10

On Reed & Simon, vol 2, chapter X, problem 4, it is asked:

Let $M$ and $N$ be closed subspaces of a separable Hilbert space. If $\dim M > \dim N$, prove that $M\cap N^{\perp} \ne \{0\}$.

Here, $\dim V$ is the cardinality of a Hilbert basis for $V$.

The solution is pretty straightforward, since we must have $\dim N = n \in \mathbb{N}$ and therefore it becomes a linear algebra problem.

My question is: does it hold for non-separable Hilbert spaces?

I couldn't prove it or think in counterexamples for it. Thanks in advance!

Eric Wofsey
  • 342,377
Rodrigo Dias
  • 4,445
  • 5
    I think the answer is yes. My sketchy idea is as follows: Consider the orthogonal projection $P(N)$ of $N$ onto $M$ and choose an orthonormal Hilbert basis for this space. Then extend the basis of $P(N)$ to an orthonormal basis for $M$. Since $\dim(P(N))<\dim(M)$, the extension is non-trivial. Any of the new basis vectors in the extension will be in $N^\perp$.

    I'd need to check with a functional analysis textbook to make sure all the operations described above are legal...

     – Yly May 04 '18 at 05:29
  • @Yly. Quite right, And nicely done. – DanielWainfleet May 04 '18 at 12:40

1 Answers1

6

More generally, suppose $M$ and $N$ are Hilbert spaces with $\dim M>\dim N$ and let $T:M\to N$ be a bounded linear map. Then the image of the adjoint map $T^*:N\to M$ cannot be dense in $M$ (the closure of the image is the closed span of the image of an orthonormal basis for $N$, and so has dimension at most $\dim N$). So there exists nonzero $v\in M$ which is orthogonal to the image of $T^*$. This means $0=\langle v,T^*w\rangle=\langle Tv,w\rangle$ for all $w\in N$, so $Tv=0$. That is, the kernel of $T$ is nontrivial.

In your setup, we now immediately conclude that $M\cap N^\perp$ is nontrivial by letting $T:M\to N$ be the orthogonal projection to $N$.

Eric Wofsey
  • 342,377