You've guessed it right but more careful treatment is needed. Furthermore, within my limited knowledge, in general, your statement is not rigorous as we can not be sure of the differentiablity of $Q^tf$.
So I decided to reformulate it as follows.
Main theorem
Theorem
if $f$ is a Lipschitz continuous function, then for almost every $(t,x) \in \mathbb{R}_+ \times \mathbb{R}^n$ (wrt to Lesbeque measure), we have:
$$ \frac{d}{dt}Q^tf(x)+ \dfrac{1}{2} \Vert \nabla f(x) \Vert^2 = 0$$
Primilinaries
Now, I present some ingredients for the demonstration.
Theorem 1
$g_t(x):= \frac{1}{2t}\Vert x \Vert^2 -Q^tf(x)$ is a convex function$\square$
Theorem 2
For a real convex function $g$ defined on $\mathbb{R}^n$, if $g$ is differentiable at $x$ then we have the following inequality for all $y$:
$$ g(y)-g(x) \ge \langle \nabla g(x), y-x) \quad \square$$
Theorem 3 (Rademacher theorem- Almost everywhere differentiability)
If $f$ is a locally Lipschitz continuous function from $\mathbb{R^n} \rightarrow \mathbb{R^m}$, then $f$ is differentiable almost everywhere.
Answer
We'll start by answering partially your question.
Theorem 4
For any $s>0$ and $x$ such that $Q^sf$ is differentiable at $x$, then we have:
$$\dfrac{d}{ds^+} Q^sf(x) =-\dfrac{1}{2} \Vert \nabla Q^sf(x) \Vert^2$$
Demonstration
For simplicity, we let $h:= Q^tf$
For any $t>0$, by choosing $y:= x-t\nabla h(x)$, we imply that
$$Q^th(x) -h(x) \le h(x-t\nabla h(x)) -h(x) +\dfrac{t}{2} \Vert \nabla h(x) \Vert^2 $$
Thus by letting $t$ converge to $0^+$, we have:
$$ \dfrac{d}{dt^+} Q^tf(x)|_{t=0} \le -\dfrac{1}{2} \Vert \nabla Q^tf(x) \Vert^2$$
Secondly, because $h$ is differentiable at $x$, $g^t$ is also differentiable at $x$, so we have:
$$ h(y) -h(x) \underbrace{=}_{\text{definition of} g^t} g^t(x)-g^t(y) -\dfrac{1}{2s}\left( \Vert x \Vert^2 - \Vert y \Vert^2 \right) \ge \langle \nabla g^t(x), x-y \rangle-\dfrac{1}{2s}\left( \Vert x \Vert^2 - \Vert y \Vert^2 \right) $$
Thus,
$$ h(y) +\frac{1}{2t}\Vert x-y \Vert^2 \ge h(x)+u(x)+ \left[v(y) +\frac{1}{2t}\Vert x-y \Vert^2\right] $$
where $u(x) := \langle \nabla g^t(x), x \rangle-\frac{1}{2s} \Vert x \Vert^2 $ and $v(y)= - \langle \nabla g^t(x),y\rangle +\frac{1}{2s} \Vert y \Vert^2$
So,
$$ Q^th(x)-h(x) \ge Q^tv(x) -v(x) $$
(Note that v(x)=u(x))
The function $v$ is clearly continously differentiable, and we can reconfirm that:
$$\dfrac{d}{dt}Q^t(v)|_{t=0^+} = -\frac{1}{2}\Vert \nabla v (x) \Vert^2 = -\frac{1}{2}\Vert \nabla h (x) \Vert^2 $$
Done.
So now what is left the regularity of $Q^tf$ and indeed, we see that $Q^tf$ is lipschitz continuous by the following theorem :
Theorem 5 (Regularity of $Q^tf$ )
The function $F(t,x) = Q^t(x) $ is $\alpha$ lipschitz continous on $\mathbb{R}_{+} \times \mathbb{R^n}$.
Hence it is differentiable almost everywhere.
Demonstration
For all points $x,y$ and $t>0$,
Let $\tilde{y}$ be the minimizer of $Q^tf(y)$, and $z:= x+\tilde{y}-y$
We see that:
$$ Q^t(x)-Q^t(y) \le f(z)+\dfrac{1}{2t}\Vert z-x\Vert^2 - f(\tilde{y})-\dfrac{1}{2t}\Vert \tilde{y}-y\Vert^2 = f(z)-f(\tilde{y}) \le \alpha \Vert x - y \Vert $$
From then we can imply that:
$$|Q^t(x)-Q^t(y)| \le \alpha \Vert x - y \Vert $$
Then we see that:
- $Q^tf(x) \le f(x)$ , and
- $Q^tf(x) = f(x) + \inf\{ f(y)-f(x)+\frac{1}{2t} |x-y|^2 \} \ge f(x)+\inf \{ -\alpha |y-x|+\frac{1}{2t} |x-y|^2 \} = f(x)-t \dfrac{\alpha^2}{2}$
Thus $F$ is Lipschitz $\square$
Remark 6 Note that $Q^sQ^t =Q^{t+s}$
From theorem 4 and theorem 5, we have the conclusion in your main theorem which is stated in the previous section.
Comments
- The main idea is to bound $Q^tf$ by a reasonable bound. While the upper bound is pretty straightforward if we know how the minimizers look like, the treatment of the lower bound is a bit trickier.
- In fact, the idea for the lower comes from another mathematical object called subdifferential. To be honest, I have written a short introduction to that object for you, however as it turned out to be not that important, I decided to leave that part aside.
- Professor Evans's PDE book also provides an excellent short chapter on this, which may be different than mine (though I learned a lot from it). It's a great reference if you are serious about PDEs.
Disclaimer I'm not an expert in PDEs
