Solution to semilinear heat equation at $t=0$: $u_t(0,x) - \Delta u(0,x) + f(x,u,u_x)= 0 \ ?$ Consider the following Cauchy problem 
$$u_t - \Delta u + f(x,u,u_x) = 0, \quad (t,x) \in (0,T) \times \mathbb{R}^n,$$
with initial condition $u(0,x) = g(x), \ x \in \mathbb{R}^n.$
Suppose that $u \in C^2((0,T) \times \mathbb{R}^n)$ is a solution of the initial value problem. 
Can we obtain (under some reasonable assumptions) that 
$$u_t(0,x) - \Delta u(0,x) + f(x,u,u_x)= 0,$$
that is, that the solution $u$ solves the PDE at time $t=0$ too?
 A: This is a well known result from diffusion PDE theory. For example [1] studied the semi-linear equation in form of 
$$u_t - \Delta u + u^\gamma = 0, \quad (t,x) \in (0,T) \times \mathbb{R}^n, \gamma>1$$ with initial value $u(0,x) = g(x), \ x \in \mathbb{R}^n.$
And with typical Green function representation we could write (2.1) in [1].
$$u(t)=e^{t\Delta}g+{\displaystyle \int_{0}^{t}e^{(t-s)\Delta}u^{\gamma}ds}$$ with $e^{t\Delta}g(x)={\displaystyle \int_{\mathbb{R}^{n}}Gr_{t}(x-y)dy}$ and $Gr_{t}(x)=\left(4\pi t\right)^{-\frac{n}{2}}e^{-\frac{\|x\|^{2}}{4t}}$. Therefore an intuition that such a solution $u$ exists if $u^\gamma,\gamma>1$satisfy regularities. And if $f(x,u,u_x)$ is somehow  regular enough then we can expand it into analytic form in terms of powers of $u,u_t$ and the Green function representation argument applied.
A rigid proof requires more work but I believe that is what I(and possibly @ Michael Renardy) had in mind about regularity on solution to the equation above.
[1]Weissler, Fred B. "Existence and non-existence of global solutions for a semilinear heat equation." Israel Journal of Mathematics 38.1 (1981): 29-40.
A: The question is vague, because the OP does not mention in which sense the PDE is to be satisfied at initial time.
In addition, the assumption that $u$ is $C^2$ for $T>0$ is very weak, if not useless. For let me assume that $f$ is a smooth function with $f(\cdot,0,0)\equiv0$, and that the initial data $g$ is $L^1\cap L^\infty({\mathbb R}^n)$. Then it is classical, because of the smoothing property of the heat semi-group, that $u$ is actually $C^\infty$ for $t>0$ ; hence the assumption is automatically fulfilled. Therefore the question asks whether, if $g\in L^1\cap L^\infty({\mathbb R}^n)$, one has
$$u_t(0)=\Delta g-f(\cdot,g,\nabla g).$$
certainly, this does not hold pointwisely, since the right-hand side may not be a function. It holds in the distributional sense ; just pass to the limit (in ${\cal D}'$) in the equation satisfied at $\{s\}\times{\mathbb R}^n$, as $s\rightarrow0+$. But then $u_t(0)$ is just defined as the limit (in ${\cal D}'$) of $u_t(s)$ as $s\rightarrow0+$, and this is a tautology. The answer will be better if we can prove that the distributional limit of $\frac{u(s)-g}s$ equals $\Delta g-f(\cdot,g,\nabla g)$. I guess that this can be done by the usual techniques of semi-groups.
