Bounded projection I have a problem proving the following:
Consider a banach space $X$. A projection is bounded if and only if $\mathrm{Im} (P)= \{ Px : x \in X\}$ nd $\mathrm{Ker}(P) = \{x \in X: Px =0\}$ are closed subspaces of $X$.
So we know that a linear operator,say P is a projection if $P^2=P$. 
Assume first that the projection is bounded. Then $\exists M$  s.t $||Px||\leq M||x||$ for all $x \in X$.
Then, since we know that $X$ is a Banach space, take a sequence say $\{x_n\}_n$ then we know that it is convergent. Say x is the limit. Apply P on $\{x_n\}_n$. Assume $\{Px_n\}$ sonverges to y. Then $x_n - Px_n = (I-P)x_n$ converges to $x - y$. It follows that $y \in \mathrm{Im}(P)$ and $x-y \in \mathrm{Ker}(P)$, so $y = Py = Px$. Hence the conclusion follows from the closed graph theorem.
Is it true what I wrote? I mean does it really proof the theorem? How to proceed next?
 A: *

*Suppose $P$ is a projection on a Banach space $X$, meaning that $P^2=P$. If $P$ is continuous, then the following subspaces are closed:
$$
              \mathcal{N}(P)=P^{-1}\{0\} \\
        \mathcal{R}(P) = (I-P)^{-1}\{0\}.
$$

*Conversely, suppose $P$ is a projection on $X$ such that $\mathcal{R}(P)$ and $\mathcal{N}(P)$ are closed. Then $P$ must be shown to be continuous. By the closed graph theorem, it suffices to show that $P$ is closed. So, suppose
$$
        x_n \rightarrow x,\;\; Px_n \rightarrow y.
$$
It must be shown that $y=Px$. To prove this, note that 
$$
      x-y\in\mathcal{R}(I-P)=\mathcal{N}(P), \;\;\;\;y\in\mathcal{R}(P)=\mathcal{N}(I-P),
$$
because these subspaces are closed by assumption. So $(I-P)y=0$ and $P(x-y)=0$, which gives $y=Py=Px$ and proves that $P$ is closed and, hence, is continuous.

A: There are some things in your proof that need to be investigated. 
First of all you do not say that you want to prove the equivalence in a banach space $X$. If you want to then please state it there. If not we cannot use neither the fact that $X$ is a banach space nor the closed graph theorem. 
Second. The last paragraph is not clear. It seems that you are assuming that $P$ is continous and then again are proving it. 
There are sequences in a banach space that do not converge so this reasoning seems off. This is one thing that you wrote which is not true. Further you do not need to assume that $Px_n$ converges. Since $x_n \to x$ you might follow by using the continuity of $P$ that $Px_n \to Px$.
Here are some hints assuming that we are talking about a banach space $X$. Assume that $P$ is continous. You need to show that both kernel and image of $P$ are closed. Remember what it means for a set $A \subset X$ to be closed. One characterisation is that for all sequences $(x_n)_{n \in \mathbb{N}} \subset A$ converging to an element $x$ in $X$ it holds that $x \in A$. 
To show that the kernel is closed consider a sequence $(x_n)_{n \in \mathbb{N}} \subset \mathrm{Ker}(P)$ converging to $x \in X$. This implies that for all $n \in \mathbb{N}$ iut holds that $Px_n = 0$. We need to show that $Px = 0$. Can you take it from here?
So far one doesn't need to use that $P$ is a projection. One can show that the kernel of every bounded linear operator is closed.
Ok, lets talk about the image. Consider a sequence $(y_n) \subset \mathrm{Im}(P) $ converging to $y \in X$. We need to show that $y \in \mathrm{Im}(P)$. This is equivalent to th existence of a $x \in X$ such that $Px = y$. Here in fact you need that $P$ is a projection. We conclude
$$ -Py_n = y_n-Py_n +y_n \to y $$
since $y_n - Py_n \in \mathrm{Ker}(P)$. This shows that 
$$y =\lim\limits_{n \to \infty} P y_n = P (\lim\limits_{n \to \infty} y_n) = Py $$
using the assumption of continuity of $P$.
Another nice way to show this is the fact that $\mathrm{Im}(P) = \mathrm{Ker}(\mathrm{Id}-P)$. Can you show this ? 
How could one conclude the closedness of $\mathrm{Im}(P)$ using this equality?
The other impilcation of above equivalence is more complicated. Here the closed graph theorem comes in handy. Assume that the mentioned spaces are closed. Consider a sequence $(x_n)_{n \in \mathbb{N}} \subset X$ converging to $x$ such that $Px_n \to y$ given $y \in X $. We need to show that $Px =y$. This is where your arguments work better. Arguing along the lines of your proof we have $x_n -Px_n \to x-y$ this shows $x-y \in \mathrm{Ker}(P)$ since $\mathrm{Ker}(P)$ is closed. Since $\mathrm{Im}(P)$ is closed as well we deduce that there exists $z \in X$ such that $Pz = y$. 
This leads to 
$$x-y + 0 = \lim\limits_{n \to \infty} x_n-Px_n +  \lim\limits_{n \to \infty} Px_n -z =  \lim\limits_{n \to \infty} x_n -z = x-z.$$ 
This shows that $z = y$ and therefore $Px = P(x-y+y) = 0+ Py = y$.
We deduce $Px = y$. The conclusion follows by the closed grph theorem.
