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Show a continuous function with $f(x)=y$ and $f(y)=x$ has a fixed point.

Show that any continuous $f:[0,1] rightarrow [0,1]$ has a fixed point $zeta$Fixed point and period of continuous functionContinuous decreasing function has a fixed pointIVT and fixed point theorem$f$ is continuous, $f : X to X$, $X$ compact, and $f$ has an $epsilon$-fixed point for each $epsilon > 0$. Show $f$ has a fixed point.Showing that $f$ has a fixed point.Prove that a continuous function has a fixed pointContinuous function and fixed pointProve that $f:[0,1] to [0,1]$ has a fixed pointFor a continuous function $f$ satisfying $f(f(x))=x$ has exactly one fixed point

3

$begingroup$

Suppose $a<b$ and $f:[a,b] to [a,b]$ be continous. Suppose that $x neq y$ in $[a,b]$ with $f(x)=y$ and $f(y)=x$. Prove that $f$ has a fixed point in $(x,y)$.

So I was thinking of considering the function $g(x)=f(x)-x$, which we know is continuous. Then we also know that because $f(a) geq a$ that $g(a)=f(a)-a geq 0$. Similarly, because $f(b) leq b$ then $g(b)=f(b)-b leq 0$.

Can we just use the fact that because $g(x)$ is continuous, $0 in [g(b),g(a)]$, the IVT says there exists $c in [a,b]$ such that $g(c)=f(c)-c=0$ so $f(c)=c$? Then we know $c$ is a fixed point.

How do we show that $c$ is in $(x,y)$??

We know that $g(x)=f(x)-x=y-x neq 0$
and $g(y)=f(y)-y=x-y neq 0$ but we don't know that those are in $(a,b)$?

real-analysis

share|cite|improve this question

edited 9 hours ago

YuiTo Cheng

2,3084937

asked 11 hours ago

big_math_boybig_math_boy

303

$endgroup$

add a comment | 

3

$begingroup$

Suppose $a<b$ and $f:[a,b] to [a,b]$ be continous. Suppose that $x neq y$ in $[a,b]$ with $f(x)=y$ and $f(y)=x$. Prove that $f$ has a fixed point in $(x,y)$.

So I was thinking of considering the function $g(x)=f(x)-x$, which we know is continuous. Then we also know that because $f(a) geq a$ that $g(a)=f(a)-a geq 0$. Similarly, because $f(b) leq b$ then $g(b)=f(b)-b leq 0$.

Can we just use the fact that because $g(x)$ is continuous, $0 in [g(b),g(a)]$, the IVT says there exists $c in [a,b]$ such that $g(c)=f(c)-c=0$ so $f(c)=c$? Then we know $c$ is a fixed point.

How do we show that $c$ is in $(x,y)$??

We know that $g(x)=f(x)-x=y-x neq 0$
and $g(y)=f(y)-y=x-y neq 0$ but we don't know that those are in $(a,b)$?

real-analysis

share|cite|improve this question

edited 9 hours ago

YuiTo Cheng

2,3084937

asked 11 hours ago

big_math_boybig_math_boy

303

$endgroup$

add a comment | 

$begingroup$

Suppose $a<b$ and $f:[a,b] to [a,b]$ be continous. Suppose that $x neq y$ in $[a,b]$ with $f(x)=y$ and $f(y)=x$. Prove that $f$ has a fixed point in $(x,y)$.

So I was thinking of considering the function $g(x)=f(x)-x$, which we know is continuous. Then we also know that because $f(a) geq a$ that $g(a)=f(a)-a geq 0$. Similarly, because $f(b) leq b$ then $g(b)=f(b)-b leq 0$.

Can we just use the fact that because $g(x)$ is continuous, $0 in [g(b),g(a)]$, the IVT says there exists $c in [a,b]$ such that $g(c)=f(c)-c=0$ so $f(c)=c$? Then we know $c$ is a fixed point.

How do we show that $c$ is in $(x,y)$??

We know that $g(x)=f(x)-x=y-x neq 0$
and $g(y)=f(y)-y=x-y neq 0$ but we don't know that those are in $(a,b)$?

real-analysis

share|cite|improve this question

edited 9 hours ago

YuiTo Cheng

2,3084937

asked 11 hours ago

big_math_boybig_math_boy

303

$endgroup$

share|cite|improve this question

edited 9 hours ago

YuiTo Cheng

2,3084937

asked 11 hours ago

big_math_boybig_math_boy

303

share|cite|improve this question

edited 9 hours ago

YuiTo Cheng

2,3084937

asked 11 hours ago

big_math_boybig_math_boy

303

edited 9 hours ago

YuiTo Cheng

2,3084937

edited 9 hours ago

YuiTo Cheng

2,3084937

asked 11 hours ago

big_math_boybig_math_boy

303

asked 11 hours ago

big_math_boybig_math_boy

303

2 Answers
2

active

oldest

votes

2

$begingroup$

You've essentially stated the argument. But, rather restrict $f$ to $[x,y]$ (where w.l.o.g $x<y$). Then, define $g:[x,y]tomathbb R$ by

$$g(t)=f(t)-t$$

for any $tin [x,y]$. As $f$ is continuous on $[a,b]$ and $[x,y]subseteq [a,b]$, $g$ is continuous on $[x,y]$. Also, you have $g(x)=f(x)-x=y-x>0$ and $g(y)=f(y)-y=x-y<0$ as $x<y$.

Thus, by the intermediate value theorem, there is a $sin (x,y)$ such that $g(s)=0$, i.e. $f(s)=s$.

share|cite|improve this answer

answered 10 hours ago

blubblub

3,241829

$endgroup$

add a comment | 

7

$begingroup$

Without loss of generality you can assume that $x < y$. Now consider $g(t) = f(t) - t$ not on the entire interval $[a, b]$ but only on $[x, y]$.

Then $ g(x) = y- x$ and $g(y) = x-y$ have opposite sign, so that you can apply the intermediate value theorem.

Note also that I have chosen a different variable name ($t$ instead of $x$) for defining $g$, in order to avoid confusion between that variable and the given (fixed) value $x$.

share|cite|improve this answer

edited 10 hours ago

answered 11 hours ago

Martin RMartin R

30.8k33560

$endgroup$

add a comment | 

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2

$begingroup$

You've essentially stated the argument. But, rather restrict $f$ to $[x,y]$ (where w.l.o.g $x<y$). Then, define $g:[x,y]tomathbb R$ by

$$g(t)=f(t)-t$$

for any $tin [x,y]$. As $f$ is continuous on $[a,b]$ and $[x,y]subseteq [a,b]$, $g$ is continuous on $[x,y]$. Also, you have $g(x)=f(x)-x=y-x>0$ and $g(y)=f(y)-y=x-y<0$ as $x<y$.

Thus, by the intermediate value theorem, there is a $sin (x,y)$ such that $g(s)=0$, i.e. $f(s)=s$.

share|cite|improve this answer

answered 10 hours ago

blubblub

3,241829

$endgroup$

add a comment | 

2

$begingroup$

You've essentially stated the argument. But, rather restrict $f$ to $[x,y]$ (where w.l.o.g $x<y$). Then, define $g:[x,y]tomathbb R$ by

$$g(t)=f(t)-t$$

for any $tin [x,y]$. As $f$ is continuous on $[a,b]$ and $[x,y]subseteq [a,b]$, $g$ is continuous on $[x,y]$. Also, you have $g(x)=f(x)-x=y-x>0$ and $g(y)=f(y)-y=x-y<0$ as $x<y$.

Thus, by the intermediate value theorem, there is a $sin (x,y)$ such that $g(s)=0$, i.e. $f(s)=s$.

share|cite|improve this answer

answered 10 hours ago

blubblub

3,241829

$endgroup$

add a comment | 

$begingroup$

You've essentially stated the argument. But, rather restrict $f$ to $[x,y]$ (where w.l.o.g $x<y$). Then, define $g:[x,y]tomathbb R$ by

$$g(t)=f(t)-t$$

for any $tin [x,y]$. As $f$ is continuous on $[a,b]$ and $[x,y]subseteq [a,b]$, $g$ is continuous on $[x,y]$. Also, you have $g(x)=f(x)-x=y-x>0$ and $g(y)=f(y)-y=x-y<0$ as $x<y$.

Thus, by the intermediate value theorem, there is a $sin (x,y)$ such that $g(s)=0$, i.e. $f(s)=s$.

share|cite|improve this answer

answered 10 hours ago

blubblub

3,241829

$endgroup$

share|cite|improve this answer

answered 10 hours ago

blubblub

3,241829

answered 10 hours ago

blubblub

3,241829

answered 10 hours ago

blubblub

3,241829

7

$begingroup$

Without loss of generality you can assume that $x < y$. Now consider $g(t) = f(t) - t$ not on the entire interval $[a, b]$ but only on $[x, y]$.

Then $ g(x) = y- x$ and $g(y) = x-y$ have opposite sign, so that you can apply the intermediate value theorem.

Note also that I have chosen a different variable name ($t$ instead of $x$) for defining $g$, in order to avoid confusion between that variable and the given (fixed) value $x$.

share|cite|improve this answer

edited 10 hours ago

answered 11 hours ago

Martin RMartin R

30.8k33560

$endgroup$

add a comment | 

7

$begingroup$

Without loss of generality you can assume that $x < y$. Now consider $g(t) = f(t) - t$ not on the entire interval $[a, b]$ but only on $[x, y]$.

Then $ g(x) = y- x$ and $g(y) = x-y$ have opposite sign, so that you can apply the intermediate value theorem.

Note also that I have chosen a different variable name ($t$ instead of $x$) for defining $g$, in order to avoid confusion between that variable and the given (fixed) value $x$.

share|cite|improve this answer

edited 10 hours ago

answered 11 hours ago

Martin RMartin R

30.8k33560

$endgroup$

add a comment | 

$begingroup$

Without loss of generality you can assume that $x < y$. Now consider $g(t) = f(t) - t$ not on the entire interval $[a, b]$ but only on $[x, y]$.

Then $ g(x) = y- x$ and $g(y) = x-y$ have opposite sign, so that you can apply the intermediate value theorem.

Note also that I have chosen a different variable name ($t$ instead of $x$) for defining $g$, in order to avoid confusion between that variable and the given (fixed) value $x$.

share|cite|improve this answer

edited 10 hours ago

answered 11 hours ago

Martin RMartin R

30.8k33560

$endgroup$

share|cite|improve this answer

edited 10 hours ago

answered 11 hours ago

Martin RMartin R

30.8k33560

answered 11 hours ago

Martin RMartin R

30.8k33560

answered 11 hours ago

Martin RMartin R

30.8k33560