Let $f:(0,1)\rightarrow [0,1]$ be a $\mathcal{C}^{\infty}$ function. We say that a function $D_r^f:\{1,...,r\}\rightarrow [0,1]$ is an $r$-discretization of $f$ if $$D_r^f(j) = \frac{1}{|a(j)-b(j)|}\cdot\int_{(a(j),b(j))} f(x)\;dx$$ where $a(j)=\frac{(j-1)}{r}$ and $b(j) = \frac{j}{r}$. In other words, we divide the interval $(0,1)$ in $r$ consecutive sub-intervals and $D^f_r(j)$ is the average of $f$ on the $j$-th sub-interval $(a(j),b(j))$.

For any two $\mathcal{C}^{\infty}$ functions $f,g:(0,1)\rightarrow [0,1]$, we let we consider the following inner product. $$\langle f,g\rangle = \int_{(0,1)} f(x)\cdot g(x)\; dx$$ On the other hand we define the inner product on $r$-discretizations as follows. $$\langle D_r^f,D_r^g\rangle = \sum_{j=1}^r D_r^f(j)\cdot D_r^g(j).$$


  1. Given an $\varepsilon \in (0,1)$, how large must $r$ be so that $|\langle D_r^f,D_r^g \rangle - \langle f,g\rangle| \leq \varepsilon$?
  2. Is it enough to assume that $f,g$ are $\mathcal{C}^{\infty}$ functions in order to be able to prove a bound on $r$ in terms of $\varepsilon$? If not, what are the weakest conditions $f$ and $g$ must satisfy?

Obs: This question was asked at mathoverflow some days ago (with no answers). Now I realize that scicomp may be the most adequate place to ask it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.