description/proof of that for \(2\) not-both-\(0\) complex numbers and non-negative real number \(p\), \(p\)-th power of absolute value of sum is smaller than (\(p\)-th power of \(2\)) times sum of \(p\)-th powers of absolute values
Topics
About: set
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Note 1
- 3: Proof
- 4: Note 2
Starting Context
- The reader knows a definition of complex numbers set.
Target Context
- The reader will have a description and a proof of the proposition that for any 2 not-both-0 complex numbers and any non-negative real number, p, the p-th power of the absolute value of the sum of the numbers is smaller than (the p-th power of 2) times the sum of the p-th powers of the absolute values of the numbers.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(c_1\): \(\in \mathbb{C}\)
\(c_2\): \(\in \mathbb{C}\)
\(p\): \(\in \mathbb{R}\) such that \(0 \le p\)
//
Statements:
\(\vert c_1 \vert \neq 0 \lor \vert c_2 \vert \neq 0\)
\(\implies\)
\(\vert c_1 + c_2 \vert^p \lt 2^p (\vert c_1 \vert^p + \vert c_2 \vert^p)\)
//
2: Note 1
We are not saying that this is the best possible evaluation: for example, for \(p = 1\), \(\vert c_1 + c_2 \vert \le \vert c_1 \vert + \vert c_2 \vert\), and for \(p = 2\), \(\vert c_1 + c_2 \vert^2 \le 2 (\vert c_1 \vert^2 + \vert c_2 \vert^2)\).
But in many cases, our concern is not having the best possible \(c\) for \(\vert c_1 + c_2 \vert^p \lt c (\vert c_1 \vert^p + \vert c_2 \vert^p)\) but to see that there is such any constant \(c\) that does not depend on \(c_1, c_2\) (but can depend on \(p\)).
This proposition is about getting a \(c\) by the arguably easiest way.
3: Proof
Whole Strategy: Step 1: see that \(\vert c_1 + c_2 \vert \le \vert c_1 \vert + \vert c_2 \vert\); Step 2: suppose that \(\vert c_2 \vert \le \vert c_1 \vert\) and let \(\vert c_2 \vert = r \vert c_1 \vert\) for an \(r \in \mathbb{R}\) such that \(0 \le r \le 1\); Step 3: evaluate \((\vert c_1 \vert + \vert c_2 \vert)^p = (\vert c_1 \vert + r \vert c_1 \vert)^p\); Step 4: see that supposing that \(\vert c_1 \vert \lt \vert c_2 \vert\) does not cause any problem.
Step 1:
\(\vert c_1 + c_2 \vert \le \vert c_1 \vert + \vert c_2 \vert\), which is a well-known fact, but to explain it succinctly, \(c_j\) can be regarded to be a vector on \(\mathbb{R}^2\) and \(\vert c_1 + c_2 \vert\) is the length of the sum of the vectors and \(\vert c_1 \vert\) and \(\vert c_2 \vert\) are the lengths of \(c_1\) and \(c_2\), and \(\vert c_1 + c_2 \vert \le \vert c_1 \vert + \vert c_2 \vert\) holds by the Euclidean geometry.
Step 2:
Let us suppose that \(\vert c_2 \vert \le \vert c_1 \vert\).
Then, \(\vert c_2 \vert = r \vert c_1 \vert\) for an \(r \in \mathbb{R}\) such that \(0 \le r \le 1\).
Step 3:
\(\vert c_1 + c_2 \vert^p \le (\vert c_1 \vert + \vert c_2 \vert)^p = (\vert c_1 \vert + r \vert c_1 \vert)^p = (\vert c_1 \vert (1 + r))^p = (1 + r)^p \vert c_1 \vert^p\).
Let us suppose that \(\vert c_2 \vert \lt \vert c_1 \vert\), which means that \(r \lt 1\).
Then, \(\lt 2^p \vert c_1 \vert^p \le 2^p (\vert c_1 \vert^p + \vert c_2 \vert^p)\).
Let us suppose that \(\vert c_2 \vert = \vert c_1 \vert\).
Then, \(\vert c_1 + c_2 \vert^p \le (\vert c_1 \vert + \vert c_2 \vert)^p = (2 \vert c_1 \vert)^p = 2^p \vert c_1 \vert^p \lt 2^p (\vert c_1 \vert^p + \vert c_2 \vert^p)\): \(0 \lt \vert c_2 \vert\), because otherwise, \(\vert c_2 \vert = \vert c_1 \vert = 0\), a contradiction against \(\vert c_1 \vert \neq 0 \lor \vert c_2 \vert \neq 0\).
So, anyway, \(\vert c_1 + c_2 \vert^p \lt 2^p (\vert c_1 \vert^p + \vert c_2 \vert^p)\).
Step 4:
Let us suppose that \(\vert c_1 \vert \lt \vert c_2 \vert\).
Then, \(\vert c_1 + c_2 \vert^p = \vert c_2 + c_1 \vert^p\), and by the previous steps, \(\vert c_2 + c_1 \vert^p \lt 2^p (\vert c_2 \vert^p + \vert c_1 \vert^p) = 2^p (\vert c_1 \vert^p + \vert c_2 \vert^p)\).
So, \(\vert c_1 + c_2 \vert^p \lt 2^p (\vert c_1 \vert^p + \vert c_2 \vert^p)\).
4: Note 2
Let us see a proof of that for \(p = 2\), \(\vert c_1 + c_2 \vert^p \le 2 (\vert c_1 \vert^2 + \vert c_2 \vert^2)\).
\(\vert c_1 + c_2 \vert^2 \le (\vert c_1 \vert + \vert c_2 \vert)^2 = \vert c_1 \vert^2 + \vert c_2 \vert^2 + 2 \vert c_1 \vert \vert c_2 \vert\).
But \(0 \le (\vert c_1 \vert - \vert c_2 \vert)^2 = \vert c_1 \vert^2 + \vert c_2 \vert^2 - 2 \vert c_1 \vert \vert c_2 \vert\), which implies that \(2 \vert c_1 \vert \vert c_2 \vert \le \vert c_1 \vert^2 + \vert c_2 \vert^2\).
So, \(\vert c_1 + c_2 \vert^2 \le \vert c_1 \vert^2 + \vert c_2 \vert^2 + \vert c_1 \vert^2 + \vert c_2 \vert^2 = 2 (\vert c_1 \vert^2 + \vert c_2 \vert^2)\).
That is the best possible evaluation, because \(\vert 1 + 1 \vert^2 = 2^2 = 4 = 2 (1 + 1) = 2 (1^2 + 1^2)\).
In order to make \(\le\) \(\lt\), we can just take any \(2 \lt c\).
