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Of the few counterexamples to this theorem that I was able to find, including this one, none of them involves an incomplete prehilbertian space and a closed incomplete convex subset.

I have another example with the space of square integrable sequences and the subspace of finitely supported sequences which again, is not closed.

I'm in search of an example of an incomplete prehilbertian space containing a closed convex subset which doesn't define a projector (hence the subset must be incomplete).

I naively tried considering the prehilbertian space of finitely supported sequences with the $l^2$ inner product and my closed convex subset was those sequences that start with a $0$. But even though the completeness assumption is not met, the projection function is still well-defined since all you have to do to project an element is replace its first term with a $0$.

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Consider the set of continuous functions on $[0,1]$ with the $L_2$ norm.

Let $f(x) = \int_0^{1 \over 2} x(t) dt$ which is bounded and let $C = f^{-1} \{1\}$ which is closed and affine (hence convex).

The element $0$ has no nearest point in $C$.

To see why there is no nearest point, first consider the solution in the completed space ($L_2[0,1]$). It is straightforward to see that $f(x) = \langle \phi, x \rangle$ where $\phi = 1_{[0,{1 \over 2}]}$ and a straightforward computation shows that $\min \{ \|x\|^2 | x \in L_2[0,1], \ f(x) = 1 \}$ is $x^*={\phi \over \|\phi\|^2}$. It is clear that $\phi$ is not equivalent to a continuous function.

Hence $\|x^*\| \le \|x\|$ for all $x \in C$, and since $x \mapsto \|x\|^2$ is strictly convex, the minimiser is unique, so $\|x^*\| < \|x\| $ for all $x \in C$ such that $x \neq x^*$.

Furthermore, since the continuous functions are dense in $L_2[0,1]$, there are continuous $x_n$ such that $x_n \to x^*$, and hence we see that $\inf \{ \|x\|^2 | x \text{ is continuous}, f(x) = 1 \} = \|x^*\|$, but for any continuous $x$ such that $f(x) = 1$ satisfies $\|x^*\| < \|x\|$.

Note: As the OP observed in the comments, it is not necessarily the case that the continuous $x_n$ are in $C$. Since $f(x_n) \to 1$ (as $x^* \in C$), we see that the sequence ${1 \over f(x_n)} x_n$ is continuous, lies in $C$ and converges to $x^*$.

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  • $\begingroup$ Would you mind elaborating on why 0 has no nearest point ? I'm trying with the negative inner product characterisation but all I could think of was that 2 is in $C$ $\endgroup$ – user493048 Jan 9 '18 at 1:08
  • $\begingroup$ I added a laborious elaboration. $\endgroup$ – copper.hat Jan 9 '18 at 2:38
  • $\begingroup$ Wonderful many thanks $\endgroup$ – user493048 Jan 9 '18 at 9:32
  • $\begingroup$ PS: Although we know there is a sequence $(x_n)_n$ of continuous functions converging to $x^*$ in $L_2$, how do we know it can be contained within $C$, that is $f(x_n)=1$ ? $\endgroup$ – user493048 Jan 9 '18 at 16:30
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    $\begingroup$ The sequence ${1 \over f(x_n)} x_n$ converges to $x_n$. $\endgroup$ – copper.hat Jan 9 '18 at 17:29
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Lemma:

Let $W$ be a subspace of an inner product space $X$, and $x_0 \in X$. If there exists a distance minimizer $x \in W$ from $W$ to $x_0$, then necessarily $x_0 - x\in W^\perp$.

The proof uses the fact that $a \perp b \iff \|a\| \le \|a + \lambda b\|, \forall \lambda \in \mathbb{C}$.

Indeed, assume that $\|x_0 - x\| = d(x_0, W)$ and take any $y \in W$, $\lambda \in \mathbb{C}$.

$$\|x_0-x + \lambda y\| = \|x_0-\underbrace{(x - \lambda y)}_{\in W}\| \le \|x_0 - x\| \implies x_0 - x \perp y$$

Therefore $x_0 - x \in W^\perp$.

Let $c_{00}$ be the space of finitely-supported sequences with $\|\cdot\|_2$ norm. Consider the subspace $$W = \left\{(x_n)_{n=1}^\infty \in c_{00} : \sum_{n=1}^\infty \frac{x_n}n = 0\right\} = \left\{(x_n)_{n=1}^\infty \in c_{00} : \left\langle (x_n)_{n=1}^\infty, \left(1, \frac12, \frac13, \ldots\right)\right\rangle = 0\right\} \le c_{00}$$

$W$ is closed in $c_{00}$. However, $W^\perp = \{0\}$.

Therefore, for any $x_0 \in c_{00} \setminus W$, the only candidate for the distance minimizer is $x_0$ itself, which is a contradiction since $x_0 \notin W$.

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