The resolvent of a Toeplitz operator may have arbitrary growth

Fix an arbitrary sequence $\{\lambda_n\}$ in the unit disc such that $\lim_n\lambda_n=1$ and any sequence $\{A_n\}$ of positive reals. Then there exists a continuous real $u$ on the unit circle such that the Toeplitz operator $T_\varphi$ (on the Hardy class $H^2$) with symbol $\varphi=e^{iu}$ satisfies
$$ \|(T_\varphi-\lambda_nI)^{-1}\|>A_n $$