Is the integral of square of a function (with parameter) positive? Suppose we have a function $f:\mathbb{R}\times\mathbb{R}_{>0}\mapsto\mathbb{R}$, with $f(x;p)\not\equiv 0$, where $p$ is some parameter. Supposing the integral is finite, I know that
$$\int_{\mathbb{R}}f(x;p)^2dx\equiv f_1(p)>0.$$
Does the inequality persist if I integrate with respect to $p$, i.e. is it true that for $\nu>0$,
$$\int_0^{\nu}f_1(p)dp \equiv g(\nu)> 0\quad ?$$
What about if I integrate again? Is it true that for $y>0$,
$$\int_0^{y}g(\nu)d\nu =\int_0^y\int_0^{\nu}f_1(p)dpd\nu\equiv h(y)> 0\quad ?$$
Is there a theorem about this, or maybe this isn't true at all.

It s definitely true. It is the monotonicity property of integrals. If you have two functions $f$ and $g$ such that $f\ge g$, then $$\int f d\mu\ge\int g d\mu.$$
Moreover, if $f>g$ in a set of positive measure, then the inequality above is also strict. Have a look at this too.
