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Let $E$ be a complex Hilbert space and $K_1,K_2$ are bounded linear operators on $E$.

Let $\omega(K_1)$ and $\omega(K_2)$ be the numerical radius of $K_1$ and $K_2$ respectively. That is \begin{align*} \omega(K_i) = \sup\Big\{|{\langle K_ix, x\rangle}|:\, \, x\in E, {\|x\|}= 1\Big\},\quad i=1,2. \end{align*}

Let $$\|(K_1,K_2)\|:=\sup\Big\{|{\langle K_1x, y\rangle}|^2+|{\langle K_2x, y\rangle}|^2:\, \, x,y\in E, {\|x\|}={\|y\|}= 1\Big\}.$$

I want to find prove whether or not the following inequality $$\omega(K_1)^2+\omega(K_2)^2\leq \|(K_1,K_2)\|,$$ holds.

It's clear that \begin{align*} \|(K_1,K_2)\| & \geq |{\langle K_1x, y\rangle}|^2+|{\langle K_2x, y\rangle}|^2, \end{align*} for all $x,y\in E, {\|x\|}={\|y\|}= 1$. In particular

\begin{align*} \|(K_1,K_2)\| & \geq |{\langle K_1x, x\rangle}|^2+|{\langle K_2x, x\rangle}|^2, \end{align*} for all $x\in E, {\|x\|}= 1$.

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$\newcommand\C{\Bbb C}\newcommand\om{\omega}$A counterexample is given by $E=\mathbb C^2$ and $K_1(x_1,x_2)=(x_1,0)$ and $K_2(x_1,x_2)=(0,x_2)$ for complex $x_1,x_2$.

Indeed, then $\om(K_1)=\om(K_2)=1=\|(K_1,K_2)\|$.

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  • $\begingroup$ Please why $1=\|(K_1,K_2)\|$? I think it's $2$. Thanks $\endgroup$ Commented Apr 19, 2024 at 12:14
  • $\begingroup$ @Student : Because (for these $K_1$ and $K_2$) $\|(K_1,K_2)\|$ is the supremum of $|x_1|^2|y_1|^2+|x_2|^2|y_2|^2$ given $|x_1|^2+|x_2|^2=1$ and $|y_1|^2+|y_2|^2=1$, so that $|x_1|^2|y_1|^2+|x_2|^2|y_2|^2\le|x_1|^2+|x_2|^2=1$. $\endgroup$ Commented Apr 19, 2024 at 13:13

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