Uniform convergence to a function

Question

I have a task:

Let the function f be infinitely differentiable on $\Bbb{R}$, and the sequence $\{f^{(n)}(x)\}$ converges uniformly to $\phi(x)$ on every finite interval $(a, b)$. Prove that $\phi(x)=Ce^x$, where $C$ is a constant.

Her are my thoughts: according to the Weierstrass' theorem, we can represent $f(x)$ as a power series. Then we can look at its n-th derivatives. After that somehow we can observe that it looks just like the Taylor series of an exponential function.

Answer 1

Let $(a_n)=\Bigl(f^{(n)}(x)\Bigl)$, so $\lim (a_n)=\lim \Bigl(f^{(n)}(x)\Bigl)=\phi(x)$. Then, $\phi'(x)={d\over dx}\lim \Bigl(f^{(n)}(x)\Bigl)\\=\lim_{h\to0}{\lim(f^{(n)}(x+h))-lim(f^{(n)}(x))\over h}=\lim_{h\to0}\lim({f^{(n)}(x+h)-f^{(n)}(x)\over h})\\=\lim\Bigl(f^{(n+1)}(x)\Bigl)=\lim(a_{n+1})$.

But if the limit exists, $\lim(a_n)=\lim(a_{n+1})$, so $\phi(x)=\lim (a_n)=\lim (a_{n+1})=\phi'(x)$. This can be solved using separation of variables to get $\phi(x)=Ce^x$.