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)$$(f+h,g)=(f,g)+(h,g)$
$(4)$ $(af,g)=a(f,g)$. I can show (3), (4) hold for polynomials, but I am unsure how to prove it for general continuous functions on $[0,1]$
 A: You could define it on the space $S = \left\{f \in C[0,1] : \sum_{n=1}^\infty f\left(\frac{1}{n}\right)^2 < +\infty\right\}$ to ensure $(\cdot, \cdot) < +\infty$.
You have already checked $(0, 0) = 0$ and $(f, g) = (g, f)$.
Linearity also holds:
\begin{align}(\alpha f + \beta g,h) &= \sum_{n=1}^\infty \left(\alpha f + \beta g\right)\left(\frac{1}{n}\right)h\left(\frac{1}{n}\right) \\
&= \sum_{n=1}^\infty \left(\alpha f\left(\frac{1}{n}\right) + \beta g\left(\frac{1}{n}\right)\right)h\left(\frac{1}{n}\right)\\
&= \alpha \sum_{n=1}^\infty f\left(\frac{1}{n}\right)h\left(\frac{1}{n}\right) + \beta \sum_{n=1}^\infty g\left(\frac{1}{n}\right)h\left(\frac{1}{n}\right)\\
&= \alpha(f, h) + \beta(g, h)
\end{align}
However, there are functions $f \in S$ such that $(f, f) = 0$, but $f \ne 0$.
An example is:
$$f(x) = \begin{cases}
0,  & \text{if $x \in \left[0, \frac{1}{2}\right]$} \\
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}$$
