Question

A homework question for a student (not me) preparing for Chinese Gao Kao.

Let \(f(x)\) be a function with domain \(\mathbb{R}\). For all \(x_1,x_2\in \mathbb{R}\), there exist a real L such that \(|f(x_1)-f(x_2)|\leq L|x_1-x_2|\). For \(0 10 years ago

Answer 1

If I am not mistaken, f(x) is Lipschitz continuous. According to wikipedia, f(x) is also a contraction.

Answer 2

The first part is simple. The second part is extraordinarily hard. And this is a homework question for students preparing for Chinese Gao Kao (A Level equivalent)...

Answer 3

Just curious, what is Lipschitz continuous?

Answer 4

A function \(f(x)\) with \(|f(x_1)-f(x_2)|\leq L|x_1-x_2|\) for all \(x_1,x_2\) in the domain of f. L is a fixed positive real number.

Answer 5

The second part reminds me of Cesaro summation.

Answer 6

I totally see that Cesaro sum thing. Here's an idea to try to alter the sum a little.

\[A_{n+1} = \frac{1}{n+1} \sum_{k=1}^{n+1} a_k = \frac{1}{n+1}\left(a_{n+1}+n* \frac{1}{n} \sum_{k=1}^{n} a_k \right)\]

\[A_{n+1} = \frac{1}{n+1}\left(a_{n+1}+nA_n \right) = \frac{a_{n+1}}{n+1} + \frac{n}{n+1} A_n\]

\[|A_{n+1}-A_n| = \left|\frac{a_{n+1}}{n+1} + \left(\frac{n}{n+1} -1 \right)A_n \right| = \frac{|a_{n+1}-A_n|}{n+1} \]

Answer 7

If I can prove \(a_{k+1}\leq a_k\) then the second part follows.

Answer 8

@Kainui Did you not try the first part or couldn't be bothered to type out the answer?

Answer 9

I didn't look at it cause you said it was simple so I assumed you solved it already.

Answer 10

Yes I solved it already because it is really simple.

Answer 11

I don't know how to solve any of this though myself, so I have no clue. I don't really care for analysis personally lol

Answer 12

For part 1, look at \(|a_k-a_{k+1}|\), then look at the definition of \(a_{k+1}\).

Answer 13

Victory is close...

Part 1
\[
\begin{align*}
&\phantom{{}={}}|a_k-a_{k+1}|\\
&=|f(a_{k-1})-f(a_k)|&\text{Definition of }a_k\\
&\leq L|a_{k-1}-a_k|&\text{Property of }f(x)\\
&\leq L^2|a_{k-2}-a_{k-1}|&\text{Do it again}\\
&\leq L^{k-1}|a_1-a_2|
\end{align*}
\]
\[
\begin{align*}
&\phantom{{}={}}\sum_{k=1}^n|a_k-a_{k+1}|\\
&\leq \sum_{k=1}^n L^{k-1}|a_1-a_2|\\
&=|a_1-a_2|\sum_{k=0}^{n-1} L^{k} &\text{Index shift}\\
&<|a_1-a_2|\sum_{k=0}^\infty L^{k}\\
&=\frac{1}{1-L}|a_1-a_2|&0<L<1
\end{align*}
\]

Part 2
\[
\begin{align*}
|a_k-a_{k+1}|&\leq L^{k-1}|a_1-a_2|\\
|a_{k-1}-a_k|&\leq L^{k-2}|a_1-a_2|\\
|a_k-a_{k+1}|+|a_{k-1}-a_k|&\leq |a_1-a_2|(L^{k-1}+L^{k-2})\\
|a_{k-1}-a_{k+1}|&\leq |a_1-a_2|(L^{k-2}+L^{k-3})\text{ Triangle inequality}\\
|a_i-a_j|&\leq|a_1-a_2|\left(\sum_{s=i-1}^{j-2} L^s\right)\text{ in general}\\
|a_i-a_j|&<|a_1-a_2|\left(\sum_{s=1}^\infty L^s\right)\text{ min. }i=1\\
|a_i-a_j|&<\frac{1}{1-L}|a_1-a_2|
\end{align*}
\]
\[
\begin{align*}
&\phantom{{}={}}|A_k-A_{k+1}|\\
&=\left|\frac{1}{k}\sum_{j=1}^k a_j-\frac{1}{k+1}\sum_{j=1}^{k+1} a_j \right|\\
&=\frac{1}{k(k+1)}\left|(k+1)\sum_{j=1}^k a_j-k\sum_{j=1}^k+1 a_j \right|\\
&=\frac{1}{k(k+1)}\left|\sum_{j=1}^k (a_j)-ka_{k+1}\right|\\
&=\frac{1}{k(k+1)}\left|\sum_{j=1}^k (a_j-a_{k+1})\right|\\
&\leq\frac{1}{k(k+1)}\sum_{j=1}^k \left|a_j-a_{k+1}\right|&\text{Triangle inequality}\\
&\leq\frac{1}{k(k+1)}\sum_{j=1}^k \left(\frac{1}{1-L}|a_1-a_2|\right)\\
&=\frac{1}{k(k+1)}\left(\frac{k}{1-L}|a_1-a_2|\right)\\
&=\frac{1}{(k+1)}\left(\frac{1}{1-L}\right)|a_1-a_2|
\end{align*}
\]

Answer 14

@ganeshie8 Help! Finished part 1 but part 2 just so hard!

Answer 15

what

Answer 16

Is it too confusing or what?

Answer 17

The question is in Chinese so I have to translate it into English.