Fixed point result for a differentiable function.

Question

Let $f$ be a differentiable function on $[0, 1]$ with $0$ and $1$ be its fixed points. Also, it is given that $$f'(0)>1, f'(1)>1.$$ Then prove that

(i) $f$ has at least one fixed point in $(0, 1)$.

(ii) $f$ has a fixed point $\xi\in (0, 1)$ such that $f'(\xi)\leq 1$.

My effort: I started by considering $g(x)=f(x)-x$. Clearly, $g$ is a differentiable function which means $g'(x)=f'(x)-1$. This means $g'(0)>0, g'(1)>0$. But I am not able to conclude anything. Please help.

Answer 1

As no one else has given an answer addressing part (ii), let me prove both parts at once (and also prove the intermediate value theorem as a corollary).

As you've reduced, what we need to do is to prove:

Suppose $g$ is a differentiable function with $g'(0)>0$ and $g'(1)>0$. Prove that there exists some $z\in (0,1)$ so that $g(z)=0$ and $g'(z)\leq 0$.

As others have noting, using the derivatives at either end tells us that least that there is some $x_0$ so that $\frac{g(x_0)}{x_0} > 0$ and some $x_1$ so that $\frac{-g(x_1)}{1-x_1} > 0$ which in particular implies $g(x_0)>0$ and $g(x_1)<0$ - or, in fact, that these inequalities hold for all $x_0$ in a neighborhood of $0$ and $x_1$ in a neighborhood of $1$, which we subtly need to ensure that we choose $x_0$ and $x_1$ so that $x_0 < x_1$.

Now, intuitively, we are looking for some place where the graph of $g$ crosses downwards - from positive to negative. A nice way to do that is to consider the last spot in $[x_0,x_1]$ where $g$ is non-negative. In particular, let $$S=\{x\in [x_0,x_1] : g(x) \geq 0\}$$ Note that $S$ is a closed (bounded) set and thus has some maximum $z\in S$. Moreover, note that $x_1$ is not in $S$ thus $x_0\leq z<x_1$. We claim that $g(z)=0$ and $g'(z)\leq 0$.

We know that the image of every interval $(z-h,z+h)$ under $g$ contains a negative number, since every input greater than $z$ gives a negative output. We also know that $g(z)\geq 0$. By continuity, we need that every neighborhood of $g(z)$ includes negative numbers. Together, this implies $g(z)=0$.

To see that the derivative is negative at this $z$, consider the quotient $$\frac{g(z+h) - g(z)}{h}=\frac{g(z+h)}h.$$ Observe that this is negative for every $0 < h \leq x_1 - z$, which implies that the limit of this as $h$ goes to $0$ must be non-positive - or, otherwise put, $g'(z)\leq 0$.

The proof that $g$ has a zero crossing suffices as a proof of the intermediate value theorem - and is a proof that I particularly like because it captures some extra geometric intuition about the crossing, which is exactly what is needed for this problem.

Answer 2

Use first the definition of $g'(0)$ and $g'(1)$ to show that $g$ assumes a strictly positive and a strictly negative value on $(0,1)$.

Answer 3

You also have $g(0) = 0$ and $g(1) = 0$ because $0, 1$ are fixed points of $f$. Now, because $g'(0) > 0$, you must have $g(x_1) > 0$ for some $x_1 \in (0, 1)$. Otherwise, for a contradiction, if $g(x) \leq 0$ were instead true for all $x \in (0, 1)$, then $g(x) - g(0) = g(x) - 0 \leq 0$ always. Hence, the fraction $$ \frac{g(x) - g(0)}{x - 0} \leq 0, \ \forall x \in (0, 1) \implies g'(0) = \lim_{x \to 0, x \in (0, 1)}\frac{g(x) - g(0)}{x - 0} \leq 0 $$ contrary to $g'(0) > 0$.

For a similar reason, there must also be some $x_2 \in (0, 1)$ such that $g(x_2) < 0$. Obviously $x_1 \neq x_2$. So now, you can conclude via IVT.

Answer 4

From the definition of derivative $$g'(0) = \lim_{h\to 0^+} \frac{g(h) - g(0)}{h} = \lim_{h\to 0^+} \frac{g(h)-0}{h} > 0$$ Since $h$ goes to $0$ from the positive side, $g(x)$ must be positive $(*)$ in some right neighbourhood of $0$ to have a positive limit. $$g'(1) = \lim_{h\to0^-} \frac{g(1+h) - g(1)}{h} = \lim_{h\to0^-} \frac{g(1+h) -0}{h} > 0$$ Since $h$ goes to $0$ from the negative side $g(1+h) < 0$, which means $g(x)$ must be negative $(**)$ in some neighbourhood of $1$ to have a positive limit. Hence, combining $(*)$ and $(**)$ there exists $c \in (0,1)$ such that $g(c) = f(c)-c=0$