Let $u$ be a solution of $\partial_t u - \Delta u =f$ with initial data $u(0,x) = 0$ on $\mathbb R^N$. How do one prove the following inequality?

$$ \int_0^T \int_{\mathbb R^N} \phi(f) \Delta u(s,x)dxds \ge 0, $$ where $\phi$ is a smooth non-decreasing function.

How can we estimate $$\int_0^T \int_{\mathbb R^N} \phi(f) \Delta u(s,x)dxds$$ from below more precisely?

Let $u$ be a solution of $\partial_t u - \Delta u =f$ with initial data $u(0,x) = 0$ on $\mathbb R^N$. How do one prove the following inequality?

$$ \int_0^T \int_{\mathbb R^N} \phi(f)(- \Delta) u(s,x)dxds \ge 0, $$ where $\phi$ is a smooth non-decreasing function.

How can we estimate $$\int_0^T \int_{\mathbb R^N} \phi(f) (-\Delta )u(s,x)dxds$$ from below more precisely?

Let $u$ be a solution of $\partial_t u - \Delta u =f$ with initialdata $u(0,x) = 0$ on $\mathbb R^N$. How do you prove the following inequality?

$$ \int_0^T \int_{\mathbb R^N} \phi(f) \Delta u(s,x)dxds \ge 0, $$ where $\phi$ is a smooth non-decreasing function.

How can we estimate $$\int_0^T \int_{\mathbb R^N} \phi(f) \Delta u(s,x)dxds$$ from below more precisely?

Let $u$ be a solution of $\partial_t u - \Delta u =f$ with initial data $u(0,x) = 0$ on $\mathbb R^N$. How do one prove the following inequality?

$$ \int_0^T \int_{\mathbb R^N} \phi(f) \Delta u(s,x)dxds \ge 0, $$ where $\phi$ is a smooth non-decreasing function.

How can we estimate $$\int_0^T \int_{\mathbb R^N} \phi(f) \Delta u(s,x)dxds$$ from below more precisely?