Show that there is $\eta\in(0,1)$ with $f''(\eta)=f(\eta)$

Suppose $f:[0,1]\to\mathbb{R}$ is twice differentiable with

$$\lim_{x\to0^+}\frac{f(x)}{x}=1,\lim_{x\to 1^-}\frac{f(x)}{x-1}=2$$

Show that there is $\eta\in(0,1)$ s.t. $f''(\eta)=f(\eta)$.

It is equivalent to prove that $f''(x)-f(x)$ has a root in $(0,1)$. Then I tried to construct function. First I noted that the ODE

$$y''-y=0$$

has solution of the form $\alpha e^x+\beta e^{-x}$, then I consider the following function :

$$(f-f')e^{x}$$

and apply mean value theorem to is. But thing does not behave that nice, and I was stuck.

The basic idea of the following is essentially the same argument as the previous answer, but it's a little shorter so I'll post it anyway.

First, I claim that the function $f : [0,1] \to \mathbb{R}$ has a strictly positive global maximum and a strictly negative global minimum, and that each is attained in an interior point of $[0,1]$.

Assuming the claim, we finish the proof. Let $a,b \in (0,1)$ be points at which the maximum and minimum of $f$ are attained. Now, $f(a) > 0 > f(b)$, and $f''(a) \leq 0, f''(b) \geq 0$, so that $$f(a) - f''(a) >0 > f(b) - f''(b)$$ It follows by the continuity of $f''$ and the intermediate value theroem that there is a point $\eta$ between $a,b$ for which $f(\eta) = f''(\eta)$.

We now prove the claim. This is not hard: $f(0+) = f(1-) = 0$, and from the left and right hand derivative limits, we know that $f$ is strictly positive in a neighborhood of $0$ and strictly negative in a neighborhood of $0$.

• Thanks for simplifying my argument. Although it was obvious I did not think that minimum / maximum can guarantee the sign of second derivative. Dec 2 '13 at 5:57
• A minor edit requested: we should change the condition $f(a) - f''(a) > f(b) - f''(b)$ to $f(a) - f''(a) > 0 > f(b) - f''(b)$. Dec 2 '13 at 6:01
• @ParamanandSingh Thank-you for the suggestion. I've made the edit. Dec 2 '13 at 18:02

From the given limits and continuity of $f$ we can see that $f(0) = 0, f(1) = 0, f'(0) = 1, f'(1) = 2$. Now we see that $f(x)$ is increasing at $x = 0$ and $f(0) = 0$ so there is an interval $(0, h)$ in which $f(x)$ is positive. Similarly $f(x)$ is increasing at $x = 1$ and hence there is an interval $(k, 1)$ with $k \in (0, 1)$ where $f(x)$ is negative. By choosing $h$ near to $0$ and $k$ near to $1$ we can ensure that $0 < h < k < 1$. Now it follows by the continuity of $f$ that $f(a) = 0$ for some $a \in (h, k)$.

Now we can see that $f(0) = f(a) = f(1) = 0$ so that $f'(x)$ vanishes at least once in $(0, a)$ and at least once more in $(a, 1)$. By continuity of $f'(x)$ there is a first value $b \in (0, a)$ such that $f'(b) = 0$. This means that $f'(x) > 0$ for $x \in [0, b)$ and $f'(b) = 0$. Similarly there is a last value $c \in (a, 1)$ such that $f'(c) = 0$. Thus $f'(x) > 0$ for all $x \in (c, 1]$ and $f'(c) = 0$.

Since $f'(x) > 0$ in $(0, b)$ it follows that $f(x) > f(0) = 0$ for all $x \in (0, b]$. Similarly $f(x) < 0$ for all $x \in [c, 1)$.

Since the argument is bit long we once revise what we have obtained so far:

1) $f'(x) > 0$ for all $x \in [0, b)$, $f'(b) = 0$ and $f'(x) > 0$ for all $x \in (c, 1]$, $f'(c) = 0$.

2) $f(x) > 0$ for all $x \in (0, b]$ and $f(x) < 0$ for all $x \in [c, 1)$.

Also by their construction $b < c$.

By mean value theorem we see that $f'(b) - f'(0) = bf''(\alpha)$ for some $\alpha \in (0, b)$ so that $f''(\alpha) < 0$. Similarly $f'(1) - f'(c) = (1 - c)f''(\beta)$ implies that $f''(\beta) > 0$ for some $\beta \in (c, 1)$. Now note that $f(x) > 0$ in $(0, b]$ so that $f(\alpha) > 0$ and similarly $f(\beta) < 0$. Now we can see that the function $g(x) = f''(x) - f(x)$ is such that $g(\alpha) < 0$ and $g(\beta) > 0$. Since $g(x)$ acts as derivative function (of $f'(x) - \int_{0}^{x}f(t)\,dt$) it has the intermediate value property and therefore there is a number $\eta \in (\alpha, \beta)$ for which $g(\eta) = 0$ i.e. $f''(\eta) = f(\eta)$.

To be explicit, I have assumed that $f'(x)$ exists and is continuous in $[0, 1]$ and differentiable in $(0, 1)$. This is what the OP probably means by "$f:[0,1]\to\mathbb{R}$ is twice differentiable."

Update: Note that the conclusion in the question is valid whenever the two limits in the question are positive. Their specific values $1$ and $2$ don't matter. I also suppose that there would be a better answer which uses mean value theorem directly by using combination of the fuctions $f, f'$ and some auxiliary functions like $e^{\pm x}$. The function suggested by OP $(f - f')e^{x}$ does give the derivative $(f - f'')e^{x}$ but the original function $(f - f')e^{x}$ fails to vanish at two points in $[0, 1]$.