## 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 $$\displaystyle \sup_{\substack{x\in\mathbb{R}\0 for each $$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?