Numerical radius of a pair of operators in Hilbert spaces Let $(C,D)$ be a pair of bounded linear operators on a complex Hilbert space $E$. The Euclidean operator radius is defined by
$$w_e(C,D)=\displaystyle\sup_{\|x\|=1}\left(|\langle Cx,x \rangle|^2+|\langle Dx,x \rangle|^2\right)^{1/2}.$$
Moreover, the following inequality holds:
$$\frac{\sqrt{2}}{4}\|C^*C+D^*D\|^{1/2}\leq w_e(C,D)\leq \|C^*C+D^*D\|^{1/2}.$$
I want to show that the constants $\frac{\sqrt{2}}{4}$ and $1$ in the above inequalities are the best possible.
For the second inequality, the following example show that we have equality:
Let $(C,D)=(B,B)$, with $B=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ (operator on $(\mathbb{C}^2,\|\cdot\|)$). Hence, I get
$w_e(C,D)=\sqrt{2}$ and $\|C^*C+D^*D\|^{1/2}=\sqrt{2}.$

I want to find $(C,D)$ such that
  $$\frac{\sqrt{2}}{4}\|C^*C+D^*D\|^{1/2}= w_e(C,D).$$

For a single operator, we have the following theorem:


Do you think that if $\text{Im}(C)\perp \text{Im}(C^*)$ and $\text{Im}(D)\perp \text{Im}(D^*)$ we have
  $$\frac{\sqrt{2}}{4}\|C^*C+D^*D\|^{1/2}= w_e(C,D)\,?$$

Thank you in advance.
 A: For your first question, consider
$$ C =  D = \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}.$$
Then the LHS of the first inequality reads
$$ \frac{\sqrt{2}}{4}\| 2\begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}\| = \frac{\sqrt{2}}{2}. $$
For the RHS it is
$$ w_e(C,D) = \sqrt{2} \sup_{\|x\|=1} |\langle \begin{pmatrix}0 \\ x_1 \end{pmatrix}, \begin{pmatrix} x_1\\x_2 \end{pmatrix}\rangle| = \sqrt{2} \sup_{\|x\|=1} |x_1x_2| = \frac{\sqrt{2}}{2}.$$
For your second question: Consider
$$ C = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, D = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},$$
then $Im(C) \perp Im(C^\ast)$ and $Im(D) \perp Im(D^\ast)$ but
$$ \frac{\sqrt{2}}{4} \| CC^\ast  + DD^\ast\|^{1/2} = \frac{\sqrt{2}}{4}, \quad w_e(C,D)) = \frac{1}{\sqrt{2}}.$$
A: In view of (arXiv):
Let $(C,D)=(\frac{1}{\sqrt{2}}A,\frac{1}{\sqrt{2}}A)$, with $A=\begin{pmatrix}0&0\\1&0\end{pmatrix}$. Hence,
$$w_e(C,D)=w(A)=\frac{1}{2},$$
but if I don't make wrong in calculus I get
$$\frac{\sqrt{2}}{4}\|C^*C+D^*D\|^{1/2}=\frac{\sqrt{2}}{4}.$$
