Suppose that $T:\,K \rightarrow K$ is a strictly pseudo-contraction where $K$ is a bounded subset of a Hilbert space, then $T$ is bounded.
The operator $T,$ is said to be a strict pseudo-contraction if there exists $k<1$ such that $$\|Tx-Ty\|^{2}\leq \|x-y\|^{2} +k\|(I-T)x-(I-T)y\|^{2}, \;\forall \;x,y\in H.$$
My trial
For each $n\in \mathbb{N}$, let $x_n\in K.$ Then, $\{x_n\}$ is bounded. Thus,
\begin{align} \|Tx_n\|^{2}&=\|Tx_n-T0+T0\|^{2}\\&\leq \|Tx_n-T0\|^{2}+2\|Tx_n-T0\|\|T0\|+\|T0\|^{2} \\&\leq 2\|Tx_n-T0\|^{2}+2\|T0\|^{2} \\&\leq 2\left[ \|x_n-0\|^{2} +k\|x_n-Tx_n-T0\|^{2}\right] +2\|T0\|^{2} \\&= 2\|x_n\|^{2}+2\|T0\|^{2} +2k\|x_n-Tx_n-T0\|^{2} \end{align}
I'm stuck here and don't know what to do. Any hints on moving forward? It might even be that my proposition is false. Counter-examples are also welcome.