Can every Banach space be densely embedded in a Hilbert space? This is clear if the Banach space is actually a Hilbert space, but much can you relax this?

If the embedding exists, is the target Hilbert space unique?

  • $\begingroup$ Is this embedding assumed to be linear? $\endgroup$ – Norbert Nov 17 '15 at 17:33
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    $\begingroup$ You can drop "densely" without changing the problem: if $T:X\to H$ is any injective bounded operator, then $\overline{T(X)}$ is itself a Hilbert space, in which $T(X)$ is dense. Second observation is that it suffices to consider $\ell_\infty(I)$ for an arbitrary index set $I$, since every Banach space is isometric to a subspace is such a space. I don't see how to define an injection from $\ell_\infty(\mathbb{R})$ into a Hilbert space... $\endgroup$ – user147263 Nov 27 '15 at 7:23
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    $\begingroup$ Note also that such an injective map $T : X \hookrightarrow H$ cannot be isometric, for $X$ does not satisfy the parallelogram rule (assuming $X$ is not already an inner product space), whereas $H$ does. $\endgroup$ – Josse van Dobben de Bruyn Nov 28 '15 at 7:13
  • $\begingroup$ Is what you care about whether the Banach space $B$ can be embedded in a Hilbert space $H$ with the same norm? $\endgroup$ – Jon Warneke Nov 29 '15 at 0:38
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    $\begingroup$ From the lack of responses it seems that the OP abandoned this question. So, as a bounty setter, I'll formulate it as follows: is it true that for every Banach space $X$ there exists an injective continuous linear operator $T:X\to H$ where $H$ is a Hilbert space? $\endgroup$ – user147263 Dec 1 '15 at 20:32

We can show that $\ell_\infty(I)$ cannot be embedded into a Hilbert space, for an uncountable index set $I$.

Here, I am interpreting the question as Normal Human suggests - where the embedding is assumed to be linear and continuous. Suppose that $f\colon \ell_\infty(I)\to H$ is an embedding into Hilbert space $H$. By continuity, there exists a $K\in\mathbb{R}$ such that $\lVert f(x)\rVert\le K\lVert x\rVert$ for all $x\in\ell_\infty(I)$.

Given any sequence $x_1,\ldots,x_n\in H$, the identity $$ \sum_{\epsilon_1,\ldots,\epsilon_n=\pm1}\left\lVert\sum_{r=1}^n\epsilon_rx_r\right\rVert^2=2^n\left(\sum_{r=1}^n\lVert x_r\rVert^2\right) $$ holds. This implies that there exists a sequence $\epsilon_r\in\{\pm1\}$ such that $$ \left\lVert\sum_{r=1}^n\epsilon_rx_r\right\rVert^2\ge\sum_{r=1}^n\lVert x_r\rVert^2. $$ Now, for each $i\in I$, let $e_i\in\ell_\infty(I)$ be defined by $(e_i)_j=0$ for $j\not=i$ and $(e_i)_i=1$. Also, for each $n\in\mathbb{Z}_{>0}$, let $S_n$ be the set of $i\in I$ such that $\lVert f(e_i)\rVert\ge1/n$. As $f$ is an embedding, we have $f(e_i)\not=0$, so $\bigcup_{n=1}^\infty S_n=I$ is uncountable. Hence, $S_n$ is infinite for some $n$. Then, for any $N > 0$, pick a sequence $i_1,\ldots,i_N$ of distinct elements of $S_n$. By what we showed above, there is a sequence $\epsilon_1,\ldots,\epsilon_N\in\{\pm1\}$ such that $x\equiv\sum_{r=1}^N\epsilon_re_{i_r}$ satisfies $$ \lVert f(x)\rVert^2\ge\sum_{r=1}^N\lVert f(e_{i_r})\rVert^2\ge N/n^2. $$ However, $\Vert x\rVert=1$, so $$ \lVert f(x)\rVert\ge n^{-1}\sqrt{N}\lVert x\rVert. $$ Choosing $N>K^2n^2$ gives a contradiction.


I will show that the answer is yes, for any separable Banach space $B$. The non-separable case is covered by George Lowther above.

So let $B$ be a separable Banach space, and consider some countable dense subset $\{x_n\}_{n\in \Bbb N} \subset B$. For each $n$ we apply the Hahn Banach theorem to find some $f_n \in B^*$ such that $f_n(x_n)=\|x_n\|$ and $\|f_n\|=1$.

We then define an inner product $\langle \cdot, \cdot \rangle_2$ on $B$ by setting $$\langle x,y \rangle_2 = \sum_{n=1}^{\infty} 2^{-n} f_n(x)f_n(y)$$ and we let $\|\cdot\|_2$ be the asociated norm.

To check that this is actually an inner product, the only nontrivial property we need to check is nondegeneracy (which means that $\|x\|_2=0 \implies x=0$). Define $g:=\sup_n f_n$. Note that $|g(x)| \leq \|x\|$ since $\|f_n\|= 1$ for all $n$. Also notice that $g$ is continuous, because for any fixed $N$ we have that $f_N(x)-\sup_nf_n(y) \leq f_N(x)-f_N(y)$ so taking the sup over all $N$ on both sides gives that $g(x)-g(y) \leq g(x-y) \leq \|x-y\|$. Symmetrically we can switch $x$ and $y$ so this actually shows that $|g(x)-g(y)| \leq \|x-y\|$, thus proving continuity. Also $g(x_n) \geq f_n(x_n)=\|x_n\|$ and thus $g(x_n)=\|x_n\|$ for all $n$. By density of the $x_n$ and continuity of $g$ it follows that $g(x) = \|x\|$ for all $x$. Hence $\|x\|=\sup_n f_n(x)$, so that if $\|x\|_2=0$ then $f_n(x)=0$ for all $n$ and thus $x=0$.

We denote by $B'$ the completion of $B$ with respect to $\|\cdot\|_2$. Then the inclusion map $i:B \to B'$ is a continuous embedding, because $$\|x\|_2^2 = \sum_n 2^{-n}f_n(x)^2 \leq \sum_n 2^{-n} \|x\|^2=\|x\|^2$$

  • $\begingroup$ So every separable Banach space can be linearly and continuously embedded into a separable hilbert space... that is in $\ell^2$, up to continuous linear isomorphism? $\endgroup$ – AIM_BLB Dec 14 '18 at 23:14
  • $\begingroup$ @AIM_BLB yes but it is only an “embedding” in a weak sense: the embedding is continuous, injective, and linear, but it is not a homeomorphism onto its image. $\endgroup$ – Shalop Dec 14 '18 at 23:28
  • $\begingroup$ Why is it not homeomorphic to it's image? Isn't that always the case? (Unless you mean that B with the weak topology is homeomorphic to its image, but not with the strong topology) $\endgroup$ – AIM_BLB Dec 14 '18 at 23:29
  • $\begingroup$ No, it is definitely not true that any continuous bijection of topological spaces is automatically a homeomorphism. The point here is that the image of the embedding lives in the hilbert space which has a strictly weaker norm. $\endgroup$ – Shalop Dec 14 '18 at 23:33
  • $\begingroup$ Oh, you're saying that the inverse isn't continuous also, my bad... However, if we pull-back the relative topology of the image across the embedding, then in that case would we have a homeomorphism? $\endgroup$ – AIM_BLB Dec 14 '18 at 23:35

Suppose that there exists an injective, bounded linear operator $T\colon X\to H$, where $H$ is a Hilbert space. Then $\|x\|_{\rm new}=\|x\|+\|Tx\|$ ($x\in X$) defines an equivalent, strictly convex norm on $X$.

Not every Banach spaces admits a strictly convex renorming -- just to mention $\ell_\infty(\omega_1)$ (Day) or $\ell_\infty / c_0$ (Bourgain).


I could be wrong, but it seems you are asking simply "can we define an inner product on an arbitrary Banach space?", which is addressed in this question: Is there a vector space that cannot be an inner product space?

Specifically, in the second answer in the question listed, by the AOC notice we can write down a Hamel basis $\{e_i\}$ for $X$, then define $(x,y)=\sum\overline{a_i}b_i$ for $x=\sum a_i e_i$ and $y=\sum b_i e_i$. This will give an inner product on $X$, and so $H=(X,\sqrt{(,)})$ will be a Hilbert space, and hence $T\colon X\hookrightarrow H$ will be such an embedding. This need not be unique since the the inner product depends on the choice of basis, and indeed as pointed out in the comments will not be isometric, and the norm will almost certainly be different.

I hope this helps.

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    $\begingroup$ This interpretation of the question is incorrect: OP is asking about a linear embedding, the norm is not preserved in any sense. However, it seems that OP want the linear embedding to be continuous, otherwise, trivially, any real vector space embeds linearly as a dense subset of a Hilbert space. $\endgroup$ – Moishe Kohan Dec 1 '15 at 20:04
  • $\begingroup$ Where does it say linear embedding? As opposed to,e.g. a homeomorphic embedding. $\endgroup$ – DanielWainfleet Dec 1 '15 at 20:18
  • $\begingroup$ @user254665 For the sake of clarity, I posted a comment with the interpretation I had in mind when setting a bounty on this. $\endgroup$ – user147263 Dec 1 '15 at 20:34
  • $\begingroup$ @studiosus, I agree this embedding need not be continuous, and it isn't particularly enlightening, but a linear embedding exists, albeit unbounded. $\endgroup$ – charlestoncrabb Dec 1 '15 at 21:19

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