Is the heat semigroup $P_t$ smooth for any function $f$ for which $P_tf$ is well-defined? Consider the heat semigroup (or heat kernel) $P_t$ which can be define on the space of non-negative functions, i.e. for any non-negative measurable function $f$
$$P_tf(x):=\frac{1}{\sqrt{2\pi t}^d}\int_{\mathbb R^d}f(y)e^{-|y-x|^2/(2t)}\,dy.$$
It can be infinite. Of course it can be defined for a much larger class of functions. My question, however, is the following:

Problem. Assume for some non-negative function $f$ we have that $P_tf(x)$ is finite for all $t\in (0,1]$ and a.e. $x\in\mathbb R^d$, i.e. one can evaluate the integral from the definition of $P_tf(x)$ and it is finite. Is it true that $P_tf(x)$ is differentiable (in a weak sense in both variables).

Actually I wanted to show that the function $u(x, t) =P_tf(x) $ solves (weakly) $$\tag{1}\partial_t u =\frac 12\Delta_x u$$
But I thought let's study the differentiability before even bothering whether the PDE (1) is satisfied. One knows that for any (regular enough $f$) the solution to the PDE (1) with boundary condition $u(x, 0)=f(x)$ is given by $P_tf(x) $. My question is like the reverse: what if I know $P_tf(x) $ is a well-defined function, then does it satisfy the PDE?
The issue is that we do not know the regularity of $f$, but we only know that it can be integrated against the heat kernel. This is related to a problem coming from stochastic processes, minimizing entropy where one gets that such quantities $P_tf(x)$ are well defined.
My attempts.  I really have no clue, I know the heat semigroup has a smoothing effect for integrable functions, or even functions growing fast, because then we can differentiate inside the integral. Here we are only given that $P_tf$ is well defined. I deep inside think that a counterexample exists. But also since the derivative of the integrand adds a polynomial term, I think that that polynomial term will not affect anything, so differentiating under the integral would be allowed.
Also, what if we add that the a.e. limit as $t\to 0^+$ exists and is equal to $f$ a.e.? Does it make the problem easier?

For the ones wondering, I added the tag 'stochastic analysis', because I would also be satisfied if we can show (or disprove) the following:
Assume $B$ is a Brownian motion. Recall the function $u(x, t) =P_tf(x) $. Do we have
$$u(B_t, 1-t) =u(B_0, 1)+\int^t_0 \nabla_x u(B_s, 1-s)\, dB_s. $$
Note that we cannot expect pointwise evaluation of $u$ so the stochastic equation may be understood equality in $L^2_{loc}(B) $. This would follow from Ito's formula if we would have $u$ satisfy the PDE (1) in a classical sense (or weakly in a nice sobolev space). As you can see both are related. I'm not sure which one is easier. The second one may use more methods from probability theory avoiding the whole differentiating, maybe? I'm clueless.
Edit. (Based on John Dawkins' answer). So the martingale in John Dawkins' answer, namely $\mathbb E[f(B_1)\mid \mathcal F_t]$ is where this question came from. Note we do not have integrability because the starting value of the Brownian motion is random. But if we are allowed to apply martingale representation theorem, then noting that $u(B_t,1-t)=\mathbb E[f(B_1)\mid \mathcal F_t]$ allows us to write
$$u(B_t,1-t)=u(B_0,0)+\int^t_0 H_s\,dB_s,$$
for some process $H$. The question is are we able to conclude that $H_s=\nabla_x u(B_s,1-s)$ in some sense?
 A: (I will stick to the 1-dimensional case, for simplicity.)
If $f$ is bounded then $u(x,t):=P_t f(x)$ is smooth on $\Bbb R\times(0,\infty)$, and you even have $\lim_{t\downarrow 0} u(x,t) = f(x)$ provided $f$ is also continuous. In this case, as you note,
$$
{\partial u\over\partial t}={1\over 2}{\partial^2u\over\partial x^2}.
$$
This in turn implies that if you fix $T>0$ then $u(B_t,T-t)$, $0\le t\le T$, is a martingale; in fact. by Ito's formula,
$$
u(B_t,T-t) = P_Tf(x)+\int_0^t u'_x(B_s,T-s)\,dB_s,
$$
where $u'_x$ is alternate notation for ${\partial u\over\partial x}$.
A more direct way to see the martingale property is to notice that $u(B_t,T-t)=\Bbb E[f(B_T)\mid\mathcal F_t]$ for $0\le t\le T$.  This last statement persists under the conditions you envision: If $f\ge 0$ is Borel measurable and $u(x,t):=P_tf(x)<\infty$ for each $t>0$ and $x\in\Bbb R$, then for each $T>0$ the process
$$
u(B_t,T-t),\qquad 0\le t\le T,
$$
is a continuous martingale, a.s. $\Bbb P^x$. (Here $\Bbb P^x$ is the law of the Brownian motion  started at $x$.) I'd guess that $u'_x$  exists in a good enough sense that the stochastic integral expression fo $u(B_t,T-t)$ above is still true in the more general setting, but a proof eludes me.
