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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
Dear user61420, if you’re truly satisfied with a particular response to your question, please accept it by clicking on the check mark beside it. This awards the owner of the response with points for the hard work that he/she has done. – Berrick Caleb Fillmore Feb 14 '15 at 23:28

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
@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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