# Vector space that can be made into a Banach space but not a Hilbert space

Are there any (real or complex) vector spaces which can be made into a Banach space given a suitable norm, but cannot be given a norm that makes it a Hilbert space?

I know that the parallelogram law tells us whether a norm comes from an inner product, and I can think of spaces which have no norm making it a Banach space (e.g. spaces of countably infinite dimension). But I can't come up with an example of a space that has a norm making it a Banach space but with no norm making it a Hilbert space.

• Which norms on $\mathbb R^n$ do you know? Every heard of the $\ell_p$ norm? If $p\ne 2$, then its not coming from an inner product. – Ittay Weiss Jan 4 '16 at 9:47
• I know that, but I am trying to find a vector space such that whichever norm we put on it, it will never be a Hilbert space. (I also want the space to have the property that it can be made into a Banach space, given a suitable norm.) – Math Maniac Jan 4 '16 at 9:54
• ah, I understand your question now. – Ittay Weiss Jan 4 '16 at 9:55
• If all is about "can be made", then all we have is the dimension of the space. So you ask for which cardinalities $a$ we can define a norm but no inner product on $\Bbb R^a$? – Hagen von Eitzen Jan 4 '16 at 10:11
• @HagenvonEitzen: That sounds like a good starting point: It can be turned into a topological vector space as product space. Next one may play around with the cardinality and first countability. What do you think? – C-Star-W-Star Jan 4 '16 at 10:43

## 1 Answer

No. We have the following result.

Proposition. Let $X$ be a real or complex vector space. Then $X$ can be furnished with a complete norm if and only if there exists an inner-product on $X$ whose corresponding norm is complete.

For a topological space $X$, denote by $d(X)$ the density of $X$, that is, the minimal cardinality of a dense set in $X$. We denote the cardinality of the continuum by $\mathfrak{c}$. Certainly, $\mathfrak{c}=\mathfrak{c}^{\aleph_0}$, which we shall need.

Proof. It is enough to prove the implication $(\Rightarrow$). Let $X$ be a Banach space. Without loss of generality $X$ we may suppose that $X$ is infinite-dimensional. We split the proof into two cases.

Case where $d(X)\leqslant \mathfrak{c}$.

$$\mathfrak{c}\leqslant b(X)\leqslant |X|\leqslant d(X)^{\aleph_0}\leqslant \mathfrak{c}^{\aleph_0}=\mathfrak{c},$$

which yields $b(X)=\mathfrak{c}$[a]. This means that $X$ is isomorphic as a vector space to the (separable!) Hilbert space $\ell_2$, so one may use any algebraic isomorphism between $X$ and $\ell_2$ to define a complete, inner-product norm on $X$.

Case where $d(X)> \mathfrak{c}$.

For Banach spaces $X$ with $d(X)> \mathfrak{c}$, the cardinality of $X$ is the same as $b(X)$. We then have $$b(X)=|X|=|\ell_2(d(X))|=b(\ell_2(d(X))),$$ so one may use any algebraic isomorphism between $X$ and the Hilbert space $\ell_2(d(X))$ to define a complete, inner-product norm on $X$. $\square$

[a]: Actually one has the equality $|X|=d(X)^{\aleph_0}$ for every infinite-dimensional Banach space but we do not need it here.

• You seem to be starting with a vector space with a topology (you talk about dense sets in it, for example) and it is quite not clear that the OP has that in mind. – Mariano Suárez-Álvarez Jan 4 '16 at 20:57
• No, he wants a space that can be made into a Banach space and not into a Hilbert space. – Mariano Suárez-Álvarez Jan 4 '16 at 21:10
• (Not all Banach space topologies on a vector space have the same density) – Mariano Suárez-Álvarez Jan 4 '16 at 21:13
• The OP wants to start with a vector space, and you are starting with a Banach space. – Mariano Suárez-Álvarez Jan 4 '16 at 21:39
• @MarianoSuárez-Alvarez it seems that you have misread my post. We start with a vector space $X$. Then we assume that it can be given a complete norm (so $X$ is a Banach space for the sake of the proof). Then we prove that it can be given a Hilbert-space norm (not related to the previous one). – Tomek Kania Jan 4 '16 at 21:46