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


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


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 $0$.

  • $\begingroup$ Thanks for simplifying my argument. Although it was obvious I did not think that minimum / maximum can guarantee the sign of second derivative. $\endgroup$
    – Paramanand Singh
    Dec 2 '13 at 5:57
  • $\begingroup$ A minor edit requested: we should change the condition $f(a) - f''(a) > f(b) - f''(b)$ to $f(a) - f''(a) > 0 > f(b) - f''(b)$. $\endgroup$
    – Paramanand Singh
    Dec 2 '13 at 6:01
  • $\begingroup$ @ParamanandSingh Thank-you for the suggestion. I've made the edit. $\endgroup$ 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 $[0, 1]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.