If a Taylor series agrees with its function in $U$, that function is analytic in $U$

Question

Assume that $f:\mathbb{R}\rightarrow\mathbb{R}$ coincides with its Taylor series centered at a point $p\in U$ on an open set $U$, i.e.

$$\exists p\in U, \forall x\in U, f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(p)}{k!}(x-p)^n.$$

Is $f$ analytic on $U$?

That is:

$$\forall p\in U, \exists \varepsilon>0, \forall x\in \mathbb{R}, |p-x|<\varepsilon\Rightarrow f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(p)}{k!}(x-p)^n?$$

Answer 1

According to Wikipedia, yes, $f$ is analytic everywhere on the interval where it is equal to its Taylor series (see the end of the introductory paragraph). I'm not sure how to prove it, but I guess convergent infinite sums of polynomials just happen to be analytic.

Answer 2

The answer to your question is yes, though it should be pointed out that the answer to your title's question is NO: "analytic at a point" does not imply "analytic on $U$" (bump functions being an obvious counterexample). But this doesn't contradict what I'm about to say because your hypothesis is doesn't just say "analytic at a point"; it is actually stronger than that.

Anyway, just to be explicit, @paul garrett's comment that "Analyticity at one point, or even on a non-trivial open, does not magically force analyticity anywhere else" is definitely correct, but again, this doesn't answer the question, and it also doesn't contradict what I'm about to say, because your assumption is stronger.

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To prove this, let us fix a point $p\in U$ as in the hypothesis. Now, define $\rho:= \sup\limits_{x\in U}|x-p|$. Then,

• $0<\rho\leq \infty$
• $U$ is contained in the open interval $(p-\rho,p+\rho)$ (which we interpret as $\Bbb{R}$ in case $\rho=\infty$). This is because if $x\in U$, then $x\in [p-\rho,p+\rho]$ definitely, but because of openness of $U$, we must actually have $x\in I=(p-\rho,p+\rho)$, otherwise it would contradict the definition of $\rho$ as a supremum.
• The formal series $S(X)=\sum\limits_{n=0}^{\infty}\frac{f^{(n)}(p)}{n!}X^n$ has radius of convergence $\geq \rho$ (why?).

So, the function $\phi:I\to\Bbb{R}$, $\phi(x):=\sum\limits_{n=0}^{\infty}\frac{f^{(n)}(p)}{n!}(x-p)^n$ is well-defined (i.e the series converges) and $\phi|_U=f|_U$ by hypothesis. So, as long as we show $\phi$ is analytic on the interval $I$, it follows that $\phi|_U$ is analytic (restriction of analytic functions to open sets is again analytic simply by looking at the definition) and hence $f|_U$ is analytic. In short, we just have to show "functions defined by convergent power series are analytic as long as we're within the radius of convergence".

The idea of the proof is just to rearrange a double sum, and we can do this as long as the series in question converge absolutely. For notational convenience, let's define $a_n=\frac{f^{(n)}(p)}{n!}$. Now, fix a point $x_0\in I$, denote $r_0=|x_0-p|$ and fix an $r$ such that $0<r<\frac{\rho- r_0}{2}$.

Recall that we can calculate the derivatives $\phi^{(n)}(x_0)$ using term-by-term differentiation. We now show that the the series centered at $x_0$ has radius of convergence $\geq r$: \begin{align} \sum_{n=0}^{\infty}\left|\frac{\phi^{(n)}(x_0)}{n!}\right|r^n&=\sum_{n=0}^{\infty}\left|\frac{1}{n!}\sum_{m=0}^{\infty}\frac{(m+n)!}{m!}a_{m+n}(x_0-p)^m\right|r^n\\ &\leq \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{(m+n)!}{n!m!}|a_{m+n}|r_0^mr^n\\ &=\sum_{k=0}^{\infty}\sum_{l=0}^k\frac{k!}{l!(k-l)!}|a_k|r_0^{k-l}r^{l}\\ &=\sum_{k=0}^{\infty}|a_k|(r_0+r)^k\\ &<\infty, \end{align} whereby in the middle I rearranged the terms (which is valid since we're dealing with non-negative terms), and I used the usual binomial theorem, and finally since $r_0+r<\rho$, it follows the series is indeed finite, and hence the formal series $\sum_{n=0}^{\infty}\frac{\phi^{(n)}(x_0)}{n!}X^n$ has radius of convergence $\geq r$.

Now, we repeat the same calculation, but this time without absolute values, and we rearrange the double sum. The triangle inequality shows that any point $x$ such that $|x-x_0|\leq r$ actually belongs to $I$. Now, for such values of $x$, we have \begin{align} \sum_{n=0}^{\infty}\frac{\phi^{(n)}(x_0)}{n!}(x-x_0)^n&=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\sum_{m=0}^{\infty}\frac{(m+n)!}{m!}a_{m+n}(x_0-p)^m\right)(x-x_0)^n\\ &=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{(m+n)!}{n!m!}a_{m+n}(x_0-p)^m(x-x_0)^n\\ &=\sum_{k=0}^{\infty}\sum_{l=0}^k\frac{k!}{l!(k-l)!}a_k(x_0-p)^{k-l}(x-x_0)^{l}\\ &=\sum_{k=0}^{\infty}a_k(x-p)^k\\ &=:\phi(x) \end{align} In the middle we were able to rearrange the double sum from $(n,m)$ to $(k,l)$ because we established absolute convergence in the previous paragraph. So, for each $x_0\in I$, we have proven the existence of an $r>0$ such that for all $|x-x_0|\leq r$, $\phi(x)$ does indeed equal its Taylor series centered at $x_0$. Hence, $\phi$ is analytic on $I$. This completes the proof.