If $P$ is a projection mapping, $\mathcal{R}(P)$ is closed and $(v_k, P(v_k))\to(v, w)\in X^2$ then why $Pw = w$?

Question

Let $X$ be a Banach space and $P$ a projection mapping. Suppose that $\mathcal{R}(P)$ (the range of $P$) is closed. Consider a sequence $(v_k, P(v_k))_{k=1}^\infty \subset X^2$ and suppose that $\lim_{k\to\infty}(v_k, P(v_k)) = (v, w)\in X^2$. I am trying to understand why this then implies that $Pw = w$. I do know that $P$ satisfies $P^2 = P$, but as I don't know beforehand whether $P$ is bounded, equivalently continuous, I can't pull out the limit $P(v) = P(\lim_{k\to\infty}Pu_k) = \lim_{k\to\infty}P^2u_k = \lim_{k\to\infty}Pu_k = v$. How should the equality be deduced?

Answer 1

If you mean exacly $Pw=w$, instead of $Pv=w$.

$Pv_k\to w$ and $Im (P)$ is closed, we know that $w\in Im(P)$, that is $w=Px$ for some $x$, which implies $Pw=P^2x=Px=w$.