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Assume you are given a function $f:\mathbb R\to \mathbb R$ with the following properties:

  • All derivatives $f^{(k)}$ exist everywhere
  • The sequence $(f^{(k)})_k$ converges pointwise to a function $h:\mathbb R \to \mathbb R$

What can be said about the function $h$ ?

I looked at a few examples and came up with functions like

$f(x) = 3e^x+7e^{-\frac x 2} +42\sin(x/7)+x^{32}$.

In this special example the limit funtion is $h(x) = 3e^x$.

What I would like to know: Is it always true that if you start with a function $f$ satisfying the assumptions above, that you end up with a function $h(x)$ which is just a scalar multiple of $e^x$ ?

Tom
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1 Answers1

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[Disclaimer: in this answer, I assume that the Taylor series for $f$, which converges everywhere, actually converges to $f$. I believe that, given the conditions stated in the question, this must be true, but I can't prove it. This shouldn't be regarded as the "final" answer to the OP's question.]

Since the sequence $f^{\left(k\right)}\left(0\right)$ converges to $h\left(0\right)$, the Taylor series for $f$ is: $$\sum_{k=0}^\infty \frac{f^{\left(k\right)}\left(0\right)}{k!}x^k=\sum_{k=0}^\infty \frac{h\left(0\right)}{k!}x^k+\sum_{k=0}^\infty \frac{\varphi_k}{k!}x^k$$ and $\varphi_k$ goes to zero. Clearly, both of the power series on the right-hand side converge everywhere, and we know that the $h(0)$ part is the exponential: $$f\left(x\right)=h\left(0\right)e^x + \sum_{k=0}^\infty \frac{\varphi_k}{k!}x^k$$

Now, differentiating this equation $n$ times: $$f^{\left(n\right)}\left(x\right)=h(0)e^x + \sum_{k=n}^\infty \frac{\varphi_k}{\left(k-n\right)!}x^{k-n}=h(0)e^x+\sum_{k=0}^\infty \frac{\varphi_{k+n}}{k!}x^k$$

Since $\varphi_k\to 0$, we know that: $$\left|\sum_{k=0}^\infty \frac{\varphi_{k+n}}{k!}x^k\right|\leqslant\sum_{k=0}^\infty \frac{\left|\varphi_{k+n}\right|}{k!}\left|x^k\right|\leqslant\varepsilon\sum_{k=0}^\infty \frac{\left|x^k\right|}{k!}=\varepsilon e^{\left|x\right|}$$ and we can make $\varepsilon$ be as small as we want by choosing $n$ big enough. This means that: $$\lim_{n\to\infty}\left|\sum_{k=0}^\infty \frac{\varphi_{k+n}}{k!}x^k\right|=0$$ pointwise, and therefore: $$\lim_{n\to\infty}f^{\left(n\right)}\left(x\right)=h(0)e^x$$ so that the answer to your question is positive.

The Phenotype
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fonini
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    Why must the Taylor series of $f$ converge to $f$? I agree that it converges, but it does not necessarily have to be $f$. – Ingix Jul 13 '17 at 12:19
  • Oops. I don't know. – fonini Jul 13 '17 at 12:36
  • If you look at questions like this (https://math.stackexchange.com/questions/859793/interchanging-pointwise-limit-and-derivative-of-a-sequence-of-c1-functions?rq=1), it turns out that counterexamples are 'messy'. So assuming that $f$ has a converging Taylor series removes lots of cases. I think that the claim is true, but when doing the proof there are always cases of exchanged limits that seem 'obviously true' but where my experience has shown me that maybe if it isn't easily proven then maybe there is a counterexample. – Ingix Jul 13 '17 at 12:41
  • The exchanged limits are fine: you can always differentiate power series term-by-term, and the (open) interval of convergence is the same. I just don't know how to get rid of the assumption that the Taylor series actually converges to $f$ – fonini Jul 13 '17 at 15:13
  • Thank you for this nice answer. Although it does not answer the question completely, at least it answers it for all real analytic functions and therefore shows that a counterexample if it may exist has to be rather ugly (not real analytic)... I decided to accept this as an answer! – Tom Jul 17 '17 at 12:30