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I'm reading a book on PDE in which the min-max theorem and the Rayleigh quotient are mentioned. In the 'compact operators' section in the given link, it seems to me that the proof employs the following proposition.

Proposition Let $H$ be an infinite dimensional Hilbert space with a countable , linearly independent subset $B=\{u_i:i=1,2,\dots\}$. Let $k$ be any positive integer.

If $B$ is in fact a basis for $H$, it is true that

(a) Let $S_k$ be any $k$-dimensional subspace of $H$ and let $S'=\overline{span\{u_i:i\geq k\}}$. Then $S'\cap S_k\neq \{0\}$.

(b) Let $V_{k-1}$ be any $k-1$-dimensional subspace of $H$ and let $V'=span\{u_i:1\leq i\leq k\}$. Then $V'\cap {V_{k-1}}^{\perp}\neq \{0\}$.

I know that this probably is very trivial but I just cannot see why they hold since I haven't touched linear algebra for a long time. If $H$ is finite-dimensional, say $\dim(H)=n$, then the proposition is clearly true because we know that $$\dim(S'\cap S_k)=\dim(S')+\dim(S_k)-\dim(S'+S_k)\geq (n-k+1)+k-n=1>0$$ $$\dim(V'\cap {V_{k-1}}^{\perp})\geq k+[n-(k-1)]-n=1>0$$

But how do we perform dimension counting arguments when $H$ is infinite-dimensional? Thank you.

  • Oh... wait... I think I may have missed the phrase "If $B$ is in fact a basis for $H$..." so I think I have to retract all of my previous comments. – Lee Mosher Jun 10 '20 at 17:04
  • @LeeMosher Sorry. I edited my question after your comment was made. You are correct in that the proposition is false without the additional assumption. – user551102 Jun 10 '20 at 17:05
  • I tried to prove (a) and (b) by contradiction, but the only formulas that I know are https://math.stackexchange.com/questions/1077537/what-is-the-orthogonal-of-an-intersection. For instance, if $S'\cap S_k={0}$, then ${S'}^{\perp}+{S_k}^{\perp}={{0}}^{\perp}=H$. I'm not sure what the last identity leads to. – user551102 Jun 10 '20 at 17:10

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