Say $\Omega\subset\mathbb{R}^3$ is a bounded, connected domain with smooth boundary. Consider the initial-boundary value problem \begin{align} \partial_t u&=\Delta u \text{ for } x\in\Omega, t>0\\ u(t,x)_{|\partial\Omega}&=\gamma(x)>0\\ u(0,x)&=g\in C^\infty(\bar\Omega), g\ge 0 \end{align} I will also go ahead and assume that $g_{|\partial\Omega}=\gamma(x)$. Then the problem has a unique solution $u$ and due to the maximum principle, we know that \begin{align} \inf g\le u(x,t)\le \sup g \quad (MP) \end{align} for all $x\in\bar\Omega$ and $t\ge 0$. However, we know from an energy method argument that in fact, if $u^*$ is the unique harmonic function that satisfies $u^*_{|\partial\Omega}=\gamma(x)$, then \begin{align} \|u(t)-u^*\|_{L^2(\Omega)}\to 0 \end{align} exponentially in time. In fact, by bootstrapping to obtain bounds on higher derivatives of $u$, we can obtain \begin{align} \|u(t)-u^*\|_{L^\infty}\to 0\quad (*) \end{align} Then an immediate consequence of the fact that $u^*$ satisfies (because $u^*$ is harmonic) \begin{align} \inf \gamma\le u^*\le \sup \gamma \end{align} is that for any $\delta>0$, we can choose $T=T(\delta)$ so that for all $t\ge T$ we have \begin{align} \inf \gamma-\delta\le u(t,x)\le \sup \gamma+\delta \quad(**) \end{align} Of course, the above is trivially true (due to (MP)) if $[\inf g,\sup g]=[\inf\gamma,\sup\gamma]$, so let's think of initial conditions such that $\sup g>\sup\gamma$ and/or $\inf g<\inf\gamma$. Now my question is the following: is there a way to obtain the conclusion (**) without appealing to $L^p$/Sobolev bounds on higher derivatives?
To elaborate, through energy methods, we can obtain for example $\|u(t)-u^*\|_{H^2(\Omega)}\to 0$, from which (*) follows, due to $H^2(\Omega)\subset L^\infty(\Omega)$.
I suppose the question is motivated by the fact that a rather simple maximum principle gives the bounds (MP), but more machinery seems to be needed to prove (**), and it feels slightly overkill. So wondering if it doesn't have to be that way.
I was thinking, an application of Harnack's inequality (or the strong max principle) implies that if $\sup g> \sup \gamma$, then the quantity $M(t)=\sup_x u(t,x)$ must be strictly decreasing until for some $t^*$ we have $M(t^*)=\sup\gamma$, at which point we have $M(t)=\sup\gamma$ for all $t\ge t^*$ (of course, in general, such $t^*$ probably doesn't even need to exist). However the fact that $M(t)$ strictly decreases does not imply that $M(T)\le\sup\gamma+\delta$ for some $T$ large enough, as, for example $M(t)$ might be asymptotically bounded below by $\sup\gamma+2\delta$ while still strictly decreasing. Of course, we know from the energy method argument that this is not the case. But...any "simpler" way to see it?