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I'm solving old exam problems in real analysis. Thus, only such methods may be used. I've been trying to solve the problem below and have encountered some issues.

Let $f\in C^\infty(\mathbb{R})$ in a neighborhood of the point $x_0$. Assume that there exists positive numbers $\delta$ and $M$ such that for any $x\in(x_0-\delta,x_0+\delta)$ one has the estimate $$\left|\frac{d^kf(x)}{dx^k}\right|\leq M\frac{k!}{\delta^k}.$$ Show that under these assumptions $$f(x)=\sum_{k=0}^\infty\frac{1}{k!}\frac{d^kf(x_0)}{dx^k}(x-x_0)^k.$$ Note that this means that the estimate above implies that $f(x)$ is analytic at $x_0$.

My first thought is to try to find something that looks what we want by dividing the estimate by $k!/\delta^k$ on both sides. Equivalently, multiply with $\delta^k/k!$ on both sides. Here $$\left|\frac{d^kf(x)}{dx^k}\right|\leq M\frac{k!}{\delta^k}\Leftrightarrow\frac{\delta^k}{k!}\left|\frac{d^kf(x)}{dx^k}\right|\leq M\frac{k!}{\delta^k}\frac{\delta^k}{k!}=M$$ for all $x\in(x_0-\delta,x_0+\delta)$. That $M$ is a fixed number means that we have a bound of the left-hand side of the equation. That $x\in[x_0-\delta,x_0+\delta]$ means that $|x_0-x|\leq\delta$. Thus $(x_0-x)^k\leq\delta^k$. Hence $$\frac{1}{k!}\left|\frac{d^kf(x)}{dx^k}\right|(x_0-x)^k\leq\frac{1}{k!}\left|\frac{d^kf(x)}{dx^k}\right|\delta^k\leq M.$$ This means that the summand is bounded.

I don't really know how to progress from this step, or if anything that I've deduced is of any value. I guess that showing pointwise or uniform convergence might give us what we want, e.g. showing that this converges to a function, but how do we know that is the function that we wanted to begin with? I feel like I'm missing something obvious here.

Laplace's Demon
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    Hint: use Taylor's theorem with Lagrange form of the remainder. – Robert Israel Aug 16 '18 at 16:09
  • The result is supposed to be derived with the estimate. If I were to use Taylor's theorem, then I could just say that the function fulfills that formula just by it being smooth. – Laplace's Demon Aug 16 '18 at 16:11
  • I don't think you understand. The remainder is the difference between $f(x)$ and the $n$'th partial sum of the series. The estimate lets you show that this difference goes to $0$ as $n \to \infty$. – Robert Israel Aug 16 '18 at 16:19
  • Possible duplicate of: Does Cauchy's estimate imply analyticity? – MSDG Aug 16 '18 at 16:19
  • Yeah! I think that you're right. I guess the error of mine was that I didn't know what the estimate was called. If we may use Taylor's theorem, then why don't we just do that straight from $f\in C^\infty(\mathbb{R})$? – Laplace's Demon Aug 16 '18 at 16:24
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    @OskarTegby You also need the estimate because not every $C^\infty$ function has an infinite power series. The standard example is $$ f(x) = \begin{cases}e^{-1/x}, & x > 0\ 0, & x \leq 0\end{cases}. $$ This function is smooth everywhere but not real-analytic at the origin since its derivatives are all 0 there. (I.e. its power series would imply that the function is identically zero in a neighbourhood of the origin, which it is clearly not.) – MSDG Aug 16 '18 at 16:28
  • @Sobi Okay. Thanks for clarifying that! – Laplace's Demon Aug 16 '18 at 16:29
  • @OskarTegby You might also find it interesting to look at wiki: Taylor's theorem and convergence of Taylor series. – MSDG Aug 16 '18 at 16:36
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    @OskarTegby "If we may use Taylor's theorem, then why don't we just do that straight from $f\in C^\infty$?". Seems like you have this straight now. It's remarkable how many students seem to think that Taylor's theorem says $f=\sum c_nx^n$. That's simply not what the theorem says... – David C. Ullrich Aug 16 '18 at 19:33

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If you take $x\in (x_0 - \rho, x_0+\rho)$ with $\rho < \delta$ then the terms of the Taylor series will be bounded by $M\frac{\rho^k}{\delta^k}$ and you can apply the Weierstrass M-test. Hence the series converges absolutely and uniformly on $(x_0 - \rho, x_0+\rho)$ and $f$ is analytic at $x_0$.

EDIT: This argument only shows that the Taylor series defines an analytic function $g$ around $x_0$. To see that $g=f$ one may fix an arbitrary point $x\in (x_0 - \rho, x_0+\rho)$ and use the Lagrange remaider for the Taylor expansion of $f$ up to order $n$ and then use the estimates to show that $|f(x)-g(x)| =0$ letting $n\to \infty$.

$$|f(x)-g(x)| = \left|\frac{f^{(n+1)}(x^\ast)}{(n+1)!}(x-x_0)^{n+1}-\sum_{k=n+1}^\infty\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k\right|\leq \\ \sum_{k=n+1}^\infty\frac{|f^{(k)}(x_0)|}{k!}|x-x_0|^k+\frac{|f^{(n+1)}(x^\ast)|}{(n+1)!}|x-x_0|^{n+1} \leq \sum_{k=n+1}^\infty M\frac{k!}{\delta^k}\frac{|x-x_0|^k}{k!} + M\frac{(n+1)!}{\delta^{n+1}} \frac{|x-x_0|^{n+1}}{(n+1)!}=\sum_{k=n+1}^\infty M\frac{|x-x_0|^k}{\delta^k} + M\frac{|x-x_0|^{n+1}}{\delta^{n+1}}$$ Since $|x-x_0|<\delta$, $\sum_{k=n+1}^\infty M\frac{|x-x_0|^k}{\delta^k}$ is the tail of a convergent geometric series, hence it tends to $0$ as $n\to \infty$. Also $M\frac{|x-x_0|^{n+1}}{\delta^{n+1}}\to 0$ as $n\to \infty$ and we conclude that $|f(x)-g(x)| \leq 0$.

Alan Muniz
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    No. Showing the Taylor series converges uniformly on compact sets is not enough. You need to show that it converges to $f$. – David C. Ullrich Aug 16 '18 at 19:17
  • Oh, that is true... Thank you. – Alan Muniz Aug 16 '18 at 19:41
  • When we insert the Cauchy estimate we get $$\sum_{k=0}^\infty\frac{|f^{(k)}(x)|}{k!}\delta^k\leq\sum_{k=0}^\infty M\frac{k!}{\delta^k}\frac{\delta^k}{k!}=\sum_{k=0}^\infty M,$$ which doesn't really converge. Should we use something else for the M-test? Also, when we compute $|f(x)-g(x)|$ we write it on out as $|(T_n(x,a)-R_n(x,a))-T_n(x,a)|$. Right? If so, then we get the same problem as with the series when showing that $\lim_{n\to\infty}|R_n(x,a)|=0$. – Laplace's Demon Aug 17 '18 at 09:38
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    I edited that to explain a bit more. Tell me if it is clear now. – Alan Muniz Aug 17 '18 at 12:18
  • @AlanMuniz Thank you for your extension of your answer! I understand everything except for why $\frac{|f^{(n+1)}(x^\ast)|}{(n+1)!}|x-x_0|^{n+1} \leq
    • \frac{\delta^{n+1}}{(n+1)!}$. Aren't we using the Cauchy estimate here and should get $\frac{|f^{(n+1)}(x^\ast)|}{(n+1)!}|x-x_0|^{n+1} \leq

    +M\frac{(n+1)!}{\delta^{n+1}}\frac{\delta^{n+1}}{(n+1)!}=M$?

     – Laplace's Demon Aug 17 '18 at 14:21
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    Just corrected that too – Alan Muniz Aug 17 '18 at 15:02
  • Okay. Thanks! Now everything is crystal clear. I'm very thankful for your answer, and your clarification. – Laplace's Demon Aug 17 '18 at 15:43