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Let $\sigma^2:[0,1]\to\mathbb{R}_+$ be Lipschitz. I want to show that: $$ \sqrt{n}\left(\sum_{i=1}^n\frac{1}{n}\sigma^2\left(\frac{i}{n}\right)-\int_0^1\sigma^2(s)ds\right)\to0\text{ as }n\to\infty $$

I know that the sum inside the parenthesis converges to the integral, but I am not sure if the convergence is fast enough.

Is Lipschitz sufficient for that? Is this limit true? How can I show it?

Guilherme Salome
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1 Answers1

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Let $L$ be the Lipschitz constant of $\sigma^2$ and $1\le i\le n$. Then $$ \Bigl|\frac{1}{n}\,\sigma^2\Bigl(\frac{i}{n}\Bigr)-\int_{(i-1)/n}^{i/n}\sigma^2(s)\,ds\Bigr|\le\int_{(i-1)/n}^{i/n}\Bigl|\sigma^2\Bigl(\frac{i}{n}\Bigr)-\sigma^2(s)\Bigr|\,ds\le\frac{L}{n^2}. $$ It follows that $$ \Bigl|\frac{1}{n}\,\sum_{i=1}^n\sigma^2\Bigl(\frac{i}{n}\Bigr)-\int_{0}^{1}\sigma^2(s)\,ds\Bigr|\le\frac{L}{n}. $$

Julián Aguirre
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  • Thanks! That is a great answer! Just to make things a little more clear, when you bound the integral using Lipschitz, you are also using the fact that $0\leq\left|\frac{i}{n}-s\right|\leq\frac{i}{n}$, in order to get that $n^2$. – Guilherme Salome Sep 13 '16 at 15:18
  • And that $\lim_{n\to\infty} x_n =0\iff\lim_{n\to\infty}\left|x_n\right|=0$. – Guilherme Salome Sep 13 '16 at 15:27
  • It should be $\Bigl|\frac{i}{n}-s\Bigr|\le\frac1n$, and the fact that the length of the interval is also $1/n$. – Julián Aguirre Sep 15 '16 at 09:20