# $a_n=\frac{1}{2^n}\sum_{k=1}^{2^n}f\biggl(\frac{k}{2^n}\biggl)$ is decreasing

We have a continuous and increasing function $$f:[0,1]\to \mathbb R$$ and the sequence $$(a_n)_{n\ge1}$$,$$a_n=\frac{1}{2^n}\sum_{k=1}^{2^n}f\biggl(\frac{k}{2^n}\biggl)$$ . Prove that $$(a_n)_{n\ge1}$$ is decreasing.

I got $$a_{n+1}=\frac{1}{2^{n+1}}\biggr( \sum_{k=1}^{2^n}f\biggl(\frac{k}{2^{n+1}}\biggl) +\sum_{k=2^n}^{2^{n+1}}f\biggl(\frac{k}{2^{n+1}}\biggl) \biggr)$$. On the solution they have$$a_{n+1}=\frac{1}{2^{n+1}}\biggr( \sum_{k=1}^{2^n}f\biggl(\frac{k}{2^n}\biggl) +\sum_{k=1}^{2^n}f\biggl(\frac{2k-1}{2^{n+1}}\biggl) \biggr)$$

and then they used $$f(\frac{2k-1}{2^{n+1}})\le f(\frac{k}{2^n})$$, which I don't understand both of them. Can somebody help me, please?

• Here is a proof that $\lim\limits_{n\rightarrow\infty}a_n=\int\limits_0^1f(x)dx$. By increasing $n$, we effectively increase the number of element in the partition of $[0,1]$. Then, you can play with Riemann/Darboux sums. – rtybase Mar 3 at 19:11

Since $$2^n$$ is even and $$2^{n+1}/2=2^n$$, we can write for any sequnce $$b_k$$

$$\sum_{k=1}^{2^{n+1}}b_k=\sum_{k=1}^{2^{n}}(\underbrace{b_{2k}}_{\text{even indexed terms}}+\underbrace{b_{2k-1}}_{\text{odd indexed terms}})$$

With $$b_{k}=f\left(\frac{k}{2^{n+1}}\right)$$ we can write

$$\sum_{k=1}^{2^{n+1}}f\left(\frac{k}{2^{n+1}}\right)=\sum_{k=1}^{2^{n}}\left(f\left(\frac{k}{2^n}\right)+f\left(\frac{2k-1}{2^{n+1}}\right)\right)$$

Can you finish now?

$$f$$ is increasing so

$$f(\frac{2k-1}{2^{n+1}})\le f(\frac{k}{2^n})$$

Because $$\frac{2k-1}{2^{n+1}}=\frac{k}{2^n}-\frac{1}{2^{n+1}}\leq\frac{k}{2^n}$$

• Ok, thank you! And why is the sum written like that? – Gaboru Mar 3 at 18:55