I am wondering if someone could help me with part (c) as I'm not sure how to approach it. I have also attached my solutions to parts (a) and (b) and would be thankful if someone could verify if what I have done is correct (especially (a) where I just assumed the ds changes to dx at the end).
Discrete Cauchy-Schwarz Proof
-1
$\begingroup$
analysis
proof-verification
summation
cauchy-integral-formula
-
0You need to use latex. It is not obvious what you want to show, and what you did – 2017-02-05
1 Answers
0
If $v(0) = 0$, with $u(x) = v(x)^2$ then applying Cauchy-Schwarz $\sup_{x \in [0,1]} v(x)^2 \le \int_0^1 |u'(x)|dx = 2\int_0^1 |v(x)v'(x)|dx \le 2 (\int_0^1 |v(x)|^2dx)^{1/2}(\int_0^1 |v'(x)|^2dx)^{1/2}$.
The discrete analogue is :
If $v(0) = 0$ : $\sup_{n \ge 1} v(n)^2 \le \sum_{n \ge 1} |v(n)^2-v(n-1)^2| = \sum_{n \ge 1} |v(n)-v(n-1)|\, |v(n)+v(n-1)|$ $ \le (\sum_{n \ge 1} |v(n)+v(n-1)|^2)^{1/2}(\sum_{n \ge 1} |v(n)-v(n-1)|^2)^{1/2}$
-
0and click on edit to see the latex I typed – 2017-02-05
-
0Can you explain where the $2$ and the $(\sum_{k \ge 1} |v(k)|^2)^{1/2}$ term comes from in the answer to part (c)? – 2017-02-05
-
0@Addy93 No, in the continuous case $|u'(x)| = |2 v(x)v'(x)|$, in the discrete case $|v(n)^2-v(n-1)^2| = |v(n)-v(n-1)| \, |v(n)+v(n-1)|$ – 2017-02-05
-
0Ok I think I understand, thank you for your help! – 2017-02-05