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I encounter a difficulty when proving the bounded solution of ancient heat equation implying constant function. Suppose $u(t,x)$ is the solution of ancient heat equation:

\begin{equation} u_{t} = \Delta u \quad \mathrm{in} ~\ (- \infty, 0] \times \mathbb{R}^{n} \end{equation}

In order to prove it is constant function, we can mimic the proof of Liouville's theorem in complex analysis by showing that $\nabla_{x} u(t,x)$ goes to zero. In usual heat equation defined in $(0, T] \times \mathbb{R}^{n}$, we have the following gradient estimation:

\begin{equation} \sup_{x \in \mathbb{R}^{n}}| \nabla_{x} u(T,x)| \leq \frac{C}{\sqrt{T}} \sup_{x \in \mathbb{R}^{n}} |u_{0}| \end{equation} where C is a constant relates to its dimension $n$ and $u_{0}$ is the initial value of $u(t,x)$. Originally, I want to change the variable $t \rightarrow -t$ such that I can apply the above gradient estimation to show that the solution of ancient heat equation $u(t,x)$ is constant function. However, there is a problem that $\sqrt{T}$ would be complex under this change of variable. Therefore, how can I combined these two concepts to show bounded solution for ancient heat equation is constant function, similar to Liouville's theorem in Complex analysis?

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  • $\begingroup$ Yes, this inequality is false if $T<0$ ... when you look at the proof, you will see that if $t<0$, it doesn't work since the disspative term is now no more negative. $\endgroup$
    – LL 3.14
    Commented Apr 13, 2020 at 8:04
  • $\begingroup$ To see that $u$ is constant you have to apply the estimates for the derivatives in the following form $$|D_t^n D_x^k u(t_0,x_0)| \le \frac{C_{n,k} \|u\|_{\infty, Q(t_0,x_0; R)}}{R^{k+2n}},$$ where $Q(t_0,x_0;R)=(t_0-R^2,t_0)\times B(x_0,R)$ adn let $R \to \infty$ for every fixed $(t_0,x_0)$ (you need only $n,k$ to be 0 or 1). $\endgroup$
    – Giorgio Metafune
    Commented Apr 13, 2020 at 8:15

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Your approach works fine, just start with negative moments of time:

\begin{equation} \sup_{x \in \mathbb{R}^{n}}| \nabla_{x} u(t,x)| \leq \frac{C}{\sqrt{T}} \sup_{x \in \mathbb{R}^{n}} |u(t-T,x)| ,\quad t\le 0,\ T>0. \end{equation}

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