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Let $X$ be a Euclidean space (i.e. a vector space over $\mathbb{R}$ or $\mathbb{C}$ with an inner product, $\textbf{not necessarily complete}$). Let $S$ be a subspace of $X$ with its orthogonal complement $S^\perp$ equal to $\{0\}$. Must we have $\overline{S} = X?$

This is certainly true whenever $\overline{S}$ is complete in $X$, since we can write $X = \overline{S} \oplus S^\perp$. In particular it is true whenever $X$ is a Hilbert space.

But in the general case that $X$ is Euclidean...?

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  • $\begingroup$ Your question doesn't make much sense? every euclidean space is a Hilbert space $\endgroup$ – JustDroppedIn May 12 at 11:33
  • $\begingroup$ The definition of a Euclidean space used in my course is that it is an inner product space - not necessarily complete. $\endgroup$ – Baran Karakus May 14 at 8:17
  • $\begingroup$ Yes but you can prove its completeness easily, you only need that $\mathbb{R}$ is complete (and this is the so called "least upper bound axiom") $\endgroup$ – JustDroppedIn May 14 at 21:03
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    $\begingroup$ This is not true. Take the space C[0, 1] of continuous functions onto the reals from [0, 1], equipped with the L2 norm, which is induced from an inner product. This space is incomplete but is a Euclidean space (i.e. an inner product space). $\endgroup$ – Baran Karakus May 18 at 17:34
  • $\begingroup$ This is not a Euclidean space, since it is not finite dimensional; the definition of a Euclidean space specifically refers to a finite dimensional space. $\endgroup$ – JustDroppedIn May 19 at 12:11
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Equip the space $X = C_c^\infty(\mathbb{R})$ of smooth compactly supported test functions with the inner product arising from $L^2(\mathbb{R})$.

Let $S = \{f \in X: \int_0^1 f(s) ds = 0\}$. I claim that $S^\perp = \{0\}$ and that $S$ is not dense.

Firstly, if $g \in S^\perp$ then $\operatorname{supp}(g) \subseteq [0,1]$. Indeed, otherwise there is some interval $[a,b]$ disjoint from $[0,1]$ such that either $g>\varepsilon$ or $g < -\varepsilon$ on $[a,b]$ for some $\varepsilon > 0$. Then, a smooth probability density function $f$ with support in $[a,b]$ lies in $S$ and has either $\int_{\mathbb{R}} fg > \varepsilon > 0$ or $\int_{\mathbb{R}} fg < - \varepsilon <0$.

As a result, if $g \neq 0$ then $g$ is not constant on $[0,1]$ so that there exist $x,y \in [0,1]$ such that $g(x) \neq g(y)$. To see that this cannot happen, let $f$ be a smooth probability density function with support in $[0,1]$ and define $f_x^\lambda(y) = \lambda^{-1}f(\lambda^{-1}(x-y))$. For $\lambda$ sufficiently small $f_x^\lambda$ and $f_y^\lambda$ have disjoint support so that $f_x^\lambda - f_y^\lambda \in S$. Hence $0 = \int_0^1 (f_x^\lambda - f_y^\lambda) g$. However, $f_x^\lambda$ is a mollifier centered at $x$ and so by standard properties of mollification $\int_0^1 f_x^\lambda g \to g(x)$ as $\lambda \to 0$ which implies that $g(x) - g(y) = 0$, a contradiction. Hence $g = 0$ and so $S^\perp = 0$.

Next I show that $S$ is not dense in $X$. Note that if $f_n \in S$ and $f_n \to f$ in $L^2(\mathbb{R})$-norm then $\int_0^1 f = \int_0^1 (f-f_n) = \int_{\mathbb{R}} 1_{[0,1]} (f-f_n) \to 0$ as $n \to \infty$ by Cauchy-Schwarz and so $\int_0^1 f = 0$. Hence, $S$ is a proper closed subset of $X$ with $S^\perp = \{0\}$.

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  • $\begingroup$ I don't see that $S$ is closed in $X$. Let $x^{(k)}$ be the vector whose first component is $1$, whose next $k$ components are $-\frac1k$, and whose remaining components are all $0$. Let $y$ be the vector whose first component is $1$ and whose other components are all $0$. All these vectors are in $X$. The difference $y-x^{(k)}$ has $k$ non-zero components, all equal to $-\frac1k$, so its $\ell^2$ norm is $1/\sqrt k$, which approaches $0$ as $k\to\infty$. So $x^{(k)}\to y$. But all of the $x^{(k)}$ are in $S$ and $y$ isn't. $\endgroup$ – Andreas Blass May 23 at 1:35
  • $\begingroup$ @AndreasBlass Shoot, thank you for pointing this out. I was quite tired and apparently very very careless when writing this. Of course, the idea of your example can easily be used to show that that $S$ was dense in the corresponding $X$. I've now updated the answer to contain the example I was really hoping to avoid when writing my previous answer. Hopefully this time I'm not missing something stupid. $\endgroup$ – Rhys Steele May 23 at 8:11

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