$f_n\to f$ then $\phi(x)=Ce^x$? 
Let $f(x)\in C_\infty(\mathbb{R})$ and a succession of derivatives $f^{(n)}(x)$ converges in $C_{[a,b]}$ to a function $\phi(x)$ in each finite interval. Show that $\phi(x)=Ce^x$, where $C$ is a constant.

I know that $Ce^x\in C_{\infty}$ but I cannot prove  $\phi(x)=Ce^x$. I think the $\max|f_n-f|=\max|{Ce^x-Ce^x}|=0 $ for $n>N\in\mathbb{N}$. And I know that $Ce^x\in C_[a,b]$, once it is continuous on the $\mathbb{R}$.
Question:
1) How do I prove $\phi(x)=Ce^x$? And not for example $\sin (x)$? 
2) What is intended on this question?
 A: Let $x\in \mathbb{R}$. We know that $f^{(n)}\to \phi$ uniformly on $[0,x]$. Thus $\phi$ is continuous on $[0,x]$, and
$$\lim_{n\to \infty}\int_0^xf^{(n)}(t)dt= \int_0^x\phi(t)dt $$
where swapping integral and limit is allowed by uniform convergence on a compact domain. On the other hand, by fundamental theorem of calculus,
$$\lim_{n\to \infty} \int_0^xf^{(n)}(t)dt=\lim_{n\to \infty}\left[ f^{(n-1)}(x)-f^{(n-1)}(0)\right] =\phi(x)-\phi(0)$$
Hence
$$\int_0^x \phi(t)dt=\phi(x)-\phi(0) $$
where $\phi:\mathbb{R}\to \mathbb{R}$ is a continuous function. Now you just need to solve the above equation for $\phi$.
A: Recall this theorem:

Let $I = [a,b]$ and $f_n : I \to \mathbb{R}$ a sequence of differentiable functions such that the sequence of derivatives $(f_n')_n$ converges uniformly to a function $g : I \to \mathbb{R}$. Also $\exists x_0 \in I$ such that the sequence $(f_n(x_0))_n$ converges. Then $(f_n)_n$ converges uniformly to a differentiable function $f : I \to \mathbb{R}$ with $f' = g$.

This theorem can be extended for functions $f_n : \mathbb{R} \to \mathbb{R}$ by restricting the domain to $[-R, R]$, say. We only lose the uniform convergence of $(f_n)_n$ to $f$ on $\mathbb{R}$, but pointwise convergence remains.
Since $f_n' \to \phi$ and  $f_n \to \phi$ uniformly, we conclude that $(f_n)_n$ converges pointwise to a differentiable function $\psi$ such that $\psi' = \phi$. But we already know that $f_n \to \phi$ uniformly so necessarily $\psi = \phi$.
Therefore
$$\phi' = \phi$$
and the only functions satisfying this are $\phi(x) = Ce^x$.

Additional details:
The assumptions from the exercise imply that $f_n \to \phi$ and $f_n' \to \phi$ in $C^\infty(\mathbb{R})$, that means uniform convergence over $\mathbb{R}$. Then in particular for any $R>0$ we have $f_n|_{[-R, R]} \to \phi|_{[-R, R]}$ and $f_n'|_{[-R, R]} \to \phi|_{[-R, R]}$  uniformly on $[-R, R]$. In particular there exists $x_0 \in [-R, R]$ such that $f_n|_{[-R, R]}(x_0) \to \phi|_{[-R, R]}(x_0)$. The theorem implies that there exists a differentiable function $\psi : [-R, R] \to \mathbb{R}$ such that $f_n|_{[-R, R]}(x_0) \to \psi$ uniformly and $\psi' = \phi|_{[-R, R]}$. But already $f_n|_{[-R, R]} \to \phi|_{[-R, R]}$ uniformly so $\psi = \phi|_{[-R, R]}$. Hence we conclude $\phi'|_{[-R, R]} = \psi' = \phi|_{[-R, R]}$.
Since $R > 0$ was arbitrary, we conclude $\phi' = \phi$. Now this implies $\psi(x) = Ce^x$. See for example here, the simplest argument is differentiating $\phi(x)e^{-x}$:
$$\frac{d}{dx}\phi(x)e^{-x} = \phi'(x)e^{-x} - \phi(x)e^{-x} =\phi(x)e^{-x} - \phi(x)e^{-x} = 0 $$
So $\phi(x)e^{-x} \equiv C$ or $\phi(x) =Ce^{-x}$.
