we are in Hilbert space $L^2 $ and we are given subspace of dimension $2K$ $$ V=Vect\{ g_k,\bar{g_k},1\le k\le K \}$$ $V$ is a sum of $K$ subspaces of dimension 2 $$ W_k=Vect \{g_k,\bar{g_k} \} $$ now we define $$ C(u,V):=\sum{}_{k=1}^{K} \|P_{W_k}u\|^2$$ where $P_{W_k}$ is orthogonal projection on $W_k$
how can i show that $ C(u,V)$ restricted on vectors $u$ from $V$ a positive definite quadratic form asociated with operator $L=\sum_{k}P_{W_k}$ , and it also satisfies $$ A\|u\|^2 \le C(u,V) \le B\|u\|^2 $$ where $A$ and $B$ are max and min eigenvalue of that operator
First show that $C^{\frac{1}{2}}(.,V)$ is a norm:
$C^{\frac{1}{2}}(\lambda u)=|\lambda|C(u,V)$ and
$\{\sum_k ||P_{W_k}(u+v)||^2\}^{\frac{1}{2}}\leq\{\sum_k ||P_{W_k}(u)||^2\}^{\frac{1}{2}}+\{\sum_k ||P_{W_k}(v)||^2\}^{\frac{1}{2}}$ if and only if
$\sum_k Re <P_{W_k}u,P_{W_k}v>\leq\{\sum_k ||P_{W_k}(u)||^2\}^{\frac{1}{2}}\{\sum_k ||P_{W_k}(v)||^2\}^{\frac{1}{2}} $, which follows from Cauchy Schwarz inequality.
Because $\sum_k||P_{W_k}(u\pm v)||^2=\sum_k ||P_{W_k}(u)||^2\pm Re<P_{W_k}u,P_{W_k}v>+||P_{W_k}v||^2$ it follows the parallelogramm identity:
$C(u+v,V)+C(u-v,V)=2(C(u,V)+C(v,V)$.
Hence C(.,V) is a quadratic form and because
$C(u,V)=0$ implies $P_{W_k}=0$ for all $k=1,...,K$ and since $u$ is in the span of these $W_k$ this implies $u=0$ and thus, that $C(.,V)$ is positive definite.
Now let $L=\sum_k P_{W_k}, u=\sum_l\lambda_l g_l+\mu_l\overline{g_l}$. Since the $P_{W_k}$ are orthogonal projections we have $P_{W_k}u=\lambda_kg_k+\mu_k\overline{g_k}$ and
$<Lu,u>=\sum_k <P_{W_k}u,\sum_l \lambda_lg_l+\mu_l\overline{g_l}>=\sum_k<P_{W_k}u,\lambda_kg_k+\mu_k\overline{g_k}>=\sum_k||P_{W_k}u||^2=C(u,V)$
and if $[A,B]$ contains the eigenvalues of $L$ we know by the min max theorem, that
$A\leq\frac{<Lu,u>}{<u,u>}\leq B$.