# Calculate the following seminorm of $A$

Consider the following operators defined on $(\mathbb{C}^2,\|\cdot\|_2)$: $$S = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, \qquad A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}.$$

I want to calculate $\|A\|_S$, with $$\|A\|_S:=\inf\left\{c\geq 0; \quad\sqrt{\langle SAx,Ax\rangle} \leq c \sqrt{\langle Sx,x\rangle},\;\forall x \in \overline{\text{Im}(S)}\right\}.$$

My attempt: Note that $\overline{\text{Im}(S)}= \text{Im}(S)= \left\{ \begin{pmatrix} x \\ x \end{pmatrix} \ : \ x\in \mathbb{C} \right\}$. We compute

$$\left\langle S A\begin{pmatrix} x \\ x \end{pmatrix}, A \begin{pmatrix} x \\ x \end{pmatrix} \right\rangle = \left\langle S \begin{pmatrix} x \\ 0 \end{pmatrix}, \begin{pmatrix} x \\ 0 \end{pmatrix} \right\rangle = \left\langle \begin{pmatrix} x \\ x \end{pmatrix}, \begin{pmatrix} x \\ 0 \end{pmatrix} \right\rangle = \vert x \vert^2 \leq 4 \vert x \vert^2$$ $$= \left\langle \begin{pmatrix} 2x \\ 2x \end{pmatrix}, \begin{pmatrix} x \\ x \end{pmatrix} \right\rangle = \left\langle S \begin{pmatrix} x \\ x \end{pmatrix}, \begin{pmatrix} x \\ x \end{pmatrix} \right\rangle.$$ Hence $$\|A\|_S\leq 1.$$

• You basically did all the work. You just showed that $\forall x \in \mathbb{C}, \left < SA \left ( \begin{matrix} x \\ x \end{matrix} \right ) \middle | A \left ( \begin{matrix} x \\ x \end{matrix} \right ) \right > = |x|^2$ and $\left < S \left ( \begin{matrix} x \\ x \end{matrix} \right ) \middle | \left ( \begin{matrix} x \\ x \end{matrix} \right ) \right > = 4|x|^2$. Now you have to find the smallest $c \in \mathbb{R_+^*}$ so that $|x| \leq 2c|x|$, which is just $\frac{1}{2}$. Mar 24, 2018 at 10:09
• @Matrefeytontias I hope that you write your comment as an answer Mar 25, 2018 at 9:41
• There you go, I did Mar 25, 2018 at 9:59

You basically did all the work. You just showed that $\forall x \in \mathbb{C}, \left < SA \left ( \begin{matrix} x \\ x \end{matrix} \right ) \middle | A \left ( \begin{matrix} x \\ x \end{matrix} \right ) \right > = |x|^2$ and $\left < S \left ( \begin{matrix} x \\ x \end{matrix} \right ) \middle | \left ( \begin{matrix} x \\ x \end{matrix} \right ) \right > = 4|x|^2$. Now you have to find the smallest $c \in \mathbb{R_+^*}$ so that $|x| \leq 2c|x|$, which is just $\frac{1}{2}$.
• Thank you for your answer. By definition $c\in \mathbb{R_+}$ Mar 25, 2018 at 10:14
• In practice $c$ can never be 0 except if the LHS is zero, since a square root can never give a negative result. Mar 25, 2018 at 10:36