I would like to proof that by using $\lVert u_0 \rVert_{L^2} \leq C \lVert \nabla u_0\rVert_{\vec{L^{2}}}$ that $$ \left| \int_{M} k u_0 dV \right| \leq C_3 \varepsilon \lVert \nabla u_0 \rVert_{\vec{L^{2}}}^{2} + \dfrac{C_4}{\varepsilon} $$ with $u_0 \in H^{1}(M) $ (1-sobolev), and $C_3$ and $C_4$ independent of $\varepsilon$. Here $(M,g)$ is a compact riemannian 2-manifold without boundary and $k$ is the Gauss curvature with respect $g$. This statement can be find in page 123 in this book (or p.105 in the first edition).
I guess I should use Holder- Inequality, i.e, $$ \left| \int_{M} k u_0 dV \right| \leq \left(\int_{M} k^{2} dV\right)^{\frac{1}{2}} \left(\int_{M}u_0^{2} dV\right)^{\frac{1}{2}} \leq \left(\int_{M} k^{2} dV\right)^{\frac{1}{2}}C \left(\int_{M} |\nabla u_0|_g^{2} dV\right)^{\frac{1}{2}} $$, and in this point I don't know what I can do with the Gauss Curvature.
Thanks for your help.
It's the simple use of Cauchy Schwarz:
$$ ab = (\sqrt\delta a)\left( \frac{1}{\sqrt\delta} b\right) \le \frac 12 \left(\delta a^2 + \frac{b^2}{\delta}\right)$$
with
$$b= \left(\int_{M} k^{2} dV\right)^{\frac{1}{2}},\ \ \ a= \left(\int_{M} |\nabla u_0|_g^{2} dV\right)^{\frac{1}{2}} $$
and $\epsilon = C\delta$ for some suitable $C$. Note $k$ is fixed so $b$ is indeed part of the constant $C_4$.