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Suppose $f$ is defined on all real numbers. $f = f''$ and $f(0)=f'(0)=0$. Then show that $f=0$ for all $x$.

The following is what I did: Since we have $f=f''$. Then, multiply $f'$ on both sides: $$f\cdot f'=f' \cdot f''$$ $$f \cdot f'-f' \cdot f''= \frac {1}{2}(f^2)'- \frac {1}{2}(f'^2)'$$ This says, $f^2-f'^2=C$, and plug in number $x=0$, we can then say $C=0$. So, $f^2-f'^2=0$.

But, what is the next step? How can I show $f=0$ for all $x$?

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up vote 3 down vote accepted

$$f'' = f \implies f(x) = c_1 e^x + c_2 e^{-x}$$ Now finish it off by plugging in values for $f(0)$ and $f'(0)$.


To finish it off, the way you have started, you get that $$f'(x) = \pm f(x)$$ If $f'(x) = f(x)$, we get that $$\exp(-x) f'(x) = \exp(-x) f(x) \implies \exp(-x) f'(x) - \exp(-x) f(x) = 0$$ This gives us $$\left(\exp(-x) f(x) \right)' = 0$$ Hence, $$\exp(-x) f(x) = c_1 \implies f(x) = c_1 \exp(x)$$ Plugging in $f(0) = 0$, we get that $f(x) = 0$.

Similarly, if $f'(x) = -f(x)$, we get that $$\exp(x) f'(x) = -\exp(x) f(x) \implies \exp(x) f'(x) + \exp(x) f(x) = 0$$ This gives us $$\left(\exp(x) f(x) \right)' = 0$$ Hence, $$\exp(x) f(x) = c_2 \implies f(x) = c_2 \exp(-x)$$ Plugging in $f(0) = 0$, we get that $f(x) = 0$.

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Well, I know this is a good approach by considering the DE. But this problem is from the real analysis, and in this course we talked nothing about the DE, so I cannot use the way of DE. Bad news, huh? – Scorpio19891119 Dec 3 '12 at 4:59
@Scorpio19891119 $$\text{DE} \subset \text{real analysis}$$ Anyway, I have continued your method to obtain the answer. – user17762 Dec 3 '12 at 5:09
I know this...But in this course, we really talked nothing about DE. But truly appreciate your help. :) – Scorpio19891119 Dec 3 '12 at 5:10
What if $f'(x)=f(x)$ for some $x$ and $f'(x)=-f(x)$ for other $x$? – N. S. Dec 3 '12 at 5:53
@N.S. Good point. Since $f \in C^{(2)}$, $f'(x) = f(x)$ for some $x$, then there exists $\epsilon > 0$ such that $f'(y) = f(y)$ for $y \in (x-\epsilon,x+\epsilon)$. Hence, we get that $f(y) =0 $ for all $y \in (x-\epsilon,x+\epsilon)$. – user17762 Dec 3 '12 at 6:01

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