Let $H$ be a complex Hilbert space
and let $B : H \to H$ be a positive bounded linear operator on $H$ ($B$ is positive means that its corresponding quadratic form $\langle Bx,x\rangle$ is non-negative)
it's clear that $\phi : H \times H \to \mathbb{C}$ defined as follows : $\phi(x,y) = \langle Bx,y\rangle$ is a sesquilinear form therefore by the Cauchy-Schwartz inequality we have that :
$|\langle Bx,y\rangle|^2 \leq \langle Bx,x\rangle \langle By,y\rangle$
the following is an inequality that we're asked to show before the construction of a special sequence that I will talk about soon :
$\forall x \in H \; \|Bx \|^2 \leq \langle Bx,x\rangle \| B\| \; \; (\star)$
this is deducible from the C-S inequality :
$$\begin{align} \| Bx\|^4 = |\langle Bx,Bx\rangle|^2 \leq \langle Bx,x\rangle \langle B^2x,Bx\rangle \leq \langle Bx,x\rangle \|B^2x\| \|Bx \| = \langle Bx,x\rangle \|B(Bx)\| \|Bx \| \implies \| Bx\|^4 \leq \langle Bx,x\rangle \|B\| \|Bx \|^2 \end{align}$$
which proves $(\star)$
I guess that will be helpful in the next part.
now let $C = \| B\| I- B$
$I$ is simply the identity operator.
now I'm told to construct a sequence $(y_n)_{n \geq1} \subset H$ such that for each $n$ we have $\| y_n \| = 1$ and $$\lim_{n \to \infty} \|Cy_n \| = 0 $$
the first issue is if I wanna use $(\star)$ then I first have to show that $C$ is postive and $C$ didn't seem to be positive at first glance
$\langle Cx,x\rangle = \|B\|\| x\|^2 - \langle Bx,x\rangle \geq \|B\|\| x\|^2 - \|Bx \| \|x\| = \|x\|(\|B\|\| x\| - \|Bx \|) \geq 0$
but as you see after writing all the stuff down it turned out that my intuition was wrong and $C$ is in fact positive.
so now I guess what I should do is plug $C$ in $(\star)$
which means that : $\forall x \in H \; \|Cx \|^2 \leq \langle Cx,x\rangle \| C\|$
so now instead of looking for a 'normalized' sequence $(y_n)_{n \geq1} \subset H$ such that $$\lim_{n \to \infty} \|Cy_n \| = 0 $$
it suffices for the sequence to satisfy the seemingly weaker condition :
$$\lim_{n \to \infty} \langle Cy_n,y_n\rangle = \lim_{n \to \infty} \| B \| \|y_n\|^2 - \langle By_n,y_n\rangle = \| B \| - \lim_{n \to \infty} \langle By_n,y_n\rangle = 0 \;\;\; \text{since the sequence is supposed to be 'normalized'}$$
my idea is the following : since $\|B \| = \sup_{\|x\| = \|y \| = 1} | \langle Bx,y\rangle|$ why not choose the $y_n$ to be a sequence which remains unchanged as $n$ increases where that only repeating term is the $z$ where $sup _{\|z\| = 1}| \langle Bz,z\rangle|$ is attained. but I have no idea how to translate that into math language nor I'm certain that what I'm saying makes sense.
any help or hints would be much appreciated. thanks!
All you need is the well-known equality $$\tag1 \|B\|=\sup\{\langle Bx,x\rangle:\ \|x\|=1\}. $$ (By including an absolute value, the equality also holds for $B$ selfadjoint).
Proof: $$ \sup\{|\langle Bx,x\rangle:\ \|x\|=1\}=\sup\{\langle B^{1/2}x,B^{1/2}x\rangle:\ \|x\|=1\}=\|B^{1/2}\|^2=\|B\|. $$
Edit: How to use $(1)$. By $(1)$, there exists a sequence $\{x_n\}$ with $\|x_n\|=1$ and $\langle Bx_n,x_n\rangle\to \|B\|$. Then \begin{align} \|Cx_n\|^2&=\|\,\|B\|x_n-Bx_n\|^2 =\langle \|B\|x_n-Bx_n,\|B\|x_n-Bx_n\rangle\\ \ \\ &=\|B\|^2+\|Bx_n\|^2-2\text{Re}\,\|B\|\,\langle x_n,Bx_n\rangle\\ \ \\ &\leq 2\|B\|^2-2\text{Re}\,\|B\|\,\langle x_n,Bx_n\rangle. \end{align} Now, since $\langle Bx_n,x_n\rangle\to\|B\|$, we get $$ \limsup_n\|Cx_n\|^2\leq2\|B\|^2-2\|B\|^2=0. $$ So $\|Cx_n\|\to0$.