Unbounded operator such that $P^2=P$

Question

Does there exist an Unbounded operator $P$ on some Banach space $X$ such that $Dom(P)=X$ and $P^2=P$?

If we don’t require $Dom(P)=X$, we can easily construct a Unbounded operator on $L^2[0,2π]$ by define $P$ which act on bases as $P\exp(in\theta)=|n|+1$ for all integers $n$.

Any help will be appreciated, thanks.

An unbounded operator is typically defined only on a (dense) subspace. — Berci, Dec 12 '19 at 12:21

score 8 · Accepted Answer · answered Dec 12 '19 at 12:24

8

Let $f \ne 0$ be an unbounded linear functional on $X$. Then there is $u \in X$ such that $f(u)=1.$ Now define $P:X \to X$ by

$$P(x):=f(x)u$$

$P$ will do the job.

answered Dec 12 '19 at 12:24

Fred

78,422

score 1 · Answer 2 · answered Dec 12 '19 at 12:22

Take a non closed vector subspace $E$ of codimension $1$ and $u$ not in $E$, for every $x\in X$, there exists unique real $c(x)$ and $P(x)$ such that $x=P(x)+c(x)u$, $P^2(x)=P(x)$ and its not bounded.

https://mathoverflow.net/questions/30868/subspaces-of-finite-codimension-in-banach-spaces

score 1 · Answer 3 · answered Dec 12 '19 at 12:25

1

Let $f$ be an unbounded linear functional and $P(x)=f(x) x_0$ where $x_0$ is a fixed vector with $f(x_0)=1$. Then $P$ is not continuous and $P^{2}=P$.

answered Dec 12 '19 at 12:25

Kavi Rama Murthy

359,332

Unbounded operator such that $P^2=P$

3 Answers3