Real analysis: if a function is differentiable does it imply that it has a power series?

Question

I would like to know if there is an analogous proof to that in complex analysis where one proves that if a function is holomorphic it is also analytic (i.e.) power series exists but for real analysis.

Moreover, for any real differentiable function can I take the analytic continuation and define it on a complex set (which includes the real numbers) to so that it is holomorphic and with which to show that it is hence analytic and that power series exist at every point (hence on the real line too)?

Answer 1

No. Although in complex analysis being holomorphic and analytic are equivalent notions, the same is not true in real analysis. Consider the example $$f(x)=\begin{cases}e^{-1/x},&x>0\\0,&x\leq0\end{cases}$$ defined on $\mathbb R$. It can be shown that this function is not only differentiable, but in fact infinitely differentiable at every point of the real line, including $x=0$. Nevertheless, it is not analytic at $0$: all of its derivatives evaluate to $0$ at $x=0$, so its Taylor series there is $0$, and thus does not agree with the function for $x>0$. You can read more about this here.

You can intuitively think about the above example like this: the function $e^{-1/x}$ decreases incredibly rapidly as $x\to0^+$ since $1/x$ blows up to $+\infty$, so that near $0^+$ all the information about the function at $x>0$ is crushed by the rapid decrease. (This is obviously not rigorous.) Thus at $x=0$, it locally appears to be exactly like the zero function, and so its Taylor series there, which can only access local information, is the zero function, despite the function not being identically zero.

Answer 2

No. In fact, even being infinitely differentiable is not sufficient: for example, define the following function:

$$f(x) = \sum\limits_{k=1}^\infty e^{-\sqrt{2^k}}\cos\left(2^kx\right).$$

This series, and each of the series $$\sum\limits_{k=1}^\infty \frac{\mathrm{d^n}}{\mathrm{d}x^n}\left(e^{-\sqrt{2^k}}\cos\left(2^kx\right)\right)$$ (which are just the same series multiplied by some power of $2^k$, with $\cos$ cyclically replaced by $-\sin$, $-\cos$, $\sin$, $\cos$, etc.) is absolutely and uniformly convergent by the Weierstrass M-test, and so $f$ is $C^\infty$: that is, at every point, we can differentiate $f$ as many times as we like.

However, if we take $x = \frac{n\pi}{2^k}$, and $m > k$, we have $$f^{(2^m)}(x) = \sum\limits_{i=m+1}^\infty e^{-\sqrt{2^i}}2^{2^mi} + \sum\limits_{i=1}^m e^{-\sqrt{2^i}}2^{2^mi}\cos(2^ix).$$ Now, we need to do some bounding. That first part is bounded below by $$e^{-\sqrt{2^{2m+2}}}2^{2^{m+1}(m+1)},$$ and the second part is bounded in absolute value by something $O(2^{kn})$, so we have $$\lim_{m\to\infty}\left(\frac{f^{(2^m)}(x)}{(2^m)!}\right)^{1/2^m} =+\infty,$$ and hence $$\limsup_{j \to \infty}\left(\frac{f^{(j)}}{j!}\right)^{1/j} = +\infty.$$ The Cauchy-Hadamard formula then implies that the radius of convergence of the Taylor series for $f$ at all such $x$ is $0$, so $f$ cannot be analytic at any such $x$. But also, domains of analyticity are open, and the set of all such points $x$ is dense in $\mathbb{R}$, so $f$ is nowhere analytic.