Let $\Bbb X$ be a Banach space, $T:\Bbb X\to \Bbb X$ be a linear map and $P:\Bbb X\to \Bbb X$ be a projection operator. Denote the closed subspace that is the range of $P$ by $\Bbb Y:=\mathcal R(P)$. Suppose that $||P||\le M$.
I was reading an article and there's a line claiming that $$ \lVert T|_{\Bbb Y}\rVert\le M\lVert T\rVert $$ with no further explaination. A first I was confused since normally we'd only have $$ \lVert T|_{\Bbb V}\rVert\le \lVert T\rVert $$ whenever $\Bbb V$ is a subspace of $\Bbb X$ but then I realized that it is possible that $T|_{\Bbb Y}$ can be view as $T|_{\Bbb Y}:(\Bbb Y,\Vert\cdot\Vert_{\Bbb Y})\to (\Bbb X,\Vert\cdot\Vert_{\Bbb X})$ where $\Bbb Y$ is endowed with a different norm from $\Bbb X$ and, similarly, $P$ is $P:(\Bbb X,\Vert\cdot\Vert_{\Bbb X})\to (\Bbb Y,\Vert\cdot\Vert_{\Bbb Y})$. If this is the case then we could have
$$\begin{align} \lVert T|_{\Bbb Y}\rVert &:=\sup_{||x||_{\Bbb Y}\le 1}|T(x)| \\ &\le \sup_{||x||_{\Bbb X}\le M}|T(x)| \\ &= \sup_{||x||_{\Bbb X}\le 1}|T(Mx)| \\ &= M\lVert T\rVert\ . \end{align}$$
My problem is that I am not quite sure if this is what the author intended or not. Is there any plausible reason that one would want a different norm on $\Bbb Y$? Is there a flaw in my reasoning?