A real-valued function $f$ which is infinitely differentiable on $[a.b]$ has the following properties:

  • $f(a)=f(b)=0$
  • $f(x)=f'(x)+f''(x)$ $\forall x \in [a,b]$

Show that $f(x)=0$ $\forall x\in [a.b]$

I tried using the Rolle's Theorem, but it only tells me that there exists a $c \in [a.b]$ for which $f'(c)=0$.

All I get is:

  • $f'(a)=-f''(a)$
  • $f'(b)=-f''(b)$
  • $f(c)=f''(c)$

Somehow none of these direct me to the solution.

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    $\begingroup$ Do you know how to solve a second-order, constant coefficient, linear differential equation like $y''+y'-y=0$? $\endgroup$ – Gerry Myerson Mar 5 '12 at 4:03
  • $\begingroup$ No. But I know solutions to $y'' \pm y=0$. Will that help? $\endgroup$ – Sidharth Iyer Mar 5 '12 at 4:10
  • $\begingroup$ No, not if you just know the answer - but if you know how to get the answer, the method should work. $\endgroup$ – Gerry Myerson Mar 5 '12 at 4:17
  • $\begingroup$ Maybe you can do a repeated application of the MVT to show that the derivative must have infinitely-many zeros: since f(a)=f(b)=0, there is c in [a,b] with f'(c)=(f(b)-f(a))/(b-a)=0. Now, repeat the process between a and c and between c and b, and so on. $\endgroup$ – AQP Mar 5 '12 at 4:19

Hint $f$ can't have a positive maximum at $c$ since then $f(c)>0, f'(c)=0, f''(c) \le 0$ implies that $f''(c)+f'(c)-f(c) < 0$. Similarly $f$ can't have a negative minimum. Hence $f = 0$.


Let $x=c$ be the $x$ coordinate of absolute max of $f(x)$ on $[a,b]$. (This point exists by the extreme value theorem). I will show that $f(c) = 0$. Since $f(a) = 0$ and $c$ is the absolute max, $f(c)\geq 0$. By Fermat's theorem, we know $f'(c) = 0$. Hence, we learn that $f(c) = f''(c)\geq 0$.

Now, assume for a contradiction that $f(c) > 0$, so $f''(c) > 0$ and $c\neq a$ and $c\neq b$. I claim that for $x$ close enough to $c$, but bigger than it, that $f(x) > f(c)$, contradicting maximality of $f(c)$.

Since $f''$ is continuous, for $x$ close enough to $c$ say, within $\delta$, we have $f''(x) > 0$. On the interval where $c<x<c+\delta$, $f'(x) \geq 0$ with equality only at $x=c$. This follows from the Mean value theorem applied to $f'$, because if $f'(x)\leq 0$ for a point $x\in(c,c+\delta)$, then by the MVT, $f''(d) \leq 0$ for some $d\in(c,c+\delta)$, giving a contradiction.

From this, it follows that $f(x)>f(c)$ for $x\in(c,c+\delta)$, because, again by the MVT, we have $\frac{f(x)-f(c)}{x-c} = f'(d) > 0$ for some $d\in(c,c+\delta)$, so, $f(x) - f(c) > 0$.

Thus, we contradict maximality of $f(c)$. From this contradiction, we deduce $f(c) = 0$ is the maximum of the function. Now, repeat a similar argument to $-f$ (changing the interval $(c,c+\delta)$ to $(c-\delta, c)$) to deduce the minimum value of $f$ is $0$. From this it follows that $f$ is identically $0$.

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    $\begingroup$ @Zev: Thanks! Sorry to be a bother. $\endgroup$ – Jason DeVito Mar 5 '12 at 4:31
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    $\begingroup$ No problem at all :) $\endgroup$ – Zev Chonoles Mar 5 '12 at 4:31
  • $\begingroup$ Do you need the whole argument with the MVT? or can you just appeal to the second derivative test - if $f'(c)=0$ and $f''(c)\gt0$ then there's a local minimum at $c$, contradicting assumption of a local maximum. $\endgroup$ – Gerry Myerson Mar 5 '12 at 5:47
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    $\begingroup$ @Gerry: I think I essentially proved the second derivative test - but you're right: If I can assume the Extreme Value theorem, Fermat's theorem, MVT, etc, then I should be able to assume the second derivative test. $\endgroup$ – Jason DeVito Mar 5 '12 at 13:15

Hint: Let $\alpha$ and $\beta$ be the roots of $X^2+X-1$.

(a) Check that $f$ satisfies $$ \left(\frac{d}{dx}-\alpha\right)\left(\frac{d}{dx}-\beta\right)f=0. $$ (b) Solve the above equation by solving two ODE of the form $y'-cy=g(x)$. (If you don't know how to solve $y'-cy=g(x)$, I'll be happy to give another hint.)

(c) Conclude.

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    $\begingroup$ Cute solution. +1 $\endgroup$ – user21436 Mar 5 '12 at 6:17
  • $\begingroup$ Dear @Kannappan: Thanks!!! $\endgroup$ – Pierre-Yves Gaillard Mar 5 '12 at 6:19

Since you know how to solve $y'' = y$ (and so I presume $y'' = ky$) here is a way to get your equation into that form.

We will use the identity

$$ (fg)'' = f'' g + 2f'g' + fg''$$

Now setting $g = e^{kx}$ gives us

$$ (fe^{kx})'' = e^{kx} (f'' + 2kf' + k^2f)$$

In order to eliminate $f''$ and $f'$, we set $k=\frac{1}{2}$, to get

$$ (f e^{x/2})'' = e^{x/2} (f'' + f' + f/4) = e^{x/2} (5f/4)$$

(using the given $f = f' + f''$)

You can now set $y = f(x)e^{x/2}$ to get

$$ y'' = \frac{5}{4} y$$


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