Direct sum of orthogonal subspaces

I'm working on the following problem set.

Let $\mathcal{H}$ be a Hilbert space and $A$ and $B$ orthogonal subspaces of $\mathcal{H}$.

Prove or disprove:

1) $A \oplus B$ is closed, then $A$ and $B$ are closed.

2) $A$ and $B$ are closed, then $A \oplus B$ is closed.

I could prove 1) and if my proof is correct, it even holds if $\mathcal{H}$ is just an inner product space.

Unfortunately, I can't manage to prove 2). Since $\mathcal{H}$ is by assumption a Hilbert space and I didn't use that fact to prove 1), I should probably use it here.

It means that $A$ and $B$ are also complete. Given some convergent sequence in $A \oplus B$, I want to show that the limit is also in $A \oplus B$. Here I'm stuck. I want to use the completeness of $A$ and $B$, but I don't see how to obtain suitable Cauchy sequences.

Can anyone drop me a hint? Or is my approach all wrong?

Solution 1:

If $(a_n+b_n)_n$ is a Cauchy sequence in $A\oplus B$ ($a_n\in A, b_n\in B$), then the identity $$\langle a_n+b_n-(a_m+b_m),a_n+b_n-(a_m+b_m)\rangle = \langle a_n-a_m,a_n-a_m\rangle+\langle b_n-b_m, b_n - b_m\rangle$$ (which follows by orthogonality of $A,B$) implies that $(a_n)_n$, $(b_n)_n$ are Cauchy sequences in $A$, $B$ respectively. Hence they converge to some $a\in A$ and $b\in B$, repectively, whence $a_n+b_n\rightarrow a+b\in A\oplus B$, so $A\oplus B$ is closed.

The same argument also works the other way around.