Let $X$ be a Banach Space and Let $(p_n)$ be a sequence of projection operators in $BL(X)$ such that $R(p_n) \subset R(P_{n+1})$ for all $n \in \mathbb{N}$. Then Show that $p_n(x) \to x$ for every $x \in X$ if and only if $(\|p_n\|)$ is a bounded sequence and $\bigcup_{n=1}^{\infty}R(p_n)$ is dense in $X$
My try:
Suppose that $p_n(x) \to x$ for every $x \in X$. Then by Banach-Steinhaus Theorem $(\|p_n\|)$ is a bounded sequence. Then since $p_n(x) \to x$, the union is dense in $X$.
On the other hand suppose that $(\|p_n\|)$ is a bounded sequence and $\bigcup_{n=1}^\infty R(p_n)$ is dense in $X$. Let $x \in X$ be fixed. Since the union is dense in $X$, there exists a $\{y_n\} \in \bigcup_{n=1}^\infty R(p_n)$ such that $y_n \to x$. Now $\|p_n\|$ is a bounded sequence. So there exists an $ \alpha \gt 0$ such that $\|p_n\| \lt \alpha$ for each $n$. Moreover each $p_n$ is a continuous linear map.Thus $p_k(y_n) \to p_k(x)$ for each $k$. Also $y_n=p_{j_n}(x_n)$(since $y_n \in \bigcup_{n=1}^\infty R(p_n)$). Thus $p_{j_{n}}(y_n)=p_{j_n}(x_n)=y_n$ which goes to $x$ as $n \to \infty$. Also $p_{j_{n}}(y_n) \to p_{j_{n}}(x)$.(since $P_{j_n}$ is continuous and $y_n \to x$)
Thus $$|p_n(x)-x| \le \|(p_n(x)-p_n(y_n))\|+\|(p_n(y_n)-p_{j_n}(y_n))\|+\|p_{j_n}(y_n)-x\|$$
The first and last terms can be made less than an arbitrary $\epsilon\gt 0$. For the middle one I make the following claim: >Since $ p_{j_n}(x_n)=y_n $, and $ p_{j_n} $ is a projection operator, $ p_{j_n}(y_n)=y_n $, I claim that $ p_{k}(y_n)=y_n $ for all $k \ge j_n $. Since the ranges are contained in one another, $ y_n=p_k (z_n) $ and $ p_k $ itself is a projection operator, $ p_k (y_n)=y_n $. Thus for a large $ n $ , the middle one vanishes. Is that right? ?
Moreover The Banach property is needed for the first implication. The second implication will go through for an arbitrary normed space.
Thanks for the help!!