let $f:\mathbb{R}\rightarrow\mathbb{R}$ has derivative of all orders at every point for some $\alpha, \beta$

Question

let $f:\mathbb{R}\rightarrow\mathbb{R}$ has derivative of all orders at every point for some $\alpha, \beta$ such that $f''(x)+\alpha f'(x)+\beta f(x)=0\; \forall x$

then how we prove that $$f(x)=\sum^{\infty}_{n=0} \frac{f^n(0)}{n!} x^n \; \forall x$$

I am really dont know where i start from

Answer 1

Since $f$ is smooth, it has a Taylor series representation of all orders and so $f(x) = \sum_{k=0}^n { f^{(k)}(0) \over n! } x^k + R_n(x)$, where $R_n(x) = \int_0^x {f^{(n+1)}(t) \over (n+1)! } (x-t)^n dt $. To show the above formula, it is sufficient to show that $R_n(x) \to 0$.

The idea is to show that the functions $f^{(k)}$ do not grow 'too fast'.

Pick some $K>0$ and let $\|g\| = \sup_{|x| \le K} |g(x)|$. Let $M= \max(\|f\|,\|f'\|)$ and $L= \max(|\alpha|+|\beta|,1)$. Note that $|f^{(k+2)}(x)| \le |\alpha| |f^{(k)}(x)| + |\beta| |f^{(k+1)}(x)|$.

Note that $\|f''\| \le L M$. Now suppose $\|f^{(n)}\| \le L^{n-1} M$, then we have $\|f^{(n+1)}\| \le L \max(\|f^{(n)}\| ,\|f^{(n-1)}\|) \le L^{n+1} M$.

Then $|R_n(x)| \le L^n M {K^{n+1} \over (n+1)!} $ from which we get the desired result.