No equivalent norm induced by inner product

In class I was asked to show that there is no inner product on $$\ell^1(\mathbb{N})$$ which gives rise to the norm $$\|\cdot\|_1$$. I was able to do so, using the parallelogram law.

Now, I am wondering if it is possible for a norm on $$\ell^1(\mathbb{N})$$ which is equivalent to $$\|\cdot\|_1$$ to be induced by an inner product. It isn't immediately obvious to me whether or not this could be the case.

So far, I have tried using the definition of equivalent norms to no avail.

Assume Banach spaces $$X$$ and $$Y$$ are isomorphic. That means there is a bounded linear bijection $$T:X\to Y.$$ Then $$T^*:Y^*\to X^*$$ is a bijection between the dual spaces.

Let $$X=(\ell^1,\,\|\cdot\|_1)$$ and $$Y=(\ell^1,\,\|\cdot\|),$$ where $$\|\cdot\|$$ denotes an equivalent norm associated with an inner product. Then the map $$Tx=x$$ for $$x\in\ell^1$$ is a bounded bijection from $$X$$ into $$Y.$$ Thus $$T^*:Y^*\to X^*=\ell^\infty$$ is a bounded bijection. The space $$Y^*$$ is separable, as it is isometrically isomorphic to $$Y$$ (this holds for every Hilbert space). Hence $$\ell^\infty$$ is separable, a contradiction.

• I know that $\ell^\infty$ is not separable. But why would your claim be the case? I don’t see how this relates to the dual space, or it being separable. May 9 at 1:23
• I have extended the answer to provide all details. May 9 at 3:40

We can show that there is no such norm in four steps:

• First, $$\ell^1$$ is not reflexive because its (continuous) dual is $$\ell^\infty$$ (every continuous linear functional over $$\ell^1$$ is of the form $$\Lambda(v) = \sum_{n = 0}^\infty u_n v_n$$ for some bounded sequence $$u$$, I'm sure this is well-documented somewhere) but the dual of $$\ell^\infty$$ is not $$\ell^1$$ (the dual of $$\ell^\infty$$ contains Hahn-Banach extensions of the limit functional $$\mathcal{L} : u \in c \mapsto \lim_{n \to \infty} u_n$$ and you can show that this leads to $$\ell^\infty$$ not being the dual of $$\ell^1$$, you can look for the keywords "Banach limit" for example, or "generalised limit").
• Reflexivity is conserved via equivalence of norms: see for example the proof given in the body of this question Norm equivalence preserves canonical evaluation map and thus reflexivity.
• Completeness is also preserved via equivalence of norms: that should be easy to see just using definitions.
• Hilbert spaces are reflexive: simply use Riesz' representation theorem twice to get the linear isomorphism that gets you that a Hilbert $$H$$ is isomorphic to $$H^{**}$$.

As such, no equivalent norm to $$\|\cdot\|_1$$ can give an inner product space, as, since it would have to be a Hilbert space since $$\ell^1$$ is complete, the existence of such a norm would mean that $$\ell^1$$ is reflexive, which it is not, absurd.

Therefore, there is no such norm.

Your argument with parallegram identity can be extended to general case. That is you should look at average of $$n$$ vectors, average over all signs. In world, by induction, one easily shows that such average is the sum of second powers of their norms. This is in inner product norm. In particular, if you take your vectors so that on k-coordinate they are $$a_k$$ and $$0$$ otherwise, you get that the average over signs should be of order of the sum of second powers of $$a_k$$. But in $$l_1$$ norm such average is equal to the sum of absolute values of $$a_k$$. So, for proper values of $$a_k$$ they can not be equivalent.

• Well, here is "more details" - one wants to show that in inner product space we have this: $\sum ||{\epsilon}_1x_1+{\epsilon}_2x_2 + { \epsilon}_3x_3 + ...|| = 2^{n-1}*(||x_1||^2 + ||x_2||^2 + ...)$. Where the sum on the left is over +1/-1 tuples. This is just parallelogram identity for any $n$. Once this is shown then chose $x_k$ as discussed above. This whole approach is well known and established trick from "local theory" of Banach spaces. May 9 at 13:22