# For bounded operators $A,B$ on a Hilbert space prove that $A^*A \leq cB^*B \Leftrightarrow \lVert A \rVert \leq \sqrt{c}\lVert B \rVert$.

For bounded operators $$A,B$$ on a Hilbert space $$\mathbb{H}$$ prove that for any $$c \geq 0$$, $$A^*A \leq cB^*B \Leftrightarrow \lVert A \rVert \leq \sqrt{c}\lVert B \rVert$$.

This is an exercise from Wolfgang Sherer's book, Mathematics of Quantum Computing. Here $$A^*$$ denotes the adjoint operator of $$A$$, $$\lVert A \rVert$$ denotes the operator norm and for operators $$A, B$$ the statement $$A \leq B$$ is defined as $$B-A \geq 0$$ (i.e for all $$|\psi\rangle \in \mathbb{H}$$, $$\langle \psi | (B-A)\psi\rangle \geq 0$$).

My work so far: Using the definitions $$A^*A \leq cB^*B$$ is equivalent to $$\langle \psi | (cB^*B-A^*A)\psi\rangle \geq 0$$ for all $$|\psi\rangle \in \mathbb{H}$$. Expanding the inner product this is equivalent to $$\lVert A\psi\rVert \leq \sqrt{c}\lVert B\psi\rVert$$. Now taking supremums over all $$|\psi\rangle$$ with $$\lVert \psi \rVert = 1$$ we get the $$\Rightarrow$$ direction.

My question: How can I proceed with the $$\Leftarrow$$ direction? In the solutions on the back of the book it states that $$\lVert A\psi\rVert \leq \sqrt{c}\lVert B\psi\rVert \Leftrightarrow \lVert A \rVert \leq \sqrt{c}\lVert B \rVert$$ citing equation $$(2.57)$$ which is found later in the book and is just the definition of trace of an operator. I don't think that this is correct though.

Here is an explicit example. Consider $$M_2(\mathbb{C}) = B(\mathbb{C}^2)$$ and $$X= \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}, \quad Y = \begin{pmatrix}1 &0 \\ 0 & 0\end{pmatrix}$$
Then $$X^*X = X$$ and $$Y^*Y = Y.$$ We have $$\|X\| = \|Y\| = 1$$ so $$\|X\| \le \|Y\|.$$ However, $$Y-X = \begin{pmatrix}0 & 0 \\ 0 & -1\end{pmatrix}$$ is not a positive matrix, so $$X \not\le Y$$.
More generally, let $$H$$ be any Hilbert space and $$X,Y$$ be distinct orthogonal projections in $$B(H)$$. Then $$\|X\| = \|Y\| = 1$$ and $$X \not\leq Y$$ or $$Y \not\le X$$ is true, so this yields large families of counterexamples.