Tychonoff's non-uniqueness example

Counterexample

There is a function $u \colon \R \times \R \to \R$, smooth and not identically zero, such that $$\partial_t u = \partial_x^2 u \ \text{on } \R \times \R \quad \text{and} \quad u(x, t) = 0 \ \text{for all } t \le 0.$$ So the Cauchy problem for the heat equation on $\R \times (0, \infty)$ with initial value $0$ has a solution other than $u = 0$. One construction: fix $\alpha > 1$, let $g(t) = \exp(-t^{-\alpha})$ for $t > 0$ and $g(t) = 0$ for $t \le 0$, and set $$u(x, t) = \sum_{k=0}^{\infty} g^{(k)}(t) \, \frac{x^{2k}}{(2k)!}.$$ Term by term, $\partial_x^2$ of the $k$-th term is the $(k-1)$-th term with $g^{(k-1)}$ replaced by $g^{(k)}$, so formally $\partial_x^2 u = \sum_k g^{(k+1)}(t) x^{2k}/(2k)! = \partial_t u$. There are constants $\theta, c > 0$ with $\abs{g^{(k)}(t)} \le k! \, (\theta t)^{-k} \exp(-c \, t^{-\alpha})$ for $t > 0$. These bounds show that the series and its differentiated series converge locally uniformly, so $u$ is a genuine solution. Since $g$ is not identically zero, neither is $u$.