Suppose that $u=u(x,t)$ is a real-valued function on $\mathbb{R}\times (0,\infty)$ with $$ u(x,t)=\int_0^t\int_{\mathbb{R}}G(x-y,t-s)f(u(y,s))\, dy\, ds\tag{1} $$ (where $G(x,t):=\frac{1}{4\pi t}e^{-\frac{\lvert x\rvert^2}{4t}}$ is the heat kernel and $f(u)$ is a bounded and continuously differentiable real function with bounded derivative.)
How can I proof whether u is differentiable?
I think I have to show that $\partial_x u=:u_x$ and $\partial_t u=:u_t$ exist and are continuous. But I have no concrete idea how to prove that.
I think one maybe should use that we also have $$ u(x,t)=\int_0^t\int_\mathbb{R}G(y,s)f(u(x-y,t-s))\, dy\, ds\tag{2} $$ and then (I hope its correct): $$ u_x(x,t)=\int_0^t\int_\mathbb{R}G(y,s)\frac{d}{dx}f(u(x-y,t-s))\, dy\, ds, $$
$$ u_t(x,t)=\int_0^t\int_\mathbb{R}G(y,s)\frac{d}{dt}f(u(x-y,t-s))\, dy\, ds+\int_\mathbb{R}G(y,s)f(u(x-y,0))\, dy $$
Can I use this to show existence and continuity of both partial derivatives? I think the problem is that I do not have information about the terms $\frac{d}{dx}f(u(x-y,t-s))$ and $\frac{d}{dt}f(u(x-y,t-s))$.
But am I in general right that the crucial point for the existence of $u_t$ and $u_x$ is to show that all these integrals are finite, i.e. the integrands are integrable functions?
Edit:
Maybe I shouldn't use (2) but stick to (1)? Then, I get $$ u_x(x,y)=\int_0^t\int_\mathbb{R}G_x(x-y,t-s)f(u(y,s))\, ds $$ I think this is a finite integral since $f$ is bounded and the heat kernel is infinitely differentiable and also integrable (again: does this imply the existence of $u_x(x,y)$?).
Moreoever, using (1), $$ u_t(x,t)=\int_0^t\int_\mathbb{R}G_t(x-y,t-s)f(u(y,s))\, dy\, ds+\int_\mathbb{R}G(x-y,0)f(u(y,t))\, dy. $$
However, $G(x-y,0)$ is a singularity. But maybe one can use $$ \int_\mathbb{R}G(x-y,0)f(u(y,t))\, dy=\int_\mathbb{R}G(y,t)f(u(x-y,t-s))\, dy. $$
Stick to definitions. Let $x,x'\in \mathbb{R}$ and $t, t'\in (0,\infty)$.
$$ u(x,t) - u(x',t') = \int_0^t \int G(x-y, t-s) f dy ds - \int_0^{t'} \int G(x'-y,t'-s) f dy ds $$
The difference you can rewrite as $I + II + III$ where
$$ I = \int_{t'}^t \int G(x-y, t-s) f dy ds $$
$$ II = \int_0^{t'} \int ( G(x-y, t-s) - G(x-y, t'-s)) f dyds $$
$$ III = \int_0^{t'} (G(x-y, t'-s) - G(x' - y, t'-s)) f dy ds $$
and your goal is to show that as $(x', t') \to (x,t)$ you have that the sum
$$ I + II + III = a (x' - x) + b(t'-t) + o(|x'-x| + |t'-t|) $$
By the definition of the heat kernel you should be able to write
$$ I = f(u(x,t)) (t-t') + o(|t-t'|) $$
and in $II$ and $III$ you can approximate
$$ G(x-y, t-s) - G(x-y, t'-s) = G_t(x-y, t-s) (t-t') + o(|t-t'|) $$
etc. (though the small $o$ need some more careful handling to ensure some degree of uniformity).
Let me do $I$ in detail for you, to show you what estimating the difference quotients entail. Here I will assume that $u$ is uniformly continuous for convenience. Therefore $f$ can just be regarded as some uniformly continuous function.
Let's estimate $$\int_{t'}^t \int_{\mathbb{R}} G(x-y, t-s) f(y,s) ~dy~ds - f(x,t) (t-t') $$ observe that this is the same as $$ \int_{t'}^t \int_{\mathbb{R}} G(x-y, t-s) (f(y,s) - f(x,t)) ~dy ~ds $$ using that for every fixed $\sigma > 0$ you have $\int_{\mathbb{R}} G(x-y, \sigma) ~dy = 1$.
So you can change variables and write as $$ \int_{0}^{t-t'} \int_{\mathbb{R}} G(y,s)( f(x-y,t-s) - f(x,t)) ~dy ~ds $$. Given $\epsilon > 0$, you can choose $\delta > 0$ such that
The second point follows from the explicit form of the heat kernel.
This tells us that for any $0 < s < \delta$
$$ \int_{\mathbb{R}} G(y,s) |f(x-y,t-s) - f(x,t)| ~dy \leq \int_{\mathbb{R}\setminus (-\delta,\delta)} + \int_{-\delta}^\delta $$ For the first integral you can bound $|f(x-y,t-s) - f(x,t)|$ by twice the sup of $f$, and use that the integral of $G$ is size $\epsilon$. For the second integral you can bound $|f(x-y,t-s) - f(x,t)|$ by $\epsilon$ and use that the integral of $G$ is no more than 1. And hence $$ \int_{\mathbb{R}} G(y,s) |f(x-y,t-s) - f(x,t)| ~dy \leq (2 \sup |f| + 1) \cdot \epsilon $$ And hence integrating in $s$ we get that $$ \int_0^{t-t'} \int_{\mathbb{R}} G(y,s) |f(x-y,t-s) - f(x,t)| ~dy \leq (2 \sup |f| + 1) \cdot \epsilon |t-t'| $$ Since $\epsilon$ is arbitrary this shows that
$$ I - f(x,t)(t-t') = o(|t-t'|)$$