Analyticity of a two variable function Let's consider a solution to heat equation in the form of power series
$$u(x, t) = \sum_{k=0}^{\infty} \frac{g^{(k)}(t)}{(2k)!}x^{2k}.$$
Let's consider some specific $g(t)$
$$g(t) = \begin{cases}\exp\{-t^{-\alpha}\}; \quad t > 0 \\ 0; \quad t \le 0 \end{cases}$$
In my opinion it's obvious that for all $t \in \mathbb{R}$ the function $u$ is analytical function of $x \in \mathbb{C}$ from the definition of being analytical. Am I right?
However I'm a bit stuck whether $u$ is an analytical function of $t$. I think that there might be a problem with $0^0$. But I can't find a specific argument.
I would appreciate any help.
 A: Your solution (or better the class of solutions), represented by the series $u(x,t)$ for every $\alpha>0$, is certainly analytic in a neighborhood of all points
$$
(x, t)\in\big\{\Bbb R\times\{t\in \Bbb R: t>0\}\}\cup \big\{\Bbb R\times\{t\in \Bbb R: t<0\}\big\}
$$
i.e. it is analytic on the union of the open upper semiplane and of the open  lower semiplane, while it is not analytic in every neighborhood of any point of the set $\Bbb R\times\{0\}$. If $\mathscr{A}(\Omega)$ is the vector space of real analytic functions defined on the domain $\Omega$, we can express this fact by saying that $u(x,t)\in\mathscr{A}(\Bbb R^2\setminus \Bbb R\times\{0\})$,  but $u(x,t)\notin\mathscr{A}(\Bbb R^2)$. This is due to the fact that $g^{(n)}\in C^\infty(\Bbb R)$ for all $k\in\Bbb N$ but
$$
g^{(k)}(t)=\frac{\mathrm{d}^k g(t)}{\mathrm{d}t^k}= 0\quad \forall t\le0
$$
since
$$
g^{(k)}(t) = 
\begin{cases}
P\big(t^{-\alpha-1}, t^{-\alpha-2},\ldots,t^{-\alpha-k}\big)\exp\{-t^{-\alpha}\} & t > 0 \\
 0 & t \le 0 
\end{cases}
$$
where $P$ is a polynomial in the $k$ variables $\big\{t^{-\alpha-1}, t^{-\alpha-2},\ldots,t^{-\alpha-k}\big\}$. Then, by explicitly computing the power series expansion for $u(t,x)$ in any neighborhood of such a point gives the following result
$$
0=\sum_{k=0}^{\infty} \sum_{j=0}^{\infty}\frac{g^{(k+j)}(0)}{j!(2k)!}t^jx^{2k}\neq u(x,t)\quad t>0.
$$
The power series obtained does not represent the function $u(x,t)$ in any neighborhood of any point of the form $(x,0)$: this means that there does not exist a power series expansion for $u(x,t)$ in any neighborhood of any point $(x,0)$, thus the function is not analytic by definition.
Note
For $\alpha=2$, this is a classical counterexample in the theory of the heat equation (see for example [1], §3.7, p. 42): it precisely shows that there are null solutions to the Cauchy problem, i.e. non trivial solution for null Cauchy data $u(x,t)=0$.
Reference
[1] Cannon, John Rozier, The one-dimensional heat equation. Foreword by Felix E. Browder. (English) Encyclopedia of Mathematics and Its Applications, Vol. 23. Menlo Park, California etc.: Addison-Wesley Publishing Company; Cambridge etc.: Cambridge University Press. pp. XXV+483 (1984), ISBN: 0-201-13522-1, MR0747979, Zbl 0567.35001.
