Compact Idempotent is of Finite Rank

Theorem

Let $H$ be a Hilbert space.

Let $T \in \map {B_0} H$ be a compact linear operator.

Let $T$ be idempotent.

Then:

$T \in \map {B_{00} } H$

that is, $T$ is a bounded finite rank operator.

Proof

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Let $\sequence {y_n}$ be a bounded sequence in the range of $T$.

Since $T$ is idempotent:

$\forall n \in \N: T y_n = y_n$

Since $T$ is compact, $\sequence {T y_n}$ and hence $\sequence {y_n}$, contains a convergent subsequence.

Thus the range of $T$ has the Bolzano-Weierstrass property.

By Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact/Sufficient Condition, the range has finite dimension.

$\blacksquare$

Sources

• 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text {II}.4$: Exercise $4$