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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.

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You can use the Riesz' lemma. – francis-jamet Jun 26 '12 at 15:39

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.)

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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.

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