An attempt at showing that function is lipschitzian

Question

Let $X$ be normed space and let $B$ denote the closed unit ball in $X$. Suppose we have map $T_0: B \rightarrow B$ which satisfies Lipschitz condition with some constant $k > 1$. We extend this map to $T_1: 2B \rightarrow B$ in the following way: $$ T_1x = \begin{cases} T_0x, & \text{for } \|x\| \leq 1, \\ (2-\|x\|)T_0(Px), & \text{for } 1 < \|x\| \leq 2, \end{cases}$$ where $Px = \frac{x}{\|x\|}$ for $\|x \| > 1$ and $Px = x$ for $\|x\| \leq 1$ ($P$ is radial projection onto unit ball). Im reading a paper where the author claims this extended map still satisfies Lipschitz condition, this time with constant $2k + 1$. I tried to verify this.

$\textbf{Case 1}:$ If $x,y \in B$, then $\|T_1x - T_1y\| = \|T_0x - T_0y\| \leq k\|x - y\|$.

$\textbf{Case 2}:$ Let $x \in B$ and $y \in 2B \setminus B$. Then \begin{equation} \begin{split} \|T_1x - T_1y\| & = \|T_0x - (2-\|y\|)T_0(Py)\| \\ &\leq \Bigl\|T_0x - T_0(Py) - T_0(Py) + \|y\|T_0(Py) \Bigr\| \\ &\leq \|T_0(Px) - T_0(Py)\| + (\|y\| - 1)\|T_0(Py)\| \\ &\leq k\|Px - Py\| + \|y\| - \|x\| \\ &\leq 2k\|x - y\| + \|x - y\| \\ & = (2k + 1)\|x - y\|. \end{split} \end{equation} Here I used the fact that in every normed space projection $P$ is Lipschitz with constant $2$.

$\textbf{Case 3}:$ Now this is where I can't finish the job. Let $x,y \in 2B \setminus B$. I only managed to show that this map is Lipschitz with constant $8k + 1$. \begin{equation} \begin{split} \|T_1x - T_1y\| & = \Bigl\|(2-\|x\|)T_0(Px) - (2-\|y\|)T_0(Py) \Bigr\| \\ & = \Bigl\|2T_0(Px) - 2T_0(Py) - \|x\|T_0(Px) + \|y\|T_0(Py) \Bigr\| \\ & \leq 2\|T_0(Px) - T_0(Py)\| + \Bigl\| \|x\|T_0(Px) - \|y\|T_0(Px) \Bigr\| + \Bigl \|\|y\|T_0(P_x) - \|y\|T_0(Py) \Bigr\| \\ & \leq 4k\|x - y\| + \Bigl | \|x\| - \|y\| \Bigr | \|T_0(Px)\| + \|y\|\|T_0(Px) - T_0(Py)\| \\ & \leq 4k\|x - y\| + \|x-y\| + 4k\|x-y\| \\ & = (8k + 1)\|x - y\|. \end{split} \end{equation}

This is all I can get at the moment. Any help will be appreciated.

$\textbf{Edit}:$ There is extra assumption on map $T_0$ which I didnt mention, because I thought its irrelevant in evaluating Lipschitz constant. Namely, we assume $\inf\{\|x - T_0x\|: x \in B\} > 0$ (such maps exist in infinite dimensional spaces).

Answer 1

Suppose, without loss of generality, that $\left\Vert x\right\Vert >\left\Vert y\right\Vert$.

Then we have $$ \begin{align} \left\Vert T_{1}\left(x\right)-T_{1}\left(y\right)\right\Vert & = \left\Vert \left(2-\left\Vert x\right\Vert \right)T_{0}\left(Px\right)-\left(2-\left\Vert y\right\Vert \right)T_{0}\left(Py\right)\right\Vert \\ & = \left\Vert \left(2-\left\Vert x\right\Vert \right)T_{0}\left(Px\right)-\left(2-\left\Vert x\right\Vert +\left\Vert x\right\Vert -\left\Vert y\right\Vert \right)T_{0}\left(Py\right)\right\Vert \\ & = \left\Vert \left(2-\left\Vert x\right\Vert \right)T_{0}\left(Px\right)-\left(2-\left\Vert x\right\Vert \right)T_{0}\left(Py\right)-\left(\left\Vert x\right\Vert -\left\Vert y\right\Vert \right)T_{0}\left(Py\right)\right\Vert \\ & \leq \left\Vert \left(2-\left\Vert x\right\Vert \right)T_{0}\left(Px\right)-\left(2-\left\Vert x\right\Vert \right)T_{0}\left(Py\right)\right\Vert +\left\Vert \left(\left\Vert x\right\Vert -\left\Vert y\right\Vert \right)T_{0}\left(Py\right)\right\Vert \\ & = \left(2-\left\Vert x\right\Vert \right)\left\Vert T_{0}\left(Px\right)-T_{0}\left(Py\right)\right\Vert +\left(\left\Vert x\right\Vert -\left\Vert y\right\Vert \right)\left\Vert T_{0}\left(Py\right)\right\Vert \end{align} $$ for $x,y \in 2B/B$.

Since $\left\Vert x\right\Vert >\left\Vert y\right\Vert$ we have $\left\Vert x\right\Vert -\left\Vert y\right\Vert =\left|\left\Vert x\right\Vert -\left\Vert y\right\Vert \right|\leq\left\Vert x-y\right\Vert $ , since $ 1 < \left\Vert x\right\Vert $ we have $ 2-\left\Vert x\right\Vert < 1 $ and since $ T_{0}\left(Py\right)\in B $ we have $ \left\Vert T_{0}\left(Py\right)\right\Vert \leq 1 $.

Therefore $$ \begin{align} \left(2-\left\Vert x\right\Vert \right)\left\Vert T_{0}\left(Px\right)-T_{0}\left(Py\right)\right\Vert +\left(\left\Vert x\right\Vert -\left\Vert y\right\Vert \right)\left\Vert T_{0}\left(Py\right)\right\Vert & < \left\Vert T_{0}\left(Px\right)-T_{0}\left(Py\right)\right\Vert +\left\Vert x-y\right\Vert \\ & \leq k\left\Vert Px-Py\right\Vert +\left\Vert x-y\right\Vert \\ & \leq 2k\left\Vert x-y\right\Vert +\left\Vert x-y\right\Vert \\ & = \left(2k+1\right)\left\Vert x-y\right\Vert \end{align} $$ as claimed.