I have this exercise:

A linear operator $P: \Omega \rightarrow \Omega$ is called a projection if both the range of P and $P^{-1}(\{0\})$ are closed and $P(P(x))=P(x) \forall x$.

Show that if $\Omega$ is a Banach space, then all projections of $\Omega$ are continuous.

This exercise is in the chapter of the open mapping theorem, and the closed graph theorem, so it is a pretty big hint that I am supposed to use one of these.

My first attempt (open mapping):

Since $P(\Omega)$ is closed, it is a Banach space. Then if we look at $P: \Omega \rightarrow P(\Omega)$, it is an open mapping. Let U be an open set of $\Omega$, then I need to show that $P^{-1}(U)$, is open. We have that $P^{-1}(U)=P^{-1}(U\cap P(\Omega))$, $U\cap P(\Omega)$ is an open set in $P(\Omega)$. So I need to show that $P^{-1}(U\cap P(\Omega))$ is open.

I don't see how it beeing an open mapping helps me, because that is sadly the other way.

I tried contradiction:

Assume for contradiction that: $P^{-1}(U\cap P(\Omega))$ is not open. Then it must also contain a limit point for its complement.So then there exists x, so that $P(x)\in U$, but $z_n \rightarrow x$, $P(z_n)\ne U$. This implies among other things, that any image of any open ball around x, is both open(open mapping theorem), and contain elements from both U and its complement. But still, this is no contradiction as far as I see.

Any tips?

My second attempt (closed graph theorem). Since we are working in a Banach space, if I can show that P is a closed operator I will be done.

So assume $\lim x_n = x$, then I must show that if $\lim T(x_n)$ converges, then $\lim T(x_n)=T(x)$.

So assume that $\lim x_n = x$ and that $\lim T(x_n)=y$. Assume for contradiction that $y \ne T(x)$, then we are in the case where $\lim T(x_n)\ne T(x)\rightarrow \lim T(x_n-x)\ne 0$. Hence we are in the case where we have a sequnce $z_n\rightarrow 0$, but $T(z_n)$ converges, but it does not converge to 0. I can't get a contradiction in the case either.

Can you guys please help me?


Let's use the closed graph theorem. We want to prove that: $${\rm gr}(P) = \{ (x,P(x)) \in \Omega \times \Omega \mid x \in \Omega \}$$is closed. Let $((x_n, P(x_n)))_{n \geq 1}$ be a sequence in the graph such that $x_n \to x$ and $P(x_n) \to y$. Since $(P(x_n))_{n \geq 1}$ is a sequence in the range of $P$, which we assume closed, we have that $y$ is in the range of $P$, so we write $y = P(\tilde{x})$ for some $\tilde{x} \in \Omega$.

Note that $x_n - P(x_n) \to x - y$, but $(x_n-P(x_n))_{n \geq 1}$ is a sequence in $P^{-1}(\{0\})$ because of the $P^{\circ 2} = P$ property. Since $P^{-1}(\{0\})$ is closed, we have $x-y \in P^{-1}(\{0\})$, so $P(x) = P(y)$. Well, using $P^{\circ 2} = P$ again we have: $$P(\tilde{x}) = y \implies P(x) = P(y) = P(\tilde{x}) = y.$$So the graph is closed, and by the closed graph theorem, $P$ is continuous.

(I really enjoyed solving this, thanks for the opportunity)

  • $\begingroup$ Of course, the closed graph theorem uses some choice (at least countable multiple choice). $\endgroup$ – Cloudscape Sep 17 '15 at 19:38

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