher, maybe somewhere around 1/4.(Seeing that for $f(x)=x^sqrt{2}$ the value of the integral is around 0.26 which is greater than 0.25).
$endgroup$
– Ban
Jan 26 at 3:13
$begingroup$
$int_0^1 x^4 dx=1/5.$
$endgroup$
– user
Jan 26 at 1:44
$begingroup$
$int_0^1 x^4 dx=1/5.$
$endgroup$
– user
Jan 26 at 1:44
$begingroup$
@user Ha. I honestly don’t know if that was a typo, or if I was just being stupid. :P
$endgroup$
– Frpzzd
Jan 26 at 1:45
$begingroup$
@user Ha. I honestly don’t know if that was a typo, or if I was just being stupid. :P
$endgroup$
– Frpzzd
Jan 26 at 1:45
1
1
$begingroup$
I would lay bets on a typo. It would be interesting to know if there are other continuous functions besides $f(x)=x^sqrt2$ which satisfy the functional equation.
$endgroup$
– user
Jan 26 at 2:05
$begingroup$
I would lay bets on a typo. It would be interesting to know if there are other continuous functions besides $f(x)=x^sqrt2$ which satisfy the functional equation.
$endgroup$
– user
Jan 26 at 2:05
1
1
$begingroup$
It never crossed my mind to use injectivity to prove that the function is strictly increasing, I understant it now, thanks. I wonder if the lower bound can be raised even higher, maybe somewhere around 1/4.(Seeing that for $f(x)=x^sqrt{2}$ the value of the integral is around 0.26 which is greater than 0.25).
$endgroup$
– Ban
Jan 26 at 3:13
$begingroup$
It never crossed my mind to use injectivity to prove that the function is strictly increasing, I understant it now, thanks. I wonder if the lower bound can be raised even higher, maybe somewhere around 1/4.(Seeing that for $f(x)=x^sqrt{2}$ the value of the integral is around 0.26 which is greater than 0.25).
$endgroup$
– Ban
Jan 26 at 3:13
add a comment |
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