Prove that an operator (with heat kernel) maps the space of $C^1$ functions with bounded derivatives to itself Let $K$ be the heat kernel. 


*

*Does the operator 
$$g \mapsto\int_0^T\int_{\mathbb{R}^N} K(t-\xi,x-\zeta) \ \  f(\xi,\zeta,g(\xi,\zeta),\nabla g(\xi,\zeta)) \, d\xi \, d\zeta$$
map the space of $C^1([0,T]\times \mathbb{R}^N)$ (for any $T>0$) functions bounded and with bounded derivatives to itself if $f: [0,T]\times \mathbb{R}^N \times \mathbb{R} \times \mathbb{R}^N \to \mathbb{R}$ is sufficiently well-behaved (for instance $C^1$)? 

*Does it do the same for the space of $C^2$ or $C^\infty$ functions with bounded derivatives?

Here $\nabla$ is used to denote the gradient with respect to the space variable $x \in \mathbb{R}^N$.
 A: For my typing convenience allow me to drop the Greek letters. For a suitable class of functions $f$ and for $g$ of class $C^k$ with all derivatives bounded, you are asking whether
$$\tilde g (t,x) = \int \limits _0 ^T \int \limits _{\Bbb R^n} K(t-u, x-y) f \big( u, y, g(u,y), (\nabla g) (u,y) \big) \ \Bbb d y \Bbb d u $$
is of class $C^k$ with all the derivatives bounded.
As you are going to see, it turns out that this has nothing to do with $K$ being the heat kernel, the problem can be posed for much more general integral kernels.
Notice that
$$\partial _t \big( K(t-u, x-y) \big) = (D_1 K) (t-u, x-y) = - \partial _u \big( K(t-u, x-y) \big)$$
and
$$\partial _{x_i} \big( K(t-u, x-y) \big) = (D_2 K) (t-u, x-y) \frac {\partial (x-y)} {\partial x_i} = \\
- (D_2 K) (t-u, x-y) \frac {\partial (x-y)} {\partial y_i} = - \partial _{y_i} \big( K(t-u, x-y) \big)$$
where $D_1 K$ and $D_2 K$ are the derivatives of $K$ with respect to the 1st, respectively the 2nd, argument ($D_2 K$ is a vector that gets paired with the vector $\dfrac {\partial (x-y)} {\partial x_i}$ in the natural way).
Using Lebesgue's dominated convergence theorem we may slip all the derivatives inside the integral, and then integration by parts gives:
$$(\partial _t \tilde g) (t,x) = \int \limits _0 ^T \int \limits _{\Bbb R^n} \partial _t K(t-u, x-y) f \big( u, y, g(u,y), (\nabla g) (u,y) \big) \ \Bbb d y \Bbb d u  = \\
- \int \limits _0 ^T \int \limits _{\Bbb R^n} \partial _u K(t-u, x-y) f \big( u, y, g(u,y), (\nabla g) (u,y) \big) \ \Bbb d y \Bbb d u = \\
\int \limits _0 ^T \int \limits _{\Bbb R^n} K(t-u, x-y) \partial _u f \big( u, y, g(u,y), (\nabla g) (u,y) \big) \ \Bbb d y \Bbb d u = \\
\int \limits _0 ^T \int \limits _{\Bbb R^n} K(t-u, x-y) (D_1 f) \big( u, y, g(u,y), (\nabla g) (u,y) \big) \ \Bbb d y \Bbb d u + \\
\int \limits _0 ^T \int \limits _{\Bbb R^n} K(t-u, x-y) (D_3 f) \big( u, y, g(u,y), (\nabla g) (u,y) \big) \frac {\partial g} {\partial u} (u,y) \ \Bbb d y \Bbb d u + \\
\int \limits _0 ^T \int \limits _{\Bbb R^n} K(t-u, x-y) (D_4 f) \big( u, y, g(u,y), (\nabla g) (u,y) \big) \frac {\partial (\nabla g)} {\partial u} (u,y) \ \Bbb d y \Bbb d u .$$
If $f$ is $C^1$ with the derivatives bounded (say, by some $M>0$), then the first term in the sum of three integrals above is bounded by
$$0 \le \left| \int \limits _0 ^T \int \limits _{\Bbb R^n} K(t-u, x-y) (D_1 f) \big( u, y, g(u,y), (\nabla g) (u,y) \big) \ \Bbb d y \Bbb d u \right| \le \\
\int \limits _0 ^T \int \limits _{\Bbb R^n} \left| K(t-u, x-y) (D_1 f) \big( u, y, g(u,y), (\nabla g) (u,y) \big)  \right| \Bbb d y \Bbb d u = \\
\int \limits _0 ^T \int \limits _{\Bbb R^n} K(t-u, x-y) \left| (D_1 f) \big( u, y, g(u,y), (\nabla g) (u,y) \big)  \right| \Bbb d y \Bbb d u \le
M \int \limits _0 ^T \int \limits _{\Bbb R^n} K(t-u, x-y) \ \Bbb d y \Bbb d u $$
which is finite.
A similar computation can be done for the other two terms, too, but for them you'll also have to use the boundedness of the derivatives of $g$.
We conclude that $\partial _t \tilde g$ is bounded with respect to both $t$ and $x$.
A similar computation can be done for $\partial _{x_i} \tilde g$ - the same idea, the same steps, just longer to write because now $x$ is a vector. You will get boundedness for $\partial _{x_i} \tilde g$ too.
When $K$ is the heat kernel, the condition of boundedness on $f$ may be relaxed, imposing instead that the map $y \mapsto (D_i f) \big( u, y, g(u,y), (\nabla g) (u,y) \big)$ be in some $L^{p_i} (\Bbb R^n)$ uniformly with respect to $u$ (because the heat kernel is in $L^q (\Bbb R^n)$ for every $q \in [1, \infty]$ it may be paired with functions from any $L^p$ to get finite results).
The same reasoning works for derivatives of arbitrary order: if $f$ and $g$ are of class $C^k$ with bounded derivatives, so will be $\tilde g$. You may do it inductively, it is just long to write. Again, when $K$ is the heat kernel, the condition on $f$ may be relaxed, for instance by imposing that its composition with $g$ and $\nabla g$ belong to some Sobolev space $W^{k,p} (\Bbb R^n)$ - again, you only want its product with $K$ to be integrable on $\Bbb R^n$. The integral on $[0,T]$ does not bother you because this is a compact space so working with it is much easier. In particular, when $K$ is of class $C^k$ (not necessarily the heat kernel) then it is known that convoluting with it will again be of class $C^k$, so in particular continuous, so in particular bounded with respect to $t \in [0,T]$, so in particular integrable with respect to $t \in [0,T]$.
