The initial value problem $$ \left\{ \begin{array}{lcl} \partial_{t}u(x,t) & = & \triangle u(x,t) + f(x,t), \ \ x \in \mathbb{R}^{n}, \ t > 0; \\ u(x,0) & = & u_{0}(x), \ \ x \in \mathbb{R}^{n}, \end{array} \right. $$ for $u_{0}$ and $f$ in some function classes.

We know that to solve such this problem above, we must divide it into two problems and one of them is the non-homogeneous problem with null initial data $$ (NH)\left\{ \begin{array}{lcl} \partial_{t}u(x,t) & = & \triangle u(x,t) + f(x,t), \ \ x \in \mathbb{R}^{n}, \ t > 0; \\ u(x,0) & = & 0, \ \ x \in \mathbb{R}^{n}, \end{array} \right. $$ I'm finding the following candidate solution the problema (NH): $$ u(x,t) = \int_{0}^{t}\int_{\mathbb{R}^{n}}K(x-y,t-s)f(y,s)dyds. $$ where $$ K(x,t) = (4\pi t)^{-n/2}e^{-\frac{|x|^{2}}{4t}} \ \ \text{and} \ \ \partial_{t}K(x,t) = \triangle K(x,t) \ \ \text{for} \ \ (x,t) \in \mathbb{R}^{n}\times \mathbb{R}^{+}. $$

**My problem:**

Considering functions $f(x,t)$ in several spaces, for example, $f \in C(\mathbb{R}_{0}^{+};L^{1}(\mathbb{R}^{n}))$ or some other space of functions (they may impose extra conditions on $f$ as they see fit). I want to show that $u$ given above is in fact a solution of (NH). **I'm having trouble justifying passing the partial derivatives with respect to $t$ into the integral.**