Let $X$ be a Banach space. A linear operator $P :X\to X$ is called a projection if $P^{2} = P$. Show that $P$ is bounded if and only if $\ker P$ and $\text{Ran } P$ are closed.
Boundedness of the projection operator.
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real-analysis
functional-analysis
operator-theory
banach-spaces
1 Answers
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Suppose that $P$ is bounded, it is equivalent to say that $P$ is continue, this implies that $I-P$ is continue and $Ker(P), Ker(I-P)=ran(P)$ are closed.
On the other side, suppose that $Ker(P)$ and $ran(P)$ are closed, let $(x_n,P(x_n))$ a converging sequence, the sequence $x_n$ converges towards $x$ and the sequence $P(x_n)$ converges towards $y$. Let $z_n=x_n-P(x_n)$, $z_n\in Ker(P)$ The sequence $z_n$ converges towards $z=x-y\in Ker(P)$ since it is closed. We deduce that $P(x-y)=P(x)-P(y)=0$ and $P(x)=P(y)=y$ since $y\in ran(P)$. So the graph is closed and the closed graph theorem implies that $P$ is continue.