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?

  • $\begingroup$ 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. $\endgroup$ – 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


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}$

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

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