# Showing that $W_0(\alpha T)=\alpha W_0( T)$.

Let $E$ be a Hilbert space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ the algebra of all bounded linear operators from $E$ to $E$. For $T\in\mathcal{L}(E)$, we consider the following set $$W_0(T)=\{\lambda\in \mathbb{C}:\;\langle T x_n\; |\;x_n\rangle\rightarrow \lambda\;\hbox{where}\;\|x_n\|=1\;\hbox{and}\;\|Tx_n\|\rightarrow \|T\|\}.$$ I want to show that $$W_0(\alpha T)=\alpha W_0( T), \;\forall \alpha\in \mathbb{C}.$$ Thank you.

The case $\alpha=0$ is trivial.
When $\alpha\ne0$, if $\langle Tx_n|x_n\rangle\to\lambda$, then $$\langle \alpha Tx_n|x_n\rangle=\alpha\langle Tx_n|x_n\rangle\to\alpha\lambda.$$ And if $\|Tx_n\|\to\|T\|$, then $$\|\alpha Tx_n\|=\alpha\|Tx_n\|\to\alpha\|T\|.$$ Since $\alpha$ is nonzero, the above steps can be reversed (by working with $1/\alpha$, so $W_0(\alpha T)=\alpha W_0(T)$.