Closed Linear Subspaces Closed under Setwise Addition

Theorem

Let $H$ be a Hilbert space.

Let $M, N$ be closed linear subspaces of $H$.

Then $M + N$ is also a closed linear subspace of $H$, where $+$ denotes setwise addition.

Proof

By Linear Subspaces Closed under Setwise Addition, $M + N$ is a linear subspace of $H$.

Now to show that it is closed.

Let $P: H \to H$ denote the orthogonal projection on $M$.

Denote by $I - P$ the complementary projection of $P$.

Define $N' := \left\{{ n - Pn: n \in N}\right\}$.

$N'$ is a closed linear subspace of $H$.

Observe $m + n = \left({m + Pn}\right) + \left({n - Pn}\right) \in M + N'$; hence, $M + N \subseteq M + N'$.

By $m + \left({n - Pn}\right) = \left({m - Pn}\right) + n \in M + N$, conclude that $M + N' \subseteq M + N$, hence equality.

Furthermore, $N' \subseteq \operatorname{ran} \left({I - P}\right) = \operatorname{ker} P$ by Range of Idempotent is Kernel of Complementary Idempotent.

That is, $N' \subseteq M^\perp$ by Properties of Orthogonal Projection, and hence $M \perp N'$.

Denote by $P'$ the orthogonal projection on $N'$.

Suppose now that $h \in M + N'$. Then:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle h$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \left({P + \left({I - P}\right)}\right)h$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \left({P + \left({I - P}\right)}\right) \left({P' + \left({I - P'}\right)}\right)h$$ $$\displaystyle$$ $$\displaystyle$$ $I$ is the identity operator $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle PP'h + \left({I - P}\right)P'h + P\left({I - P'}\right)h \left({I - P}\right)\left({I - P'}\right)h$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle 0 + P'h + Ph + 0$$ $$\displaystyle$$ $$\displaystyle$$

For this last equality, observe $M \perp N'$, hence $M \subseteq N'^\perp$, $N' \subseteq M^\perp$ and $\left({M + N'}\right)^\perp \subseteq M^\perp \cup N'^\perp$.

The conclusion is that every $h \in M + N'$ can be uniquely decomposed as $P'h + Ph$, with $P'h \in N', Ph \in M$.

So suppose there is a sequence $h_n \to h$ in $M + N'$. Then $Ph_n$ and $P'h_n$ are sequences in $M, N'$, respectively.

As those are closed linear subspaces of $H$, there are $m \in M, n \in N'$ with $Ph_n \to m, P'h_n \to n$.

It follows that $h = m + n \in M + N'$.

That is, $M + N' = M + N$ is a closed linear subspace of $H$.

$\blacksquare$