Show that $f$ is the zero function if $f''(x)=f(x)$ and $f(0)=f'(0)=0$

Suppose that $f''(x)=f(x)$ for all real numbers $x$, and that $f(0)=f'(0)=0$. Show that $f$ is the zero function.

I know that $f''(0)=0$ from the assumptions listed. I want to consider the Taylor series for $f$ centered at $x=0$, which is $\sum$ $[f^{(n)}(0)/n!]x^n$. I am not sure where to go from here.

• If $f''=f$ for all $x$, then $f(x) = A e^x +Be^{-x}$ in general. Now apply the initial conditions to find the $A=B=0$. Dec 2, 2016 at 19:52
• @Dr.MV: nice; +1. Perhaps OP also needs to be convinced that this solution is unique. Dec 2, 2016 at 19:54
• @Dr.MV I have seen the proof you suggested before. I want to prove the result using the Taylor series. Dec 2, 2016 at 19:56
• @GitGud: good point; +1. I have edited my wording. Dec 2, 2016 at 19:57

5 Answers

Because $f'' = f$ on $\mathbb R,$ we see that $f\in C^2(\mathbb R).$ Let $M= \sup_{[0,1]}|f|.$ Then $M=|f(x_0)|$ for some $x_0 \in [0,1].$ By Taylor, there is $c\in (0,x_0)$ such that

$$f(x_0) = f(0) + f'(0)x_0 + f''(c)x_0^2/2 = f''(c)x_0^2/2 = f(c)x_0^2/2.$$

Taking absolute values then gives $M\le Mx_0^2/2 \le M/2.$ That implies $M=0.$ Thus $f\equiv 0$ on $[0,1].$ This argument can be continued to the right to give $f\equiv 0$ on $[0,\infty).$ The argument also works to the left, so we have $f\equiv 0$ on $\mathbb R$ as desired.

• Who said $f$ (and hence $f''$) was continuous? Dec 3, 2016 at 0:27
• @EricTowers $f$ is given to be twice differentiable. Therefore $f''=f$ is twice differentiable. That implies $f''$ is continuous.
– zhw.
Dec 3, 2016 at 0:36
• @EricTowers Never. But if $f$ is differentiable, it is continuous. So if $f''$ is differentiable, then $f''$ is continuous.
– zhw.
Dec 3, 2016 at 0:45
• @EricTowers We have $f=f''.$ $f$ is differentiable, hence continuous. Therefore $f''$ is continuous.
– zhw.
Dec 3, 2016 at 1:16
• @EricTowers Does one need to spell out the obvious?? Dec 3, 2016 at 4:12

Suppose $f''=f$ and consider $g(x)=(f'(x)+f(x))e^{-x}$. Then $$g'(x)=(f''(x)+f'(x))e^{-x}-(f'(x)+f(x))e^{-x}=0$$ Therefore $g(x)$ is constant. Since $$g(0)=0$$ we have $f'(x)+f(x)=0$, for every $x$. Therefore $f'=-f$. Consider $$h(x)=f(x)e^{x}$$ Then $h'(x)=f'(x)e^x+f(x)e^x=0$ so also $h$ is constant. Since $h(0)=0$, we are done.

• Nice solution. +1
– zhw.
Dec 2, 2016 at 23:16
• Very nicely done! +1 Dec 2, 2016 at 23:35

For a real function the fact that the Taylor series is $0$ to any order does not mean the function is $0$; the well known example is

$$f(x)=\begin{cases} e^{-1\over x^2}&x\neq 0\\0&x=0\end{cases}$$

One can check it is infinitely derivable at $0$ with $\forall n,\,f^{(n)}(0)=0$.

No to solve your problem you need to use unicity theorems related to ODE (e.g Cauchy Lipschitz) or equivalently knowing that the solution of a linear second order ODE is uniquely determined by two parameters ($y(0)$ and $y'(0)$) and noticing that the zero function verifies the $y''+y=0$, $y(0)=0$ and $y'(0)=0$ and so is the unique solution.

• My professor specifically told me not to use differential equations, but to use the Taylor series instead. So, I know there has to be some way around the ODE method, but I don't know how to do that. Dec 2, 2016 at 20:08
• Look at @Dr. MV solution if you know about analytic functions you can use Taylor series. Dec 2, 2016 at 20:11
• This is well-written. +1 Dec 2, 2016 at 21:02

We assume that $f$ is twice differentiable. Then, if $f''=f$, then we see inductively that $f$ is $C^\infty$. We assume that it can be represented in terms of its Taylor series as

$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}\,x^n \tag 1$$

Furthermore, in its interval of convergence, the series in $(1)$ may be differentiated term-by-term. Differentiating twice, we obtain

\begin{align} f''(x)&=\sum_{n=2}^\infty\frac{f^{(n)}(0)}{(n-2)!}\,x^{n-2}\\\\ &=\sum_{n=0}^\infty\frac{f^{(n+2)}(0)}{n!}\,x^n \tag 2 \end{align}

Since $f=f''$, then by uniqueness of the Taylor series we have from equating $(1)$ and $(2)$

$$f^{(n)}(0)=f^{(n+2)}(0) \tag 3$$

for all $n$.

Furthermore, we have from the initial conditions

\begin{align} \left.\left(\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}\,x^n\right)\right|_{x=0}&=f^{(0)}(0)\\\\ &=0 \tag 4 \end{align}

\begin{align} \left.\left(\sum_{n=0}^\infty\frac{f^{(n+1)}(0)}{n!}\,x^n\right)\right|_{x=0}&=f^{(1)}(0)\\\\ &=0 \tag 5 \end{align}

Putting together $(3)-(5)$, we find that $f^{(n)}(0)=0$ for all $n$ and we are done!

NOTE:

The function $f(x)= e^{-1/x^2}$ for $x\ne 0$ and $f(0)=0$ is $C^\infty$. But its Taylor series is $0$ and therefore does not represent $f(x)$ anywhere. So, the assumption that $f(x)$ can be represented by its Taylor series was a key here.

• Why is $f$ analytic?
– zhw.
Dec 2, 2016 at 20:29
• $C^\infty$ does not imply real analytic. Recall $\exp(-1/x^2).$
– zhw.
Dec 2, 2016 at 20:31
• How are you dividing by $f,g$ when they could be $0?$ Also, $\log f(0)$ isn't defined.
– zhw.
Dec 2, 2016 at 20:49
• Don't worry about it my friend.
– zhw.
Dec 2, 2016 at 21:04
• I don't understand why you can assume the function is analytic? It's certainly smooth, but analytic? If you need that assumption, can you show why it fails without? Or if you don't, then can you show a proof that doesn't require it? Dec 2, 2016 at 22:45

Since you wanted to extract the result from a series we can claim that if an analytic solution, $A(x)$, exists, it will be of the form $$A(x) = \sum_{n=0}^{\infty} a_nx^n$$ now we have $$A''(x) = \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n$$ and of course $A(0) = A'(0) = 0$.

So our differential equation tells us that \begin{align*} \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n - \sum_{n=0}^{\infty} a_nx^n &= 0 \\ \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n - a_nx^n &= 0 \\ \sum_{n=0}^{\infty}\left [ (n+2)(n+1)a_{n+2}- a_n \right ]x^n &= 0 \end{align*}

Equating each term with $0$, we have $$a_{n+2} = \frac{a_n}{(n+2)(n+1)}$$

Which with the initial conditions $a_0 = A(0) = 0$ and $a_1 = A'(0) = 0$, clearly we have $a_n = 0$.

• How does this differ from my approach in any substantive way? Dec 3, 2016 at 4:10