In the Wikipedia article on scalar curvature, it is noticed that the Hölder inequality implies $$Y(g) \geq - \left(\int_M |R(g)|^{n/2} \mathrm{d}V_g\right)^{n/2}$$ for the Yamabe functional $Y(g)$ and the scalar curvature $R(g)$. It is now concluded that this implies that the scalar curvature is bounded from below.
How can one conclude that the $L^{n/2}$ metric of the scalar curvature is bounded on a conformal class? Is this somehow obvious?
It's not obvious, but it follows by a straightforward argument using formula (1.1) in the Bulletin survey article by Lee and Parker. The case $n = 2$ is trivial, so assume $n > 2$. If a metric $\bar{g}$ is conformal to the metric $g$, then there exists a positive function $u$ such that $$ \bar{g} = u^{\frac{2}{n-2}}g $$ A straightforward calculation using (1.1) from Lee-Parker shows that $$ \int S(\bar{g})\,dV(\bar{g}) \ge \int uS(g)\,dV(g), $$ where $S(g)$ is the scalar curvature of $g$. The inequality now follows by substituting this into the definition of the Yamabe functional of $\bar{g}$ and applying Holder's inequality.
The infimum is taken when varying f, g is fixed. So $Y(g)=\inf_f \mathcal E(e^{2f}g)\geq -(\int|R_g|^\frac{n}{2}dV_g)^\frac{2}{n}$ is also finite.
