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If $S$ and $T$ are subspaces of some finite-dimensional inner product space then $$S^\bot+T^\bot = (S\cap T)^\bot.$$ See, for example, this post or this post

Does it hold in infinite-dimensional inner product spaces? What about Hilbert spaces?

My attempt:

I have noticed that $$S^\bot+T^\bot \subseteq (S\cap T)^\bot \subseteq (S^\bot+T^\bot)^{\bot\bot}$$ holds in any inner product space. So to find a counterexample, I need that $W=S^\bot+T^\bot$ fulfills $W\subsetneq W^{\bot\bot}$. A standard example I know, which fulfills this, is the set of all sequence with finite support in the inner product space $\ell_2$. However, I do not think that this space can be expressed as $S^\bot+T^\bot$ for some two subspaces.

I have tried to play around with some similar subspaces of $\ell_2$, but I did not find a counterexample.

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Martin Sleziak
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1 Answers1

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It does not hold in general.

Recall two basic facts: for any subspace $E$ of a Hilbert space, we have that

  1. $E^{\perp}$ is closed;

  2. $E^{\perp\perp} = \bar{E}$, the closure of $E$. In particular, if $E$ is closed then $E^{\perp\perp}=E$.

Now in The direct sum of two closed subspace is closed? (Hilbert space), you can find an example of two closed subspaces $X_1, X_2$ of a Hilbert space $H$, such that $X_1 \cap X_2 = 0$ but $X_1 + X_2$ is not closed. Taking $S = X_1^\perp$, $T = X_2^\perp$, we have by fact 2 that $S^\perp = X_1$ and $T^\perp = X_2$. So $S^{\perp} + T^{\perp} = X_1 + X_2$ is not closed. On the other hand, by fact 1 $(S \cap T)^\perp$ is necessarily closed. In this case, since $S \cap T = 0$, we have $(S \cap T)^\perp = H$ which is definitely closed. So $S^\perp + T^\perp \ne (S \cap T)^\perp$.

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Nate Eldredge
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