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If $T:H\to B$ is isomorphism of Banach spaces and $H$ is Hilbert, must $B$ necessarily be Hilbert?

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Which of the various available notions of "isomorphism" do you mean? –  Chris Eagle Feb 7 '13 at 22:51

2 Answers 2

I suppose that by ‘isomorphism’, you mean ‘bi-continuous linear mapping’. In what follows, $ \mathbb{F} $ shall denote either $ \mathbb{R} $ or $ \mathbb{C} $.


Let us first make a definition.

Definition Let $ (X,\| \cdot \|) $ be a normed vector space over $ \mathbb{F} $. We say that $ \| \cdot \| $ arises from an inner product if and only if there exists an inner product $ \langle \cdot,\cdot \rangle: X \times X \to \mathbb{F} $ such that $ \| x \|^{2} = \langle x,x \rangle $ for all $ x \in X $.

The following theorem gives a necessary and sufficient condition for a Banach space to underlie a Hilbert space.

Theorem Let $ (X,\| \cdot \|) $ be a Banach space. Then $ \| \cdot \| $ arises from an inner product if and only if the following identity, called the Parallelogram Law, holds: $$ \forall x,y \in X: \quad \| x + y \|^{2} + \| x - y \|^{2} = 2 \| x \|^{2} + 2 \| y \|^{2}. $$ If $ \| \cdot \| $ arises from an inner product $ \langle \cdot,\cdot \rangle $, then $ \langle \cdot,\cdot \rangle $ is necessarily unique and $ (X,\langle \cdot,\cdot \rangle) $ is a Hilbert space.

A special case where the answer is ‘yes’

Let $ (\mathcal{H},\langle \cdot,\cdot \rangle) $ be a Hilbert space and $ (X,\| \cdot \|) $ a Banach space. If $ T: \mathcal{H} \to X $ is an isometric isomorphism, then $ \| \cdot \| $ satisfies the Parallelogram Law. It immediately follows from the theorem that $ \| \cdot \| $ arises from an inner product $ \langle \cdot,\cdot \rangle $ such that $ (X,\| \cdot \|) $ underlies the Hilbert space $ (X,\langle \cdot,\cdot \rangle) $.

A counterexample that shows that the answer is ‘no’ in general

Any norm on $ \mathbb{R}^{2} $ is complete, and any two norms on $ \mathbb{R}^{2} $ are equivalent. Hence, $ \mathbb{R}^{2} $ equipped with any norm is a Banach space.

We already know that the $ \| \cdot \|_{2} $-norm on $ \mathbb{R}^{2} $ arises from the standard dot product $ \bullet $ on $ \mathbb{R}^{2} $. Hence, $ (\mathbb{R}^{2},\| \cdot \|_{2}) $ underlies the Hilbert space $ (\mathbb{R}^{2},\bullet) $.

On the other hand, the $ \| \cdot \|_{p} $-norm on $ \mathbb{R}^{2} $, where $ p \in (1,2) \cup (2,\infty) $, does not satisfy the Parallelogram Law. Hence, by the theorem, $ \| \cdot \|_{p} $ does not arise from an inner product. In other words, $ (\mathbb{R}^{2},\| \cdot \|_{p}) $ does not underlie any Hilbert space.

Consider now the identity mapping $ \text{id}: (\mathbb{R}^{2},\| \cdot \|_{2}) \to (\mathbb{R}^{2},\| \cdot \|_{p}) $. This is a non-isometric isomorphism, where the isomorphism is due to the equivalence of $ \| \cdot \|_{2} $ and $ \| \cdot \|_{p} $.

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Strictly speaking, the answer has to be "no". Consider the case where $B=H$, but we forget the inner product and only give $B$ the structure of a normed space. Then the identity map $I: H\to B$ is a Banach space isomorphism, but $B$ is not a Hilbert space.

Of course if you require that $T$ is an isometric isomorphism, then the answer is obviously "yes - from a certain point of view" since $B$ will satisfy the parallelogram law and will thus admit an inner product under which it's a Hilbert space. See: Norms Induced by Inner Products and the Parallelogram Law.

The only tricky case is when all we know is that $T: H\to B$ is bi-continuous. I'll have to think a bit more about that$\ldots$

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The last paragraph fails even for two dimensional spaces. –  Martin Feb 7 '13 at 23:13
    
@Martin: Your comment actually allowed me to construct the counterexample that I gave. :) –  Haskell Curry Feb 8 '13 at 0:54
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@Haskell: Very good, that was the point :-) Notice that you can use the same idea to give an infinite-dimensional example. Fix an orthonormal basis in a separable Hilbert space and consider the equivalent norm $$\lVert x \rVert^2 = \sum_{n=0}^\infty \left[ \left(\lvert x_{2n}\rvert^p + \lvert x_{2n+1}\rvert^p\right)^{1/p}\right]^2$$ on it. –  Martin Feb 8 '13 at 9:48

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