# Question about an unusual inner product

Consider the space $C([0,1])$, given by the inner product $(f,g)=\sum_{n=1}^{\infty}f(\frac{1}{n})g(\frac{1}{n})$. I want to show this is not an inner product, but I want to check which inner product axioms fail so I might be able to modify it.

Certainly if $f=0$, then $(f,f)=0$. Now if we let $f,g\in C([0,1])$, with $f,g>0$, then certainly the inner product $(f,g)>0$ because $(f,g)=f(1)g(1)+f(1/2)g(1/2)+...$ are all strictly positive.

However, this only works if $f,g$ are exclusively positive on $[0,1]$. Consider the constant functions $f=-1$ and $g=1$ with $f,g:[0,1]\rightarrow \mathbb{R}$. We get a sum with exclusively negative terms, and the series diverges tends to $-\infty$, so $(f,g)<0$ in some cases.

$(f,g)=(g,f)$ because $\sum_{n=1}^{\infty}f(1/n)g(1/n)$ certainly can be rearranged to $\sum_{n=1}^{\infty}g(1/n)f(1/n)$ by commutativity.

However, I am unsure how to rigorously check

(3)(f+h,g)=(f,g)+(h,g) (4) (af,g)=a(f,g). I can show (3), (4) hold for polynomials, but I am unsure how to prove it for general continuous functions on [0,1] • A macroscopic problem is that (f,g)=+\infty most of the times. – user228113 Sep 18, 2017 at 14:35 • Can you construct a function f\in C[0,1] such that f(1/n)=0 and f\ne 0? (Of course, as @G.Sassatelli said it is not well defined.) – mfl Sep 18, 2017 at 14:35 • It's best not to call it an inner product at the beginning if the goal is to prove it's not an inner product. – zhw. Sep 18, 2017 at 14:37 • Per @zhw, we can call this a bilinear form rather than an inner product, to avoid self-contradiction. Sep 18, 2017 at 14:41 • @hardmath It's not a bilinear form either, is it? – zhw. Sep 18, 2017 at 15:20 ## 1 Answer You could define it on the space S = \left\{f \in C[0,1] : \sum_{n=1}^\infty f\left(\frac{1}{n}\right)^2 < +\infty\right\} to ensure (\cdot, \cdot) < +\infty. You have already checked (0, 0) = 0 and (f, g) = (g, f). Linearity also holds: \begin{align}(\alpha f + \beta g,h) &= \sum_{n=1}^\infty \left(\alpha f + \beta g\right)\left(\frac{1}{n}\right)h\left(\frac{1}{n}\right) \\ &= \sum_{n=1}^\infty \left(\alpha f\left(\frac{1}{n}\right) + \beta g\left(\frac{1}{n}\right)\right)h\left(\frac{1}{n}\right)\\ &= \alpha \sum_{n=1}^\infty f\left(\frac{1}{n}\right)h\left(\frac{1}{n}\right) + \beta \sum_{n=1}^\infty g\left(\frac{1}{n}\right)h\left(\frac{1}{n}\right)\\ &= \alpha(f, h) + \beta(g, h) \end{align} However, there are functions f \in S such that (f, f) = 0, but f \ne 0. An example is:f(x) = \begin{cases} 0, & \text{ifx \in \left[0, \frac{1}{2}\right]$} \\ x - \frac{1}{2}, & \text{if$x \in \left\langle\frac{1}{2}, \frac{3}{4}\right]$}\\ 1 - x, & \text{if$x \in \left\langle\frac{3}{4}, 1\right]$} \end{cases}$$• question about the details: For$x\in (1/2,1)$if we use$(1/n)^2-1/2$and$1-\frac{1}{n^2}$for$x\in (3/4,1)$, I don't see how$\sum_{n}f(1/n)^2$goes to$0$when we combine these sums. (Of course,$\sum_{n}f(1/n)^2=0$on$(0,1/2)$Sep 18, 2017 at 15:53 • @Kernel_Dirichlet I think you're confused about something. This particular$f$has$f(1/n)=0$for$n=1,2,\dots.\$
– zhw.
Sep 18, 2017 at 16:13
• I see it now, thanks! Sep 18, 2017 at 16:40