Suppose $u(x,t)$ satisfies the following conditions:
(i) $u$ is continuous on the closure of the upper half-plane.
(ii) $u$ satisfies the heat equation $\frac{\partial{u}}{\partial t}=\frac{\partial^2{u}}{\partial x^2}$ for $t>0$.
(iii) $u$ satisfies the boundary condition $u(x,0)=0$.
(iv) $u(\cdot,t)\in\mathcal{S(\mathbb{R})}$ uniformly in $t$, that is $\displaystyle \sup_{\substack{x\in\mathbb{R}\\0<t<T}}|x|^k\left|\frac{\partial^l}{\partial x^l}u(x,t)\right|<\infty$ for each $k,l\geq0$.
Then we can conclude $u=0$.
In the proof, it let $E(t)=\int_{\mathbb{R}}|u(x,t)|^2\,dx$ and assert that it can be differentiate under the integral sign due to the above properties. I guess it may be the properties (iv), but how I don't know why? Is it beacause $\displaystyle\int_{\mathbb{R}}[\partial_tu(x,t)\bar{u}(x,t)+u(x,t)\partial_t\bar{u}(x,t)]\,dx=\int_{\mathbb{R}}[\partial_x^2u(x,t)\bar{u}(x,t)+u(x,t)\partial_x^2\bar{u}(x,t)]\,dx$, which is uniformly bounded in $t$ as $(iv)$ claims?
I'll note $$a_{k,l} = \sup_{\substack{x\in\mathbb{R}\\0<t<T}} \lvert x\rvert^k \left\lvert \frac{\partial^l}{\partial x^l}u(x,t) \right\rvert$$
Then $$ \lvert \partial_x^2u(x,t)\bar{u}(x,t)+u(x,t)\partial_x^2\bar{u}(x,t)\rvert \leq 2 a_{0,2}a_{0,0} $$ And $$ \lvert \partial_x^2u(x,t)\bar{u}(x,t)+u(x,t)\partial_x^2\bar{u}(x,t)\rvert \leq 2 a_{2,2}a_{2,0}\lvert x\rvert^{-2} $$ Therefore $$ \lvert \partial_x^2u(x,t)\bar{u}(x,t)+u(x,t)\partial_x^2\bar{u}(x,t)\rvert \leq 2\min( a_{0,2}a_{0,0}, a_{2,2}a_{2,0}\lvert x\rvert^{-2} )$$
So the norm of $\partial_t \lvert u(x,t)\rvert^2$ is dominated by an integrable function, which justifies the differentiation under the integral sign.