$\textbf{Question}$
Let $(\mathcal{H}, \langle \cdot \, | \, \cdot \rangle)$ be a real Hilbert space with induced norm $\|\cdot\| = \sqrt{\langle \cdot \, | \, \cdot \rangle}$ and let $$\mathscr{B}(\mathcal{H},\mathcal{H}) \equiv \mathscr{B}(\mathcal{H}) = \{ T : \mathcal{H} \rightarrow \mathcal{H} \, | \, T ~\textrm{is linear and continuous} \}.$$ Additionally, let $T \in \mathscr{B}(\mathcal{H})$ be 1-Lipschitzian, i.e., $$(\forall x \in \mathcal{H})(\forall y \in \mathcal{H}) ~~ \| T x - T y \| \leq \| x - y \|.$$ Show that $\textrm{Fix} \, T = \textrm{Fix} \, T^{*},$ where $T^{*}$ is the adjoint of $T$ (i.e., $T^{*} \in \mathscr{B}(\mathcal{H})$ is the unique operator that satisfies $\langle T x \, | \, y \rangle = \langle x \, | \, T^{*} y \rangle$ for every $x$ and $y$ in $\mathcal{H}$) and $\textrm{Fix} \, T = \{ x \in \mathcal{H} \, | \, T x = x \}.$
$\textit{Remark}$: $0_{\mathcal{H}}$ designates the zero element in $\mathcal{H}.$
$\textbf{Incomplete Solution Attempt}$
Since $T$ is 1-Lipschitzian, one can show (which I have already done) that $\| T \| \leq 1$ (in fact, this is an equivalent characterization in this particular context). Now, take $x \in \textrm{Fix} \, T.$ Then $$ \| x \| = \|T x\| \leq \| T \| \| x \| \leq \| x \|. $$ Hence, $\| T \| \| x \| = \| x \|,$ or, equivalently, $(\| T \| - 1) \| x \| = 0.$ Therefore, either $\| T \| = 1$ or $\| x \| = 0.$ If $\| x \| = 0,$ then $x = 0_{\mathcal{H}}$ and consequently, $\textrm{Fix} \, T = \{ 0_{\mathcal{H}} \}$ (this follows from the linearity of $T;$ indeed, $(\forall z \in \mathcal{H}) ~~ T 0_{\mathcal{H}} = T (0 z) = 0 T z = 0_{\mathcal{H}}$). Thus, $x \in \textrm{Fix} \, T \Rightarrow \| T \| = 1$ or $\textrm{Fix} \, T = \{ 0_{\mathcal{H}} \}.$
Similarly, since $\| T \| = \| T^{*} \|,$ $x \in \textrm{Fix} \, T^{*} \Rightarrow \| T^{*} \| = 1$ or $\textrm{Fix} \, T^{*} = \{ 0_{\mathcal{H}} \}.$
$\textbf{Issue(s)}$
By the above logic, I have only proved the following: either $\textrm{Fix} \, T = \textrm{Fix} \, T^{*} = \{ 0_{\mathcal{H}} \}$ or $\| T \| = \| T^{*} \| = 1.$ I am having trouble resolving the latter consequence ($\| T \| = \| T^{*} \| = 1$), as the former is supposed to be the only true possibility. Any help is appreciated.
The assumption $\|Tx-Ty\|\le \|x-y\|$ is equivalent to $\|T\|\le 1.$ Thus $\|T^*\|\le 1.$ Let $Tv=v,$ for $v\neq 0.$ We obtain $$\|T^*v-v\|^2=\|T^*v\|^2+\|v\|^2-\langle T^*v,v\rangle-\langle v,T^*v\rangle\\ \le 2\|v\|^2-\langle v,Tv\rangle -\langle Tv,v\rangle =0$$ Hence $T^*v=v.$ By switching the roles of $T$ and $T^*$ we get that $T^*v=v$ implies $Tv=v.$ Therefore $\ker(I-T)=\ker (I-T^*).$