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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?

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    $\begingroup$ One possible way might be to let $g(x) = \lim_{n \to \infty}f^{(n)}(x)$. Try to show that $g'(x) = g(x)$. for any $x \in \mathbb{R}$. Then show that $g(x)/e^{x}$ is constant by showing the derivative is zero. $\endgroup$ Jul 17, 2018 at 21:14
  • $\begingroup$ Think like this. $\phi$ is the limit of the derivatives of $f$. So, when you derive a high derivative, you still get a little closer to $\phi$. But when you take $\phi$ and derive it, it will be $\phi$ itself (which one has to prove). But the only functions $\phi$ that satisfy $\phi'=\phi$ are of the form $Ce^x$. $\endgroup$
    – amsmath
    Jul 17, 2018 at 21:28

2 Answers 2

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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 $\phi$.

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  • $\begingroup$ How do I solve the equation for $\phi$? Is it supposed to give me $Ce^x$?Thanks in advance! $\endgroup$ Jul 18, 2018 at 20:59
  • $\begingroup$ Since $\phi$ is continuous, by fundamental theorem of calculus the left-hand side of the last equality is differentiable, hence the right-hand side is too and differentiating term by term you get $\phi'(x)=\phi(x)$ for all $x$. Can you solve this differential equation? $\endgroup$
    – Lorenzo Q
    Jul 18, 2018 at 21:35
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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 $\phi(x) =Ce^{-x}$.

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  • $\begingroup$ Thanks for your answer! I understood the theorem but I fail to see the connection that you made to the exercise, like $\phi(x)=Ce^x$. Could you develop a little bit more! Thanks in advance! $\endgroup$ Jul 18, 2018 at 20:48
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    $\begingroup$ @PedroGomes Have a look now, I added some details. $\endgroup$ Jul 18, 2018 at 21:10

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