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


$(4)$ $(af,g)=a(f,g)$. I can show (3), (4) hold for polynomials, but I am unsure how to prove it for general continuous functions on $[0,1]$

  • 2
    $\begingroup$ A macroscopic problem is that $(f,g)=+\infty$ most of the times. $\endgroup$
    – user228113
    Sep 18, 2017 at 14:35
  • 1
    $\begingroup$ 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.) $\endgroup$
    – mfl
    Sep 18, 2017 at 14:35
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    $\begingroup$ 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. $\endgroup$
    – zhw.
    Sep 18, 2017 at 14:37
  • $\begingroup$ Per @zhw, we can call this a bilinear form rather than an inner product, to avoid self-contradiction. $\endgroup$
    – hardmath
    Sep 18, 2017 at 14:41
  • $\begingroup$ @hardmath It's not a bilinear form either, is it? $\endgroup$
    – zhw.
    Sep 18, 2017 at 15:20

1 Answer 1


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:

$$f(x) = \begin{cases} 0, & \text{if $x \in \left[0, \frac{1}{2}\right]$} \\ x - \frac{1}{2}, & \text{if $x \in \left\langle\frac{1}{2}, \frac{3}{4}\right]$}\\ 1 - x, & \text{if $x \in \left\langle\frac{3}{4}, 1\right]$} \end{cases}$$

  • $\begingroup$ 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 $(0,1/2)$ $\endgroup$ Sep 18, 2017 at 15:53
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    $\begingroup$ @Kernel_Dirichlet I think you're confused about something. This particular $f$ has $f(1/n)=0$ for $n=1,2,\dots.$ $\endgroup$
    – zhw.
    Sep 18, 2017 at 16:13
  • $\begingroup$ I see it now, thanks! $\endgroup$ Sep 18, 2017 at 16:40

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