How to prove the non-existence of continuous solution for a functional equation?
$begingroup$
Let us consider the following functional equation
$$
f(f(x))=-x^3+sin(x^2+ln(1+|x|))
$$
How to prove that there is NO continuous function $f:mathbb{R}tomathbb{R}$ satisfies the equation above?
First I notice that
$$
F(x):=-x^3+sin(x^2+ln(1+|x|))
$$
has only one fixed point $0$. From this it follows that
$$
f(0)=0.
$$ Otherwise, if we assume that $a:=f(0)>0$, then $f(a)=F(0)=0$. It is easy to see that $f(x)$ has a fixed point in $(0,a)$ by intermediate value theorem. Similarly, we can rule out the case $f(0)<0$. Therefore we have $f(0)=0$
I want to show that $f$ has a nonzero fixed point $x^*$, from which we find a contradiction. This is my initial idea. But I don't how to find such non-zero fixed point.
If I find a closed interval $[a,b]$, which does not include 0, such that $f([a,b])subset [a,b]$. Then we know that $f$ has at least one fixed point in $[a,b]$ by intermediate value theorem.
Maybe, my idea is wrong. Are there some other approaches?
What follows is the plot of function $F$, which may give you some hints:
calculus real-analysis
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
$endgroup$
add a comment |
$begingroup$
Let us consider the following functional equation
$$
f(f(x))=-x^3+sin(x^2+ln(1+|x|))
$$
How to prove that there is NO continuous function $f:mathbb{R}tomathbb{R}$ satisfies the equation above?
First I notice that
$$
F(x):=-x^3+sin(x^2+ln(1+|x|))
$$
has only one fixed point $0$. From this it follows that
$$
f(0)=0.
$$ Otherwise, if we assume that $a:=f(0)>0$, then $f(a)=F(0)=0$. It is easy to see that $f(x)$ has a fixed point in $(0,a)$ by intermediate value theorem. Similarly, we can rule out the case $f(0)<0$. Therefore we have $f(0)=0$
I want to show that $f$ has a nonzero fixed point $x^*$, from which we find a contradiction. This is my initial idea. But I don't how to find such non-zero fixed point.
If I find a closed interval $[a,b]$, which does not include 0, such that $f([a,b])subset [a,b]$. Then we know that $f$ has at least one fixed point in $[a,b]$ by intermediate value theorem.
Maybe, my idea is wrong. Are there some other approaches?
What follows is the plot of function $F$, which may give you some hints:
calculus real-analysis
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
$endgroup$
add a comment |
$begingroup$
Let us consider the following functional equation
$$
f(f(x))=-x^3+sin(x^2+ln(1+|x|))
$$
How to prove that there is NO continuous function $f:mathbb{R}tomathbb{R}$ satisfies the equation above?
First I notice that
$$
F(x):=-x^3+sin(x^2+ln(1+|x|))
$$
has only one fixed point $0$. From this it follows that
$$
f(0)=0.
$$ Otherwise, if we assume that $a:=f(0)>0$, then $f(a)=F(0)=0$. It is easy to see that $f(x)$ has a fixed point in $(0,a)$ by intermediate value theorem. Similarly, we can rule out the case $f(0)<0$. Therefore we have $f(0)=0$
I want to show that $f$ has a nonzero fixed point $x^*$, from which we find a contradiction. This is my initial idea. But I don't how to find such non-zero fixed point.
If I find a closed interval $[a,b]$, which does not include 0, such that $f([a,b])subset [a,b]$. Then we know that $f$ has at least one fixed point in $[a,b]$ by intermediate value theorem.
Maybe, my idea is wrong. Are there some other approaches?
What follows is the plot of function $F$, which may give you some hints:
calculus real-analysis
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
$endgroup$
Let us consider the following functional equation
$$
f(f(x))=-x^3+sin(x^2+ln(1+|x|))
$$
How to prove that there is NO continuous function $f:mathbb{R}tomathbb{R}$ satisfies the equation above?
First I notice that
$$
F(x):=-x^3+sin(x^2+ln(1+|x|))
$$
has only one fixed point $0$. From this it follows that
$$
f(0)=0.
$$ Otherwise, if we assume that $a:=f(0)>0$, then $f(a)=F(0)=0$. It is easy to see that $f(x)$ has a fixed point in $(0,a)$ by intermediate value theorem. Similarly, we can rule out the case $f(0)<0$. Therefore we have $f(0)=0$
I want to show that $f$ has a nonzero fixed point $x^*$, from which we find a contradiction. This is my initial idea. But I don't how to find such non-zero fixed point.
If I find a closed interval $[a,b]$, which does not include 0, such that $f([a,b])subset [a,b]$. Then we know that $f$ has at least one fixed point in $[a,b]$ by intermediate value theorem.
Maybe, my idea is wrong. Are there some other approaches?
What follows is the plot of function $F$, which may give you some hints:
calculus real-analysis
calculus real-analysis
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
edited Dec 1 '18 at 16:37
Yuhang
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
asked Dec 1 '18 at 16:12
YuhangYuhang
835118
835118
add a comment |
add a comment |
1 Answer
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$begingroup$
Another approach: look to large values. Let $g(x)$ be that expression on the right, and note that $lim_{xtoinfty} g(x)=-infty$ and $lim_{xto-infty} g(x)=infty$. Also, there is some $M$ such that $g$ is strictly decreasing on both $(-infty,-M)$ and $(M,infty)$. Suppose there is a continuous $f$ with $fcirc f = g$. Then, since $g$ is injective on $(-infty,-M)$, $f$ is also injective on $(-infty,-M)$, hence monotone. Let $L=lim_{xtoinfty} f(x)$, in the extended real line. If $L$ is finite, then by continuity $f(L)=lim_{xtoinfty}f(f(x))=-infty$. That's impossible, so either $L=infty$ or $L=-infty$. If $L=infty$, then $lim_{xtoinfty} f(f(x)) =lim_{ytoinfty} f(y) = infty$, a contradiction. If $L=-infty$, then $-infty = lim_{xtoinfty} f(f(x))=lim_{yto-infty} f(y)$, and we'll have the same problem on the other side ($lim_{xto-infty} f(f(x))=lim_{ytoinfty} f(y)=-infty$). Either way, we can't have both $lim_{xtoinfty} f(f(x))=-infty$ and $lim_{xto-infty} f(f(x))=infty$ on the same globally continuous $f$. There is no such function.
answered Dec 1 '18 at 17:16
jmerryjmerry
3,737514
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
Another approach: look to large values. Let $g(x)$ be that expression on the right, and note that $lim_{xtoinfty} g(x)=-infty$ and $lim_{xto-infty} g(x)=infty$. Also, there is some $M$ such that $g$ is strictly decreasing on both $(-infty,-M)$ and $(M,infty)$. Suppose there is a continuous $f$ with $fcirc f = g$. Then, since $g$ is injective on $(-infty,-M)$, $f$ is also injective on $(-infty,-M)$, hence monotone. Let $L=lim_{xtoinfty} f(x)$, in the extended real line. If $L$ is finite, then by continuity $f(L)=lim_{xtoinfty}f(f(x))=-infty$. That's impossible, so either $L=infty$ or $L=-infty$. If $L=infty$, then $lim_{xtoinfty} f(f(x)) =lim_{ytoinfty} f(y) = infty$, a contradiction. If $L=-infty$, then $-infty = lim_{xtoinfty} f(f(x))=lim_{yto-infty} f(y)$, and we'll have the same problem on the other side ($lim_{xto-infty} f(f(x))=lim_{ytoinfty} f(y)=-infty$). Either way, we can't have both $lim_{xtoinfty} f(f(x))=-infty$ and $lim_{xto-infty} f(f(x))=infty$ on the same globally continuous $f$. There is no such function.
answered Dec 1 '18 at 17:16
jmerryjmerry
3,737514
$endgroup$
add a comment |
$begingroup$
Another approach: look to large values. Let $g(x)$ be that expression on the right, and note that $lim_{xtoinfty} g(x)=-infty$ and $lim_{xto-infty} g(x)=infty$. Also, there is some $M$ such that $g$ is strictly decreasing on both $(-infty,-M)$ and $(M,infty)$. Suppose there is a continuous $f$ with $fcirc f = g$. Then, since $g$ is injective on $(-infty,-M)$, $f$ is also injective on $(-infty,-M)$, hence monotone. Let $L=lim_{xtoinfty} f(x)$, in the extended real line. If $L$ is finite, then by continuity $f(L)=lim_{xtoinfty}f(f(x))=-infty$. That's impossible, so either $L=infty$ or $L=-infty$. If $L=infty$, then $lim_{xtoinfty} f(f(x)) =lim_{ytoinfty} f(y) = infty$, a contradiction. If $L=-infty$, then $-infty = lim_{xtoinfty} f(f(x))=lim_{yto-infty} f(y)$, and we'll have the same problem on the other side ($lim_{xto-infty} f(f(x))=lim_{ytoinfty} f(y)=-infty$). Either way, we can't have both $lim_{xtoinfty} f(f(x))=-infty$ and $lim_{xto-infty} f(f(x))=infty$ on the same globally continuous $f$. There is no such function.
answered Dec 1 '18 at 17:16
jmerryjmerry
3,737514
$endgroup$
add a comment |
$begingroup$
Another approach: look to large values. Let $g(x)$ be that expression on the right, and note that $lim_{xtoinfty} g(x)=-infty$ and $lim_{xto-infty} g(x)=infty$. Also, there is some $M$ such that $g$ is strictly decreasing on both $(-infty,-M)$ and $(M,infty)$. Suppose there is a continuous $f$ with $fcirc f = g$. Then, since $g$ is injective on $(-infty,-M)$, $f$ is also injective on $(-infty,-M)$, hence monotone. Let $L=lim_{xtoinfty} f(x)$, in the extended real line. If $L$ is finite, then by continuity $f(L)=lim_{xtoinfty}f(f(x))=-infty$. That's impossible, so either $L=infty$ or $L=-infty$. If $L=infty$, then $lim_{xtoinfty} f(f(x)) =lim_{ytoinfty} f(y) = infty$, a contradiction. If $L=-infty$, then $-infty = lim_{xtoinfty} f(f(x))=lim_{yto-infty} f(y)$, and we'll have the same problem on the other side ($lim_{xto-infty} f(f(x))=lim_{ytoinfty} f(y)=-infty$). Either way, we can't have both $lim_{xtoinfty} f(f(x))=-infty$ and $lim_{xto-infty} f(f(x))=infty$ on the same globally continuous $f$. There is no such function.
answered Dec 1 '18 at 17:16
jmerryjmerry
3,737514
$endgroup$
Another approach: look to large values. Let $g(x)$ be that expression on the right, and note that $lim_{xtoinfty} g(x)=-infty$ and $lim_{xto-infty} g(x)=infty$. Also, there is some $M$ such that $g$ is strictly decreasing on both $(-infty,-M)$ and $(M,infty)$. Suppose there is a continuous $f$ with $fcirc f = g$. Then, since $g$ is injective on $(-infty,-M)$, $f$ is also injective on $(-infty,-M)$, hence monotone. Let $L=lim_{xtoinfty} f(x)$, in the extended real line. If $L$ is finite, then by continuity $f(L)=lim_{xtoinfty}f(f(x))=-infty$. That's impossible, so either $L=infty$ or $L=-infty$. If $L=infty$, then $lim_{xtoinfty} f(f(x)) =lim_{ytoinfty} f(y) = infty$, a contradiction. If $L=-infty$, then $-infty = lim_{xtoinfty} f(f(x))=lim_{yto-infty} f(y)$, and we'll have the same problem on the other side ($lim_{xto-infty} f(f(x))=lim_{ytoinfty} f(y)=-infty$). Either way, we can't have both $lim_{xtoinfty} f(f(x))=-infty$ and $lim_{xto-infty} f(f(x))=infty$ on the same globally continuous $f$. There is no such function.
answered Dec 1 '18 at 17:16
jmerryjmerry
3,737514
answered Dec 1 '18 at 17:16
jmerryjmerry
3,737514
answered Dec 1 '18 at 17:16
jmerryjmerry
3,737514
answered Dec 1 '18 at 17:16
jmerryjmerry
3,737514
3,737514
add a comment |
add a comment |
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