# Banach Space that is NOT Hilbert

1. Oct 2, 2008

### Old Guy

I know that all Hilbert spaces are Banach spaces, and that the converse is not true, but I've been unable to come up with a (hopefully simple!) example of a Banach space that is not also a Hilbert space. Any help would be appreciated!

2. Oct 2, 2008

### Anthony

Hint: a necessary and sufficient for a Banach space $$(\mathcal{B},\|\cdot\|)$$ to be a Hilbert space is for the norm to satisfy the parallelogram identity:

$$\| a+b \|^2 + \|a-b \|^2 = 2 \|a\|^2 + 2\| b\|^2$$

for each $$a,b \in \mathcal{B}$$. Now think of some simple Banach spaces and check the above.

3. Oct 2, 2008

### Old Guy

Thank you, Anthony; I was aware of this condition but could not come up with anything. I believe I'm looking for a complete normed vector space (Banach) that is NOT complete under the inner product norm (as required for Hilbert), but I can't quite get my brain around how this would look.

4. Oct 2, 2008

### jostpuur

Speaking about spaces that are not Hilbert spaces is not quite clear always. Strictly speaking, if you don't define any inner product onto a given Banach space, then it is not a Hilbert space! But if you are asked to give an example of a Banach space that is not an Hilbert space, most surely it means that you want a kind of space that cannot even be made into a Hilbert space. That means that an inner product that would give the original norm, doesn't exist.

5. Oct 2, 2008

### jostpuur

I got stuck there. How do you prove the bilinearity of

$$(x|y) = \frac{1}{2}(\|x+y\|^2 - \|x\|^2 - \|y\|^2)$$

by using the parallelogram only? (I'm dealing with real vector spaces first.)

6. Oct 2, 2008

### jostpuur

This completeness business is little bit misdirection, since dealing with finite dimensional spaces, where completeness is trivial, is enough. One can device a two dimensional norm space which is not an inner product space, and then you have a Banach space which is not a Hilbert space.

7. Oct 2, 2008

### Old Guy

This is exactly what I'm seeking; can someone please provide an example? Thank you.

8. Oct 2, 2008

### cliowa

Think of a space of functions: Say you fix some interval, look at the continuous functions...

9. Oct 2, 2008

### morphism

Or just consider the p-norms on R^2, and use the parallelogram law. It's very easy to prove that the only p-norm that comes from an inner product is the 2-norm.

10. Oct 3, 2008

### Anthony

If $$\| \cdot \|$$ satisfies the parallelogram identity, then the induced inner product is given by the polarization identity:

$$(a,b) = \frac{1}{4} \left( \| a+b\|^2 - \|a-b\|^2\right)$$

The only tough thing to prove is linearity. Here's a hint that should lead you on the right direction:

\begin{align*} \| (a+b)+c\|^2 + \| (a+b)-c\|^2 &= 2\left( \| a+b\|^2 + \|c\|^2 \right) \\ \| (a-b)+c\|^2 + \| (a-b)-c\|^2 &= 2\left( \| a-b\|^2 + \|c\|^2 \right)\end{align*}

for $$a,b,c\in\mathcal{B}$$, which both follow from the parallelogram identity. Now subtract them and use the definition of $$(\cdot, \cdot)$$ given by the polarization identity.

Old Guy: I would follow morphism's advice and try out some norms you know of in $$\mathbf{R}^n$$.

11. Oct 3, 2008

### HallsofIvy

Staff Emeritus
Why do you want to prove it? Your original problem was to find a Banach space that is not a Hilbert space. It was pointed out that a norm that does not satisfy the "parallelogram" inequality is not an innerproduct space. It was also pointed out that any Lp[/sup] for p other than 2 is a Banach space that is not a Hilbert space. There is no need to prove that "if the parallelgram inequality is satisified, a Banach space is a Hilbert space.

12. Oct 3, 2008

### Anthony

HallsofIvy, it was Old Guy, not jostpuur, who posed the original problem. :)