Why $E(t)=\int_{\mathbb{R}}|u(x,t)|^2\,dx$ can be dfferentiated under integral sign?

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?