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With Axiom of choice it is possible to construct an inner product on $C(\mathbb R)$.

My question is, is it possible to explicitly construct an inner product on $C(\mathbb R)$? I.e. to give a closed formular to calculate the inner product? I know it is straight-forward to write down a scalar product using a Hamel basis. This is not the answer I am looking for.

This question came to me, when a student asked me in the lecture today 'whether there are vector spaces without inner products'. So I tried to find scalar produces for function spaces. I think I managed to write one down for $L^1((0,1))$. But I failed to construct one for $C(\mathbb R)$.

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    $\begingroup$ Also if you found an inner product for $L^1((0,1))$ then we would have $L^1((0,1))$ a hilbert space and the riesz representation theorem would tell us that the dual can be identified with the space itself, but we know the dual of $L^1$ is identified with $L^\infty$ not $L^1$ itself. $\endgroup$ – DanZimm May 30 '14 at 12:19
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    $\begingroup$ @DanZimm He wanted an inner product, but not necessarily completeness with respect to the inner product, nor that the inner product generates the given topology. So $\langle x,y \rangle = \sum_{n=1}^\infty 2^{-n} x_n y_n$ is an inner product on $l^1$. $\endgroup$ – Stephen Montgomery-Smith May 30 '14 at 12:22
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    $\begingroup$ I don't think the first statement is true. Assuming AC, $C(\mathbb{R})$ has cardinality $\mathfrak{c}$, hence Hamel dimension $\mathfrak{c}$, hence is isomorphic as a vector space to $\ell^2(\mathbb{N})$, hence admits a complete inner product. Or have I made a mistake? $\endgroup$ – Nate Eldredge Nov 12 '14 at 4:22
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    $\begingroup$ I suspect that the case is analogous to the existence of discontinuous linear maps in infinite-dimensional Banach spaces, where AC is required: .en.wikipedia.org/wiki/Discontinuous_linear_map#Axiom_of_choice. $\endgroup$ – Martín-Blas Pérez Pinilla Nov 12 '14 at 7:58
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    $\begingroup$ "There are vector spaces without inner products" could refer to vector spaces over fields which are not ${\bf R}$ or ${\bf C}$, or topological vector spaces which do not admit a norm compatible with the topology (like $C^\infty([0,1])$). Using axiom of choice, one can just take any basis and declare it to be the orthonormal basis. The resulting inner product will be incomplete, but otherwise fine. And for spaces of dimension outside $[\aleph_0,\mathfrak c)$, we can find (using choice) a linear isomorphism with a Hilbert space. $\endgroup$ – tomasz Nov 13 '14 at 16:54
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(Update. Added a self-contained proof that under ZF+DC+BP there is no norm.)

Martín-Blas Pérez Pinilla is on the right track. You can't even put a norm on $C(\mathbb{R})$ without using the axiom of choice in an essential way. Dependent choice is not enough.

Claim. It is consistent with ZF+DC that there does not exist any norm on $C(\mathbb{R})$.

Recall that a subset $E$ of a topological space is said to have the Baire property if it can be written as a symmetric difference $E = U \triangle M$ where $U$ is open and $M$ is meager (a countable union of nowhere dense sets).

A celebrated theorem of Shelah says that consistent with ZF+DC is the statement BP: "Every subset of $\mathbb{R}$ has the Baire property." From BP it follows that in fact every subset of any Polish space has the Baire property.

Let $\tau$ be the usual topology on $C(\mathbb{R})$ (uniform convergence on compact sets). It is induced by a translation-invariant metric:

$$d(f,g) := \sum_{n=1}^\infty 2^{-n} \min\left(1, \sup_{[-n,n]} |f-g|\right) $$

The metric $d$ is complete (this comes from the fact that a uniform limit of continuous functions is continuous). And it's not hard to see that $\tau$ is separable (the polynomials with rational coefficients are $\tau$-dense, by the Weierstrass approximation theorem). So $(C(\mathbb{R}),\tau)$ is a Polish space.

Suppose now that $\|\cdot\|$ is a norm on $C(\mathbb{R})$. Let $B$ be the closed unit $\|\cdot\|$-ball. We will show that $B$ does not have the Baire property with respect to $\tau$. Specifically, let $U$ be any $\tau$-open set; we will show that $B \triangle U$ is $\tau$-nonmeager.

Suppose first that $U = \emptyset$ so that $B \triangle U = B$. Since $\bigcup_{n=1}^\infty nB = C(\mathbb{R})$, by the Baire category theorem $B$ is $\tau$-nonmeager.

Now suppose that $U$ is nonempty. We will show $U \setminus B$ is $\tau$-nonmeager. Let us begin by showing $U \setminus B$ is nonempty. Let $f \in U$ and let $g \in C(\mathbb{R})$ be your favorite nonzero continuous function which is supported in $[0,1]$. Let $g_n(x) = g(x-n)$ be translates of $g$. For each $n$, let $a_n$ be a real number sufficiently large that $\|f + a_n g_n\| > 1$, so that $f + a_n g_n \notin B$. Then $a_n g_n \to 0$ uniformly on compact sets (i.e. in the $\tau$ topology), so for some $N$ we have $h := f + a_N g_N \in U$. Thus $h \in U \setminus B$. (To say this another way, every nonempty $\tau$-open set is unbounded, but $B$ is bounded, so $U$ cannot be a subset of $B$.)

Now for any $u \in C(\mathbb{R})$, we have $h + \frac{1}{n} u \to h$ in both the $\tau$ and $\|\cdot\|$ topologies. So for sufficiently large $n$, we have $h + \frac{1}{n} u \in U$ and $h + \frac{1}{n} u \in B^c$ (since $B$ is $\|\cdot\|$-closed). That is, $u \in n((U \setminus B)-h)$. Since $u$ was arbitrary we have shown $\bigcup_{n=1}^\infty n((U \setminus B)-h) = C(\mathbb{R})$. By the Baire category theorem, $U \setminus B$ is $\tau$-nonmeager, hence so is $B \triangle U$.

So we have shown that if $C(\mathbb{R})$ has a norm, then it has a set that lacks the Baire property with respect to $\tau$. Under ZF+DC+BP there is no such set and hence no norm.

(Credit where credit is due: This proof is loosely based on the idea of the proof of the Garnir-Wright closed graph theorem from Theorem 27.45 of Eric Schechter's Handbook of Analysis and its Foundations.)

Incidentally, the only property of the vector space $C(\mathbb{R})$ we used is that it admits a Polish topology in which every nonempty open set is unbounded. So the same argument would apply to other vector spaces with this property, such as $\mathbb{R}^{\mathbb{N}}$, $C^\infty(\mathbb{R}^d)$, etc.

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  • $\begingroup$ Maybe this is an answer to this question as well? $\endgroup$ – Asaf Karagila Nov 13 '14 at 15:43
  • $\begingroup$ @AsafKaragila: Yes, I think it is indeed. I will post an answer there with a link to this question. If I had known about the other question, I probably should have answered it and closed this one as duplicate. Not sure of the best way to arrange things now. $\endgroup$ – Nate Eldredge Nov 13 '14 at 16:42
  • $\begingroup$ Well, both are somewhat old, and you couldn't have closed an active bounty question anyway. I think the right thing to do is to post a short answer outlining the argument and a link here. $\endgroup$ – Asaf Karagila Nov 13 '14 at 16:51

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