Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I consider $X$ to be an infinite dimensional Banach space and $P\in P(X)$, that is, $P$ is a continuous linear projection. How does one prove that $P$ is compact if and only if $\dim R(P)$ is finite?

Thank you.

share|improve this question
    
You can use the Riesz' lemma. –  francis-jamet Jun 26 '12 at 15:39
add comment

2 Answers

Here is an outline of one way to show this.

  • Prove that $P(X)$ is closed, hence a Banach space.

  • Note that the restriction of $P$ to $P(X)$ is the identity operator on the Banach space $P(X)$.

  • Note that the identity operator on a Banach space is compact if and only if the closed unit ball of the space is compact.

  • Prove that the closed unit ball of a Banach space is compact if and only if the space is finite dimensional. (francis-jamet and Norbert have mentioned Riesz's lemma, which is useful for this, perhaps the most important part.)

share|improve this answer
add comment

We will prove implication $\Longrightarrow$ ad absurdum. Assume that $\mathrm{dim}(R(P))=+\infty$, then $R(P)$ is infinite dimensional subspace of $X$. Using Riesz's lemma about almost perpendicular show that $R(P)$ contains sequence $\{x_n:n\in\mathbb{N}\}$ in the unit ball of $R(P)$ such that $$ m\neq n\Longrightarrow \Vert x_n-x_m\Vert>1/2 $$ Since $P$ acts as identity on $R(P)$ we see that unit ball of $X$ after applying projection $P$ contains this sequence. Note that relatively compact set can't contain such subsets, since they have no limit points. Thus image of the unit ball under projection $P$ is not relatively compact. Hence $P$ is not a compact operator.

Implication $\Longleftarrow$ is easier. Assume that $\mathrm{dim}(R(P))<+\infty$, consider image of unit ball under projection $P$. This bounded subset since $P$ is bounded. Moreover this is subset of finite dimensional subspace $R(P)$. We know that bounded subsets of finite dimensional spaces are relatively compact. Thus $P$ is a compact operator.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.