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

EDIT

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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    @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 $y \in (x-\epsilon,x+\epsilon)$.2012-12-03