# Isomorphism of Banach space

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
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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