Why is this semi-inner product not an inner product?

Question

I am just starting to learn about inner products and am working through an example in "A Course on Functional Analysis" by John Conway. The problem is the following:

Let $X$ be the collection of all sequences $\{a_n:n\geq1\}$ of scalars $a_n$ from $\mathbb{F}$ such that $a_n = 0$ for all but a finite number of values $n$. If addition and scalar multiplication are defined on $X$ by $$ \{a_n\} + \{b_n\} \equiv \{a_n + b_n\} $$ $$ \alpha\{a_n\} \equiv \{\alpha a_n\} $$ Then $X$ is a vector space over $\mathbb{F}$. If $u(\{a_n\},\{b_n\}) \equiv \sum_{n=1}^{\infty} a_{2n}\overline{b_{2n}}$ then $u$ is a semi-inner product that is not an inner product.

According to the book it says that to be an inner product in addition to the standard conditions the function $u$ must satisfy $u(x,x) = 0 \implies x=0$.

I don't see how this function fails to satisfy this condition. Here is what I tried:

$$ u(a_n,a_n) = \sum_{n=1}^\infty a_{2n}\overline{a_{2n}} $$

If $\mathbb{F} = \mathbb{R}$ then we have $\sum_{n=1}^\infty a_{2n}^2$. It is evident to see that this will be zero only if $a_{2n} = 0$ for all $n$. Then if $\mathbb{F} = \mathbb{C}$, consider $a_n = e^{i\theta_{2n}}$, thus we get the product:

$$ u(a_n,a_n) = \sum_{n=1}^\infty e^{i\theta_{2n}}e^{-i\theta_{2n}} = \sum_{n=1}^\infty 1 $$

Which is a divergent series and hence can't be equal to zero. This text only considers these two fields, so how is it that $u$ fails to be an inner product on $X$?

$\textbf{Edit:}$ Is it simply that since we are only considering the even terms, there may be non-zero terms in the sequence at the odd indices?

Answer 1

Yes, your edit is indeed correct. The fact that we are taking only the even terms means that the odd terms can literally be anything (i.e., not zero), and we can still get $u(x, x) = 0$. Hence, $u(x,\ x)=0$ doesn't imply $x = 0$.

As a simple example, take $x = (1,\ 0,\ 1,\ 0,\ 1,\ 0,\ \cdots)$ (where eventually we trail off and no longer have $1$'s, by the requirement of the elements of $X$).

With regards to the convergence, yes the inner product is defined as an infinite sum, but it's an infinite sum where infinitely many of the terms in the sum are 0 (as required by elements of $X$). Thus there are only finitely many nonzero terms, and so the sum is finite. Divergence, in a sense, has nothing to do with this.

Edit: To clarify, your sequence $a_n = e^{i\theta_{2n}}$ isn't necessarily nonzero for a finite number of terms, and so wouldn't necessarily be in $X$. You would have to restrict $\theta_{2n}$ by requiring that it is only nonzero for finitely many $n$. In this case, $a_n$ is also nonzero for finitely many $n$. Let's say $N$ is the last integer such that $\theta_{N}$ is nonzero. Then your sum is really $$ u(a_n,\ a_n) = \sum_{n=1}^N 1 = N < \infty $$

Answer 2

Your edit is exactly correct. In particular for the following sequence $\{a_n\}$ given by $$ (1,0,0,0,\ldots) $$ we would find that $u(\{a_n\},\{a_n\}) = 0$, despite the sequence being nonzero.

Also note that convergence has nothing to do with it, and your example with the divergent series does not apply; remember that the sequences in this space can only have finitely many nonzero terms. Therefore the semi-inner product of two sequences in this space is always a finite sum and convergence is not an issue.