In the [Wikipedia article on scalar curvature](https://en.wikipedia.org/wiki/Yamabe_invariant), 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?



$L^{n/2}$ norm of scalar curvature
It's not obvious, but it follows by a straightforward argument using formula (1.1) in the [Bulletin survey article by Lee and Parker](https://projecteuclid.org/euclid.bams/1183553962). 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.


