Alternative Proofs: Closed proper subspace of Hilbert space has non-empty orthogonal complement

Question

Let $(H,\langle \cdot, \cdot \rangle)$ be a Hilbert space. I am asking for people to present any short/basic/fundamental proofs they know of the following theorem.

Theorem 1: Let $V \subseteq H$ be a closed, proper vector subspace of $H$. Then $V^\perp$ contains a non-zero vector.

Theorem 1 follows pretty easily as a corollary of either of the following theorems.

Theorem 2: A vector subspace $U$ of $H$ is dense if and only if $U^\perp$ is trivial.

Notes on Theorem 2: The proof may be found here, requiring many pages of prior results. These results include $U^\perp = (\overline{U})^\perp$ and $(U^\perp)^\perp = \overline{U}$ for vector subspaces $U$ of $H$, which in turn require the "Hilbert projection theorem" and the continuity of the inner product.

Theorem 3: $H = U \oplus U^\perp$ for any closed vector subspace $U$ of $H$.

Notes on Theorem 3: A proof of this may be found here. I believe this proof requires less machinery than the proof of Theorem 2. In particular, only the construction of the orthogonal projection and some basic properties of the projection are required.

Allow me to explain my thoughts a bit further. Theorem 2 and 3 are definitely not long proofs which make use of important, fundamental facts about orthogonality in Hilbert space. However, Theorem 2 and 3 seem to be much stronger results than Theorem 1. I have a gut feeling that Theorem 1 may be proven without having to resort to the "machinery" required for Theorem 2 or 3 (if you could really call it machinery). I guess another way to put it is that I believe there should be much more basic proofs of Theorem 1.

However, I have not been able to make progress on such a proof myself. If I were to guess, a proof by contradiction making clever use of the continuity of the inner product might work. But, again, I've had no such success. Please feel free to drop your proof approaches below or any reasoning as to why you think Theorem 2 or 3 are the shortest / most fundamental methods.

I hope this post allows for a better understanding for the true difficulty or simplicity of Theorem 1. So, although this is not strictly what I am looking for, also feel free to drop any super sleek proofs of Theorem 1 that use really heavy machinery ahaha Maybe such proofs will enlighten why Theorem 1 cannot be proven in a more fundamental manner. Thanks!

Answer 1

You can prove it without heavy machinery, but a bit of computation, using distance-minimising sequences.

Suppose $V \le H$ is closed and proper. Choose any point $x_0 \in H \setminus V$. Let $$r = \inf_{x \in V} \|x - x_0\| > 0,$$ since $V$ is closed. Suppose $(x_n)_{n \ge 1} \in V$ such that $\lim_{n \to \infty} \|x_n - x_0\| = r$. I claim that $(x_n)$ is Cauchy.

Fix $\varepsilon > 0$. Let $$\delta = \sqrt{r^2 + \frac{\varepsilon}{4}} - r > 0.$$ Since $\|x_n - x_0\| \to r$, there exists some $N$ such that $$n \ge N \implies r \le \|x_n - x_0\| < r + \delta = \sqrt{r^2 + \frac{\varepsilon}{4}}.$$

Suppose $m, n \ge N$. Parallelogram law states that: $$\|(x_m - x_0) - (x_n - x_0)\|^2 + \|(x_m - x_0) + (x_n - x_0)\|^2 = 2\|x_m - x_0\|^2 + 2\|x_n - x_0\|^2,$$ or equivalently, $$\|x_m - x_n\|^2 = 2\|x_m - x_0\|^2 + 2\|x_n - x_0\|^2 - 4\left\|\frac{x_m + x_n}{2} - x_0\right\|^2.$$ Note that $\frac{x_m + x_n}{2} \in V$, so $$\left\|\frac{x_m + x_n}{2} - x_0\right\|^2 \ge r^2,$$ by definition of $r$. Therefore, $$\|x_m - x_n\|^2 < 2\left(r^2 + \frac{\varepsilon}{4}\right) + 2\left(r^2 + \frac{\varepsilon}{4}\right) - 4r^2 = \varepsilon.$$

Thus, $(x_n)$ is Cauchy. It has a limit $x_\infty$, by completeness. I claim that $x_\infty - x_0 \in V^\perp \setminus \{0\}$.

First, \begin{align*} \|x_\infty - x_0\| &= \left\|\lim_{n \to \infty} x_n - x_0\right\| \\ &= \lim_{n \to \infty} \|x_n - x_0\| \\ &= r > 0, \end{align*} so $x_\infty - x_0 \neq 0$.

To see that $x_\infty - x_0 \in V^\perp$, fix $v \in V \setminus \{0\}$, and suppose $t \in \Bbb{C}$ (or just $\Bbb{R}$, if we're in the reals) to be decided shortly. Then $x_\infty + tv \in V$, so $$\|x_\infty + tv - x_0\| \ge r = \|x_\infty - x_0\|.$$ Then \begin{align*} 0 &\le \|x_\infty - x_0 + tv\|^2 - \|x_\infty - x_0\|^2 \\ &= |t|^2\|v\|^2 - 2\operatorname{Re}(t \langle x_\infty - x_0, v \rangle). \end{align*} In particular, let $$t = \frac{\langle v, x_\infty - x_0\rangle}{\|v\|^2},$$ so that the above becomes: $$0 \le -\frac{|\langle x_\infty - x_0, v\rangle|^2}{\|v\|^2},$$ which implies $\langle v, x_\infty - x_0\rangle = 0$ for all $v \in V \setminus \{0\}$ (and hence all $v \in V$). Thus $x_\infty - x_0 \in V^\perp$.

Answer 2

Since $V$ is a proper subspace, there exists $w\notin V$. Let $v_i$ be an orthonormal basis for $V$. Consider $w^* = w - \sum_i \langle w, v_i\rangle v_i$. Because $w\notin V$, $w^* \ne 0$. However, $\langle w^*, v_i\rangle = 0$ for all $v_i$. Therefore $w^*$ is an element of $V^\perp$.

(This proof may need a slight modification if $V$ is infinite dimensional and separable, but the same idea should apply. I have no idea if it would work for $V$ inseparable, but no one cares about inseparable Hilbert spaces anyways.)