1

Assume $H$ is a Hilbert space and $H_1\subset H$ and $H_2 \subset H$ are (closed) subspaces with $H_1 \cap H_2 = \{0\}$.

Is there an $H_3 \subset H$, such that $H = H_1 \oplus (H_2 \oplus H_3)$ ?

If $H$ was finite dimensional, $H_3$ could be chosen as the orthogonal complement to $H_1 \cup H_2$. Is this also legit in infinite dimensions?

Jan
  • 1,008
  • 1
    It can happen that $H_1+H_2$ is not closed. – GEdgar May 25 '13 at 16:13
  • 1
    If your $\oplus$ means orthogonal sum or topological direct sum, then not always as $H_1\oplus H_2$ need not even be closed. If it simply means algebraic direct sum, then yes by basis completion. – Julien May 25 '13 at 16:18

0 Answers0