Let f : $\mathbb{R} \to \mathbb{R}$ be differentiable function and suppose that $|f'(x)|\le 0.49$ for all $x \in \mathbb{R}$. Prove that the equation $f(x) =\frac{2x+\sin(x)}{2}$ has a unique solution in $\mathbb{R}$.

I started by defining a function $g$ where $g(x)= f(x)-\frac{2x+\sin(x)}{2}$ and applying the mean value theorem. I don't know how to move on from there, how should I proceed?

  • 3
    $\begingroup$ Suppose $g(x)$ has at least $2$ roots, say $a$ and $b$. Then there exists a $c\in]a;b[$ with $g'(c)=0$, i.e... and find a lower and upper bound for $\frac{\mathrm d}{\mathrm d x}\frac{2x+\sin(x)}{2}$ $\endgroup$ – Bart Michels Dec 30 '12 at 14:38
  • $\begingroup$ Have you calculated the derivative of $g$? $\endgroup$ – hmakholm left over Monica Dec 30 '12 at 14:43
  • $\begingroup$ @barto, good hint. They'll also have to show $g$ has at least one root, but your hint might suggest the right idea for that, too. $\endgroup$ – Michael E2 Dec 30 '12 at 14:47
  • $\begingroup$ Isn't it enough to show $g$ has one root? or am I missing something? $\endgroup$ – user10444 Dec 30 '12 at 14:58
  • $\begingroup$ You have to show that $g$ has exactly one root. $\endgroup$ – Bart Michels Dec 30 '12 at 15:03

Hint: Use the Banach fixed point theorem.


First, write the function as

$$ x = f(x)-\frac{1}{2}\sin(x) \implies x_{n+1} = f(x_n)-\frac{1}{2}\sin(x_n),\quad n\in \mathbb{Z^{+}}. $$

The above is an iteration procedure to find the root which implies

$$ |x_{n+1} - x_n |=| f(x_n)-\frac{1}{2}\sin(x_n)-(f(x_{n-1})-\frac{1}{2}\sin(x_{n-1})) | $$

$$ \leq | f(x_n)-f(x_{n-1}) |+\frac{1}{2}| \sin(x_n)-\sin(x_{n-1}) | $$

$$ \leq |f'(\eta)|| x_n-x_{n-1} |+\frac{1}{2}| x_n-x_{n-1} |$$

$$|x_{n+1}-x_{n}| \leq 0.49| x_n-x_{n-1} |+\frac{1}{2}| x_n-x_{n-1} | = 0.99|x_n-x_{n-1}|.$$

Now, you can prove that the sequence converges. Can you finish the proof now?

Note: we used the mean value theorem in the above derivations.

  • $\begingroup$ @user10444: You are welcome. $\endgroup$ – Mhenni Benghorbal Dec 30 '12 at 15:49
  • $\begingroup$ I did something similar in the answer I posted, I used a different sequence though. $\endgroup$ – user10444 Dec 30 '12 at 23:08
  • $\begingroup$ @user10444: Good that you got the point and benefited from the article. $\endgroup$ – Mhenni Benghorbal Dec 30 '12 at 23:10
  • $\begingroup$ thank you very much for pointing me to the article , I appreciate the help. $\endgroup$ – user10444 Dec 30 '12 at 23:15
  • $\begingroup$ @user10444: You are welcome. $\endgroup$ – Mhenni Benghorbal Dec 30 '12 at 23:19

To show that there is a solution:

Let $a=f(0)$. Let $\ell_1$ be the line of slope $.49$ passing through the point $(0,a)$ and let $\ell_2$ be the line of slope $-.49$ passing through the point $(0,a)$. These two lines divide the plane into four regions. let $A$ be the "left" region and $B$ be the "right" region. Using the fact that $|f'(x)|\le.49$, it follows from the Mean Value Theorem that the graph of $f$ is contained in the regions $A$ and $B$.

Now, let $h(x)={2x+\sin x\over 2}$. Note that $h(n\pi)=n\pi$ for each integer $n$. Since the slopes of the lines $\ell_1$ and $\ell_2$ are $.49$ and $-.49$ respectively, it follows that there is an $n$ with $h(n\pi)>f(n\pi)$ and an $m$ with $h(m\pi)<f(m\pi)$. (Drawing the picture will be helpful here. Note that the graph of $h$ wiggles along the line $y=x$; and thus its graph eventually lies atop $B$ for large positive $x$ and below the region $A$ for "large" negative $x$).

But then we have $g(n\pi)=f(n\pi)-h(n\pi)<0$ and $g(m\pi)>0$. So, as $g$ is continuous, it follows from the Intermediate Value Theorem that $g$ has at least one zero.

To show that this zero is unique:

Use Barto's hint: If $g$ had two distinct zeroes, then there would be a point $c$ at which $g'(c)=0$. But one may compute that $h'(x)\ge 1/2$ for all $x$. With the hypothesis that $|f'(x)|\le.49$ for all $x$, it follows that $g'(x)$ can never be $0$.


I reread the article Mhenni Benghorbal mentioned and was able to solve it in the following way: Define the function $g:\mathbb{R}\to \mathbb{R} $ by $g(x)= f(x) - \frac{\sin(x)}{2}$ , showing $g$ has a fixed point solves the problem.

Let $x_n$ be a sequence defined recursively by $x_{n+1} = g(x_n)$ by the mean value theorem $\exists$ $c$ $\in$ $[x_n,x_n+1]$(in reverse order if $x_n$ is decreasing) such that $|g(x_n)-g(x_{n-1})|=|g'(c)|.|x_n-x_{n-1}|$ then $|x_{n+1}-x_n|\le0.99|x_n-x_{n-1}|$ by induction on $n$ we get $|x_{n+1}-x_n|\le0.99^{n-1}|x_2-x_1|$

$x_n$ is a Cauchy sequence since $|x_p-x_q|\le|x_p-x_{p-1}|+|x_{p-1}-x_{p-2}|+...+|x_{q+1}-x_q|$ $=|x_2-x_1|(0.99^{q-1}+0.99^q+...+0.99^{p-2})\le|x_2-x_1|\large{\frac{0.99^{q-1}}{1-0.99}}$. Hence for any $\epsilon>0$ $\exists$ $N=[\log_{0.99}\large{\frac{\epsilon(1-0.99)}{|x_2-x_1|}}]+2$ such that $|x_p-x_q|<\epsilon$ when $p>q\ge N$. Then $x_n$ converges to a value x.
Since $g$ is continuous $\displaystyle\lim_{n\to\infty}x_{n+1}=g(\displaystyle\lim_{n\to\infty}x_n)$ $\implies$ $x=g(x)$ i.e. $x= f(x) -\frac{\sin(x)}{2}$ then $f(x) =\frac{2x+\sin(x)}{2}$

The point is unique since if not there are two points $x_1,x_2$ such that $g(x_1)=x_1$ and $g(x_2)=x_2$ by the mean value theorem we get $\frac{g(x_1)-g(x_2)}{x_1-x_2}=1$ but $f'(x)<0.49$ then $g'(x)<0.99$ which is a contradiction hence the fixed point is unique.


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