We are in Hilbert space $L^2$ we are given a subspace of dimension K as $$ V=\{ g_k,1 \le k \le K \} $$ everything that folows is defined on $V$
we define map $$ x \mapsto Q(x):= \sum_{k=1}^{K} |\langle x,g_k \rangle|^2$$ how can i show that this map defines positive definite quadratic form ?
We are also given that eigenvalues of Gram matrix $G=(\langle g_k,g_l \rangle)_{k,l} $ of family $(g_k)_{k=1}^{K}$ ( its linear independent family) lie in interval $[A,B]$ ($A>0$)
how can i show that
$$ A\|x \|^2 \le Q(x) \le B \| x \|^2$$
First check, that $Q^{\frac{1}{2}}$ is a norm. The triangle inequality needs some calculation (... $Q(x+y)=\sum_{k=1}^K |<x,g_k>|^2+2Re<x,g_k>\overline {<y,g_k>} +|<y,g_k>|^2$ and $(\sum_{k=1}^K Re<x,g_k>\overline {<y,g_k>})^2\leq\sum_{k=1}^K|<x,g_k>|^2\sum_{k=1}^K|<y,g_k>|^2$, which follows from Cauchy Schwartz). The same calculation yields $Q(x+y)+Q(x-y)=2(Q(x)+Q(y))$, the parallelogram identity. Thus $Q$ is a quadratic form. For every $x\in V$ there are unique $x_1,...,x_K\in\mathbb{C}$ such that $x=\sum_{l=1}^Kx_lg_l$ . Then $Q(x)=\sum_{k,l,m=1}^Kx_l\overline{x_m}<g_l,g_k>\overline{<g_m,g_k>}$ and if $G$ is the operator on $V$, that belongs to the Gram matrix ,then $Gx=\sum_{k,l=1}^Kx_l<g_k,g_l>g_k$ and $<Gx,x>=\sum_{k,l,m=1}^K x_l\overline{x_m}<g_k,g_l><g_k,g_m>$ For the Rayleigh quotient $\frac{|<Gx,x>|}{<x,x>} $it is known by the min max theorem, that $A\leq\frac{|<Gx,x>|}{<x,x>}\leq B$.