Prove that $f(x) = 0$ for all $x \in \mathbb{R}$ (Analysis)

Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that $f(x) = f^{(4)}(x)$ with $f(0) = f’(0) = f’’(0) = f’’’(0) = 0.$ Prove $f(x) = 0$ for all $x \in \mathbb{R}$

My Attempts:

Suppose $x \in \mathbb{R}$. Note that $\displaystyle f'(0) = \lim_{x\to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x\to 0} \frac{f(x)}{x} = \lim_{x\to 0} \frac{f'(x)}{1}=0$. (L'Hôpital's Rule was used in the second to last limit due to the form $\frac{0}{0}$). With this approach, I am not necessarily finding if $x = 0$ on the whole real line. This led me to a different approach:

Suppose $x \in [0,b]$. By Mean Value Theorem, there exists $c \in (0,b)$ so that $\displaystyle \frac{f(b)-f(0)}{b-0} = f'(c)$. This approach doesn't bring me anywhere either, even if I repeatedly use Mean Value Theorem. Any suggestions on how to proceed and conclude? (I am currently reading/finishing the chapter on Differentiation in baby Rudin.)

Answer 1

The general solution to the ode is the sum $$ c_1e^x+c_2e^{-x}+c_3\sin x+c_4\cos x=f(x) $$ using the usual methods.

This gives us the system of equations $$ f(0)=c_1+c_2+c_4=0\\ f'(0)=c_1-c_2+c_3=0\\ f''(0)=c_1+c_2-c_4=0\\ f'''(0)=c_1-c_2-c_3 $$ which has unique solution $c_1=c_2=c_3=c_4=0$.

Answer 2

Here is one proof which avoids differential equations. Let us assume that $|x|\leq 2$. By Taylor's theorem we have $$f(x) =\frac{x^4}{4!}f^{(4)}(c_1)=\frac{x^4}{4!}f(c_1), x\neq 0$$ where $0<|c_1|<|x|$ and applying this repeatedly and noting that at each step we have $0<|c_n|<|x|$ we get $$|f(x)| \leq \left(\frac{x^4}{4!}\right)^{n}|f(c_n) |\leq \left(\frac{x^4}{24}\right)^{n}M$$ where $M$ is the maximum value of $|f|$ between $0$ and $x$. Since $|x|\leq 2$ by Squeeze Theorem we can see that $f(x) =0$ for all $x\in[-2,2]$. The proof is extended to other values of $x$ by using functions $g_1(x) =f(x-1),h_1(x)=f(x+1)$ and applying the above argument repeatedly to $g_1, h_1$ and their descendents $g_n, h_n$ given recursively as $$g_n(x) =g_{n-1}(x-1),h_n(x)=h_{n-1}(x+1)$$ One may also use the function $g(x) =f(kx) $ to extend the proof to all values of $x\in[-2k,2k]$ with much less effort.

Incidentally the same technique gives another proof of the following important theorem:

Theorem: If $f:\mathbb{R} \to\mathbb {R} $ is such that $f'(x) =f(x), f(0)=0$ then $f(x) =0$ for all $x \in\mathbb {R} $.

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Update: As mentioned in my comment to the question, one can also use mean value theorem four times instead of Taylor's theorem to get the simpler inequality $|f(x) |\leq x^4|f(c)|$ for some $c$ between $0$ and $x$ and the proof can be continued with this inequality in a similar manner.

Answer 3

Notice that $f'(x)=f^{(5)}(x),\,f''(x)=f^{(6)}(x),\,f'''(x)=f^{(7)}(x),\,f^{(4)}(x)=f^{(8)}(x)=f(x)$ (assuming f posseses first, second, third derivatives for all x, that is true because we have $f(x)=f^{(4)}(x)$). So that if $i\in \{5, 6, 7,...\}\,$$$f^{(i)}=\begin{cases} f(x), & \text{if $i = 4,8,12,...$} \\ f'(x), & \text{if $i=5,9,13,...$} \\ f''(x), & \text{if $i=6,10,14,...$} \\ f'''(x), & \text{if $i=7,11,15,...$} \end{cases}$$ Then we claim that f is $C^\infty$, and since it's defined for x = 0, we have the McLaurin's power series of f given by $$f(x)=\sum_{n=0}^{inf}\frac{f^{(n)}(0)x^n}{n!}=\sum_{n=0}^{inf}\frac{(0)x^n}{n!}=0$$

Answer 4

So $f^{(4)}(x)-f(x)=0$, solving $y^{(4)}-y=0$, or the characteristic equation with $m^{4}-1=0$, then $(m+1)(m-1)\left(m+i\right)\left(m-i\right)=0$, so $f(x)=C_{1}e^{-x}+C_{2}e^{x}+C_{3}\cos x+C_{4}\sin x$.