f'(1)>1 implies f has a fixed point in (0,1)

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f'(1)>1 implies f has a fixed point in (0,1)

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Source: ISI(BS) 2010 #4

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Sayan

#1 h May 17, 2012, 4:17 AM • 2 Y

Y by Adventure10, Mango247

A real valued function $f$ is defined on the interval $(-1,2)$ . A point $x_0$ is said to be a fixed point of $f$ if $f(x_0)=x_0$ . Suppose that $f$ is a differentiable function such that $f(0)>0$ and $f(1)=1$ . Show that if $f'(1)>1$ , then $f$ has a fixed point in the interval $(0,1)$ .

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#2 h May 17, 2012, 4:34 AM • 5 Y

Y by Abhinandan18, ring_r, Severus, Adventure10, and 1 other user

Define $g(x)=f(x)-x$ . Since $g'(1)>0$ and $g(1)=0$ there is a $0<\xi <1$ such that $g(\xi )<0$ . But $g(0)>0$ , so by the intermediate value theorem there is an $x_0\in (0,\xi)\subset (0,1)$ such that $g(x_0)=0\Leftrightarrow f(x_0)=x_0$ .

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#3 h May 17, 2012, 8:02 AM • 6 Y

Y by chipgiboy, ring_r, integrated_JRC, Jaggoseni, Adventure10, Mango247

JoeBlow wrote:

Define $g(x)=f(x)-x$ . Since $g'(1)>0$ and $g(1)=0$ there is a $0<\xi <1$ such that $g(\xi )<0$

maybe this little thing have to be clarified:
$0< g'(1) = \lim_{x \to 1^{-}} \frac{g(x)}{x-1}$
then there exists $c$ in $]0,1[$ such that $\frac{g(x)}{x-1} >0$ $\forall x \in ]c,1[$

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Aranya

#4 h May 1, 2014, 10:33 AM • 1 Y

Y by Adventure10

My solution: By the Mean Value Theorem, $\dfrac {f(1)-f(0)}{1-0}=f'(c)$ for some c belonging to (0,1).

That gives $f'(c)<1$ for some c in (0,1).

Now, let $f(x)>x$ for all x in (0,1). Then, $f'(x)>1$ for all x in (0,1) which contradicts our result.

Taking $f(x)<x$ for all x belonging to (0,1) implies $f'(x)<1$ for all x lying in (0,1) which is false by hypothesis.

Therefore, there must be at least one $x_0$ so that $f(x_0)=x_0$ for $x_0$ belonging to (0,1). QED.

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kd7

#5 h May 12, 2017, 5:08 AM • 1 Y

Y by Adventure10

Aranya wrote:

My solution: By the Mean Value Theorem, $\dfrac {f(1)-f(0)}{1-0}=f'(c)$ for some c belonging to (0,1).

That gives $f'(c)<1$ for some c in (0,1).

Now, let $f(x)>x$ for all x in (0,1). Then, $f'(x)>1$ for all x in (0,1) which contradicts our result.

Taking $f(x)<x$ for all x belonging to (0,1) implies $f'(x)<1$ for all x lying in (0,1) which is false by hypothesis.

Therefore, there must be at least one $x_0$ so that $f(x_0)=x_0$ for $x_0$ belonging to (0,1). QED.

No. Not necessary if $f(x)>x$ that implies $f'(x)>1$

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#6 h Aug 25, 2020, 12:29 PM

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Well, an alternate solution.

Just consider the line $y=x$ .

We claim that $f(x)$ intersects this line at some $x=\beta$ , where, $\beta \in (0,1)$

Proof:

We know that the slope of $y=x$ at $x=1$ is $1$ .

Given that $f(1)=1$ and $f'(1)>1$ . This implies that at the point $1$ , $f(x)$ intersects $y=x$ from under the line to above.

Converting this into formal language we may state that $f(1-\epsilon)<1$ and $f(1+\epsilon)>1$ for some $\epsilon \rightarrow 0^+$

Therefore, when $x \rightarrow 1^-$ , $f(x)$ lies below $y=x$ ......... $(\zeta_1)$

Now, given that $f(0)>0$ . This clearly states that $f(x)$ lies above $y=x$ when $x=0$ .......... $(\zeta_2)$

Combining $\zeta_1$ and $\zeta_2$ we may conclude that $f(x)$ must have intersected the line $y=x$ at some point in the interval of $(0,1)$ .

This proves our claim. $\blacksquare$

$\boxed{\therefore f(\beta)=\beta \text{ for some } \beta \in (0,1)}$ , which is just what was to be proved.

Q.E.D. $\square$

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#7 h Apr 20, 2021, 8:28 PM

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Sayan wrote:

A real valued function $f$ is defined on the interval $(-1,2)$ . A point $x_0$ is said to be a fixed point of $f$ if $f(x_0)=x_0$ . Suppose that $f$ is a differentiable function such that $f(0)>0$ and $f(1)=1$ . Show that if $f'(1)>1$ , then $f$ has a fixed point in the interval $(0,1)$ .

Can Someone Please check This...

Now $f(0)>0, f(1)=1\implies$ By LMVT There exist a constant $k\in(0,1)$ such that $\frac{f(1) -f(0)}{1-0}=f'(k)\implies$

either $1>f'(k)\geq 0$ or $f'(k)<0$
FTSOC Assume that doesn't exist a fixed point $\in (0, 1)$ Then we again by LMVT We can't have any $x_1\in (0, 1)$ Such that $f'(x_1)=1$ But this is never possible As there exist a $k\in (0, 1)$ such that either $f'(k)<0$ or $1>f'(k)\geq 0$ and $f'(1)>1,f(x)$ is continuous So there will always exist a $x_1\in (0, 1)$ such that $f(x_1)=1$ Hence Contradiction.

This post has been edited 1 time. Last edited by Maths_1729, Apr 20, 2021, 8:29 PM

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#8 h Jan 30, 2023, 12:18 PM

Y by

Storage(same as #2)

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#9 h Jan 30, 2023, 6:56 PM

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MathsSolver007 wrote:

Storage(same as #2)

This one is a worth mentioning fact(an often overlooked one)

$\textbf{Fact:}$ $\text{For a function f, if}~ \nolinebreak \exists c \text{ such that } f'(c)>0(\text{or} < 0) \text{ then f is not necessarily increasing(or decreasing) on any neighbourhood around } c$

Take $f(x) = \begin{cases} \frac{x}{2} + x^2 sin(\frac{1}{x}) ~&\text{if} ~ x \neq 0 \\ 0 ~ &\text{if} ~ x = 0 \end{cases}$
One can check that $f'(0) >0$ but the function is not increasing on any open interval containing 0.

@above your solution is therefore incorrect

This post has been edited 1 time. Last edited by Agsh2005, Jan 30, 2023, 6:56 PM

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#10 h Jan 30, 2023, 7:28 PM

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A proper proof for this problem would be the following.

Consider the function $g(x) = f(x) -x$ on the interval $[0,1]$ . By the Extreme Value theorem we can say that $\exists ~ m \in [0,1]$ such that
$g(m) \leq g(x) ~ \forall x \in [0,1]$ and $\exists M \in [0,1]$ such that $g(M)\geq g(x)~ \forall x \in [0,1]$ Now we have that $g(1)=0$ which gives us $g(m)\leq 0$ Since $g(0)>0$ we must have that $M \neq 1$ .

$\text{Claim:}$ $m \in (0,1)$ .
$\text{Proof:}$ Let us assume FTSoC that $m=1$ .
But that would give us $g'(1)= \lim_{x \to 1^{-}} \frac{g(x)- g(1)} {x-1} < 0$ which is a contradiction.

We can now conclude by the Intermediate Value Theorem.

This post has been edited 1 time. Last edited by Agsh2005, Jan 30, 2023, 7:28 PM

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#11 h Jan 30, 2023, 9:15 PM

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Agsh2005 wrote:

A proper proof for this problem would be the following.

Consider the function $g(x) = f(x) -x$ on the interval $[0,1]$ . By the Extreme Value theorem we can say that $\exists ~ m \in [0,1]$ such that
$g(m) \leq g(x) ~ \forall x \in [0,1]$ and $\exists M \in [0,1]$ such that $g(M)\geq g(x)~ \forall x \in [0,1]$ Now we have that $g(1)=0$ which gives us $g(m)\leq 0$ Since $g(0)>0$ we must have that $M \neq 1$ .

$\text{Claim:}$ $m \in (0,1)$ .
$\text{Proof:}$ Let us assume FTSoC that $m=1$ .
But that would give us $g'(1)= \lim_{x \to 1^{-}} \frac{g(x)- g(1)} {x-1} < 0$ which is a contradiction.

We can now conclude by the Intermediate Value Theorem.

Agreed that my wording was a little bit unclear.

I didn't mean to say that $f$ is increasing or decreasing in the $\varepsilon$ nbd.

The definition of the derivative (the left hand limit) gives us that $\frac{g(1)-g(h)}{1-h} >0$ which in turn implies the existence of some $h$ such that $g(h)>0$ . We finish by IVT

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lifeismathematics

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lifeismathematics

#13 h Apr 28, 2023, 10:30 AM • 1 Y

Y by MathSigma

we define $g(x)=f(x)-x$

we have $g(0)>0 , g(1)=0$ and $g'(1)>0$

we propose a $\textbf{\textcolor{red}{Lemma}}:-$ if $h'(c)>0$ , then $\exists$ a $\delta>0$ s.t. $h(x)<h(c)$ for $x \in (c-\delta,c)$

we use the lemma to get for $x^{*} \in (1-\delta,1) , g(x^{*})<g(1)=0$ so we get $g(x^{*})<0$ and $g(0)>0$ by $\textbf{IVT}$ it follows that $\exists$ a $x_{0}$ s.t. $f(x_{0})=x_{0}$ $\blacksquare$

This post has been edited 1 time. Last edited by lifeismathematics, Apr 28, 2023, 10:31 AM

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#14 h Oct 28, 2023, 6:22 AM

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Okay, very standard and doable problem. I'm not sure, but prolly irrelevant compared to today's standards? Editing on July 14, 2024: Okay this is not that bad, I should not speak bad about problems

Solution: Consider the function $g(x) \coloneqq f(x) - x$ . The problem conditions now translate to $g(0) > 0$ , $g(1) = 0$ with $g$ differentiable on the interval. We also have that $g'(1) > 0$ with $g$ being continuous since all differentiable functions are continuous. It suffices to show that $g$ has a zero in $(0,1)$ .

Lemma: If $g'(1) > 0$ , then there is some $\xi$ such that $g(\xi) < 0$ for $\xi \in (0,1)$ .

Proof: We will only consider left-hand limit of $g$ at $1$ . By the $\epsilon-\delta$ definition of limits, we can conclude that for some $\delta > 0$ , we must have
$1-\delta < x < 1 \implies \left| \frac{g(x)-g(1)}{x-1} -g'(1)\right| < g'(1) \iff -2g'(1) < \frac{g(x)}{1-x} < 0$ which is basically by picking $\epsilon = g'(1)$ . Since $1-x > 0$ , we can say that there exists there is some $\xi \in (0,1)$ such that $g(\xi) < 0$ which proves the lemma. $\square$

Now note that
$g'(1) = \lim_{x \to 1}\frac{g(x)-g(1)}{x-1} > 0 \implies \exists \, a \colon g(a) < 0 \text{ for some $a \in (0,1)$.}$ Applying Intermediate Value Theorem on $(0,g(0))$ and $(a,g(a))$ , we get that $g(\alpha)$ must be $0$ for some $\alpha \in (0,1)$ . This finishes the proof. $\blacksquare$

This post has been edited 1 time. Last edited by SatisfiedMagma, Jul 14, 2024, 11:29 AM
Reason: proving the crucial lemma with "rigor"

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#15 h Apr 27, 2025, 7:12 PM

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g(x) =f(x)-x
Then, g(0)=f(0)>0
g(1)=f(1)-1=0
Since g is continuous and differentiable, we get
g'(x)=f'(x)-1
g'(1)=f'(1)>1, so g is increasing at the point 1
Now we take a k(very small) neighborhood around 1 say (k-1,k+1) the there exist a point c belong to (k-1,k+1) s.t. g(c) <0
So we got g(c)<0 & g(0)>0 , Then by IVT, there exist a point t belong to (0,c) which is subset of (0,1) s.t. g(t)=0
So, g(t)=0
f(t)-t=0
f(t)=t . Done!

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