Prove that a function having a derivative bounded by 0.49 has a unique solution $\frac{2x+\sin(x)}{2}$

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?

Solution 1:

Hint: Use the Banach fixed point theorem.

Added:

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.

Solution 2:

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 $0$.