Consider the Hopf-Lax semigroup given by $$ Q_tf(x) = \inf_{y \in \mathbb{R}^d} \:\{f(y) +\frac{1}{2t}|x-y|^2\} $$ for $f \in C_b(\mathbb{R}^d)$.
Prove that
$$ \frac{d}{dt} Q_tf(x) + \frac{1}{2} |\nabla Q_t f(x)|^2 = 0 $$
So far I only showed that $Q_t f \to f$ as $t\to 0^+$ but I have problems with computing the derivatives because of the infimum. I also thought of dealing with the infimum by thinking of the minimizers being in the direction of the steepest descent (given by the opposite of the gradient) but could not make it work. Any help?
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:
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
Disclaimer I'm not an expert in PDEs