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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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    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.2012-03-05

4 Answers 4

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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 $f = 0$.

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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, 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 $0$.

  • 1
    @Gerry: I think I essenti$a$lly proved the second derivative test - $b$ut 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.2012-03-05
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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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    Dear @Kannappan: Thanks!!!2012-03-05
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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