If \(X\) and \(Y\) are independent real random variables with densities \(f\) and \(g\) then \(X+Y\) has density \(f*g\). This density is bounded as soon as \(f\) or \(g\) is bounded since \[\left\Vert f*g\right\Vert_\infty\leq \min(\left\Vert f\right\Vert_\infty,\left\Vert g\right\Vert_\infty).\] One may ask if \(XY\) has similarly a bounded density. The answer is unfortunately negative in general. To see it, we note first that when \(X\) and \(Y\) are non negative then \(XY\) has density \[t\in\mathbb{R}_+\mapsto \int_0^\infty\!\frac{f(x)}{x}g\left(\frac{t}{x}\right) dx.\] Now if for instance \(X\) and \(Y\) are uniform on \([0,1]\) then \(XY\) has density \[t\mapsto -\log(t)\mathbf{1}_{[0,1]}(t)\] which is unbounded… If \(X\) is non negative with density \(f\) then \(X^2\) has density \[t\in\mathbb{R}_+\mapsto \frac{f(\sqrt{t})}{2\sqrt{t}}.\] For instance when \(X\) is uniform then \(X^2\) has (unbounded) density \[t\mapsto \frac{1}{2\sqrt{t}}\mathbf{1}_{[0,1]}(t).\] The density of \(X^2\) is bounded if \(f\) is bounded and \(f(t)=O(t)\) as \(t\to0\) (imposes \(f(0)=0\)).