Range of Orthogonal Projection

Theorem

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Let $H$ be a Hilbert space.

Let $K$ be a closed linear subspace of $H$.

Let $P_K$ denote the orthogonal projection on $K$.

Then:

$P_K \sqbrk H = K$

where $P_K \sqbrk H$ denotes the image of $H$ under $P_K$.

Proof

We first show that $P_K \sqbrk H \subseteq K$.

Let $k \in P_K \sqbrk H$.

Then there exists $h \in H$ such that:

$\map {P_K} h = k$

From the definition of the orthogonal projection, we have:

$\map {P_K} h \in K$

so:

$h \in K$

giving:

$P_K \sqbrk H \subseteq K$

$\Box$

We now show that:

$K \subseteq P_K \sqbrk H$

Let $k \in K$.

Then, from Fixed Points of Orthogonal Projection:

$\map {P_K} k = k$

So:

$k \in P_K \sqbrk H$

and so:

$K \subseteq P_K \sqbrk H$

$\Box$

Hence, by definition of set equality:

$K = P_K \sqbrk H$

$\blacksquare$

Sources

• 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Theorem $2.7 \text{(d)}$