First we define Rayleigh quotient as $$R(u)=\frac{\int_M|\nabla u|^2 dV_g}{\int_M |u|^2d V_g}$$ We can deduce that the first nonzero eigenvalue of Laplacian operator is $$\lambda_1=\inf\{R(u): \int_M u dV_g=0\}$$ Later I found that some people write it as $$\lambda_1=\inf\{\sup\{R(u):u\in L\}:\dim L=2\}$$ Where $L$ is a subspace of $C^{\infty}(M)$. Here $M$ is a riemannian manifold, but I think it doesn't really matter in my question, or just think it as a compact domain in $R^n$.
My question: I guess they are equivalent, but it is not that obvious for me, can any one show some ideas how to prove the equivalence between these two definitions or show me some references about it?
Remark: After thinking, I have some ideas: (1) if we can show that every 2-dimensional subspace $L_2$ of $C^{\infty}(M)$ contains a function $u$ with property $\int_M udV_g=0$, then in every $L_2$ subspace we have $$\sup\{R(u):u\in L_2\}\geq R(u).$$ Naturally we can get "no less" inequality.
(2) For any $u$ with property $\int_M udV_g=0$, we can construct a $L_2$ subspace as $L_2=\{au+b\}$, then for any $v\in L_2$ $$R(v)=R(au+b)=\frac{\int_M a^2|\nabla u|^2dV_g}{\int_M a^2u^2+b^2 dV_g}=\frac{\int_M |\nabla u|^2dV_g}{\int_M u^2dV_g+b^2/a^2V(g) }\leq R(u).$$
From here can we get "no bigger" inequality?